RD Sharma Solutions for Class 8 Chapter 1– Rational Numbers

In this chapter, we provide RD Sharma Solutions for Class 8 Chapter 1– Rational Numbers for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 8 Chapter 1– Rational Numbers pdf, free RD Sharma Solutions for Class 8 Chapter 1– Rational Numbers book pdf download. Now you will get step by step solution to each question.

Feature Contents

Access Answers to Maths RD Sharma Solutions for Class 8 Chapter 1 Rational Numbers

EXERCISE 1.1 PAGE NO: 1.5

1. Add the following rational numbers:

(i) -5/7 and 3/7

(ii) -15/4 and 7/4

(iii) -8/11 and -4/11

(iv) 6/13 and -9/13

Solution:

Since the denominators are of same positive numbers we can add them directly

(i) -5/7 + 3/7 = (-5+3)/7 = -2/7

(ii) -15/4 + 7/4 = (-15+7)/4 = -8/4

Further dividing by 4 we get,

-8/4 = -2

(iii) -8/11 + -4/11 = (-8 + (-4))/11 = (-8-4)/11 = -12/11

(iv) 6/13 + -9/13 = (6 + (-9))/13 = (6-9)/13 = -3/13

2. Add the following rational numbers:

(i) 3/4 and -5/8

Solution: The denominators are 4 and 8

By taking LCM for 4 and 8 is 8

We rewrite the given fraction in order to get the same denominator

3/4 = (3×2) / (4×2) = 6/8 and

-5/8 = (-5×1) / (8×1) = -5/8

Since the denominators are same we can add them directly

6/8 + -5/8 = (6 + (-5))/8 = (6-5)/8 = 1/8

(ii) 5/-9 and 7/3

Solution: Firstly we need to convert the denominators to positive numbers.

5/-9 = (5 × -1)/ (-9 × -1) = -5/9

The denominators are 9 and 3

By taking LCM for 9 and 3 is 9

We rewrite the given fraction in order to get the same denominator

-5/9 = (-5×1) / (9×1) = -5/9 and

7/3 = (7×3) / (3×3) = 21/9

Since the denominators are same we can add them directly

-5/9 + 21/9 = (-5+21)/9 = 16/9

(iii) -3 and 3/5

Solution: The denominators are 1 and 5

By taking LCM for 1 and 5 is 5

We rewrite the given fraction in order to get the same denominator

-3/1 = (-3×5) / (1×5) = -15/5 and

3/5 = (3×1) / (5×1) = 3/5

Now, the denominators are same we can add them directly

-15/5 + 3/5 = (-15+3)/5 = -12/5

(iv) -7/27 and 11/18

Solution: The denominators are 27 and 18

By taking LCM for 27 and 18 is 54

We rewrite the given fraction in order to get the same denominator

-7/27 = (-7×2) / (27×2) = -14/54 and

11/18 = (11×3) / (18×3) = 33/54

Now, the denominators are same we can add them directly

-14/54 + 33/54 = (-14+33)/54 = 19/54

(v) 31/-4 and -5/8

Solution: Firstly we need to convert the denominators to positive numbers.

31/-4 = (31 × -1)/ (-4 × -1) = -31/4

The denominators are 4 and 8

By taking LCM for 4 and 8 is 8

We rewrite the given fraction in order to get the same denominator

-31/4 = (-31×2) / (4×2) = -62/8 and

-5/8 = (-5×1) / (8×1) = -5/8

Since the denominators are same we can add them directly

-62/8 + (-5)/8 = (-62 + (-5))/8 = (-62-5)/8 = -67/8

(vi) 5/36 and -7/12

Solution: The denominators are 36 and 12

By taking LCM for 36 and 12 is 36

We rewrite the given fraction in order to get the same denominator

5/36 = (5×1) / (36×1) = 5/36 and

-7/12 = (-7×3) / (12×3) = -21/36

Now, the denominators are same we can add them directly

5/36 + -21/36 = (5 + (-21))/36 = 5-21/36 = -16/36 = -4/9

(vii) -5/16 and 7/24

Solution: The denominators are 16 and 24

By taking LCM for 16 and 24 is 48

We rewrite the given fraction in order to get the same denominator

-5/16 = (-5×3) / (16×3) = -15/48 and

7/24 = (7×2) / (24×2) = 14/48

Now, the denominators are same we can add them directly

-15/48 + 14/48 = (-15 + 14)/48 = -1/48

(viii) 7/-18 and 8/27

Solution: Firstly we need to convert the denominators to positive numbers.

7/-18 = (7 × -1)/ (-18 × -1) = -7/18

The denominators are 18 and 27

By taking LCM for 18 and 27 is 54

We rewrite the given fraction in order to get the same denominator

-7/18 = (-7×3) / (18×3) = -21/54 and

8/27 = (8×2) / (27×2) = 16/54

Since the denominators are same we can add them directly

-21/54 + 16/54 = (-21 + 16)/54 = -5/54

3.Simplify:

(i) 8/9 + -11/6

Solution: let us take the LCM for 9 and 6 which is 18

(8×2)/(9×2) + (-11×3)/(6×3)

16/18 + -33/18

Since the denominators are same we can add them directly

(16-33)/18 = -17/18

(ii) 3 + 5/-7

Solution: Firstly convert the denominator to positive number

5/-7 = (5×-1)/(-7×-1) = -5/7

3/1 + -5/7

Now let us take the LCM for 1 and 7 which is 7

(3×7)/(1×7) + (-5×1)/(7×1)

21/7 + -5/7

Since the denominators are same we can add them directly

(21-5)/7 = 16/7

(iii) 1/-12 + 2/-15

Solution: Firstly convert the denominator to positive number

1/-12 = (1×-1)/(-12×-1) = -1/12

2/-15 = (2×-1)/(-15×-1) = -2/15

-1/12 + -2/15

Now let us take the LCM for 12 and 15 which is 60

(-1×5)/(12×5) + (-2×4)/(15×4)

-5/60 + -8/60

Since the denominators are same we can add them directly

(-5-8)/60 = -13/60

(iv) -8/19 + -4/57

Solution: let us take the LCM for 19 and 57 which is 57

(-8×3)/(19×3) + (-4×1)/(57×1)

-24/57 + -4/57

Since the denominators are same we can add them directly

(-24-4)/57 = -28/57

(v) 7/9 + 3/-4

Solution: Firstly convert the denominator to positive number

3/-4 = (3×-1)/(-4×-1) = -3/4

7/9 + -3/4

Now let us take the LCM for 9 and 4 which is 36

(7×4)/(9×4) + (-3×9)/(4×9)

28/36 + -27/36

Since the denominators are same we can add them directly

(28-27)/36 = 1/36

(vi) 5/26 + 11/-39

Solution: Firstly convert the denominator to positive number

11/-39 = (11×-1)/(-39×-1) = -11/39

5/26 + -11/39

Now let us take the LCM for 26 and 39 which is 78

(5×3)/(26×3) + (-11×2)/(39×2)

15/78 + -22/78

Since the denominators are same we can add them directly

(15-22)/78 = -7/78

(vii) -16/9 + -5/12

Solution: let us take the LCM for 9 and 12 which is 108

(-16×12)/(9×12) + (-5×9)/(12×9)

-192/108 + -45/108

Since the denominators are same we can add them directly

(-192-45)/108 = -237/108

Further divide the fraction by 3 we get,

-237/108 = -79/36

(viii) -13/8 + 5/36

Solution: let us take the LCM for 8 and 36 which is 72

(-13×9)/(8×9) + (5×2)/(36×2)

-117/72 + 10/72

Since the denominators are same we can add them directly

(-117+10)/72 = -107/72

(ix) 0 + -3/5

Solution: We know that anything added to 0 results in the same.

0 + -3/5 = -3/5

(x) 1 + -4/5

Solution: let us take the LCM for 1 and 5 which is 5

(1×5)/(1×5) + (-4×1)/(5×1)

5/5 + -4/5

Since the denominators are same we can add them directly

(5-4)/5 = 1/5

4. Add and express the sum as a mixed fraction:

(i) -12/5 and 43/10

Solution: let us add the given fraction

-12/5 + 43/10

let us take the LCM for 5 and 10 which is 10

(-12×2)/(5×2) + (43×1)/(10×1)

-24/10 + 43/10

Since the denominators are same we can add them directly

(-24+43)/10 = 19/10

19/10 can be written as (1frac{9}{10}) in mixed fraction.

(ii) 24/7 and -11/4

Solution: let us add the given fraction

24/7 + -11/4

let us take the LCM for 7 and 4 which is 28

(24×4)/(7×4) + (-11×7)/(4×7)

96/28 + -77/28

Since the denominators are same we can add them directly

(96-77)/28 = 19/28

(iii) -31/6 and -27/8

Solution: let us add the given fraction

-31/6 + -27/8

let us take the LCM for 6 and 8 which is 24

(-31×4)/(6×4) + (-27×3)/(8×3)

-124/24 + -81/24

Since the denominators are same we can add them directly

(-124-81)/24 = -205/24

-205/24 can be written as (-8frac{13}{24}) in mixed fraction.

(iv) 101/6 and 7/8

Solution: let us add the given fraction

101/6 + 7/8

let us take the LCM for 6 and 8 which is 24

(101×4)/(6×4) + (7×3)/(8×3)

404/24 + 21/24

Since the denominators are same we can add them directly

(404+21)/24 = 425/24

425/24 can be written as (17frac{17}{24}) in mixed fraction.


EXERCISE 1.2 PAGE NO: 1.14

1. Verify commutativity of addition of rational numbers for each of the following pairs of rational numbers:

(i) -11/5 and 4/7

Solution: By using the commutativity law, the addition of rational numbers is commutative ∴ a/b + c/d = c/d + a/b

In order to verify the above property let us consider the given fraction

-11/5 and 4/7 as

-11/5 + 4/7 and 4/7 + -11/5

The denominators are 5 and 7

By taking LCM for 5 and 7 is 35

We rewrite the given fraction in order to get the same denominator

Now, -11/5 = (-11 × 7) / (5 ×7) = -77/35

4/7 = (4 ×5) / (7 ×5) = 20/35

Since the denominators are same we can add them directly

-77/35 + 20/35 = (-77+20)/35 = -57/35

4/7 + -11/5

The denominators are 7 and 5

By taking LCM for 7 and 5 is 35

We rewrite the given fraction in order to get the same denominator

Now, 4/7 = (4 × 5) / (7 ×5) = 20/35

-11/5 = (-11 ×7) / (5 ×7) = -77/35

Since the denominators are same we can add them directly

20/35 + -77/35 = (20 + (-77))/35 = (20-77)/35 = -57/35

∴ -11/5 + 4/7 = 4/7 + -11/5 is satisfied.

(ii) 4/9 and 7/-12

Solution: Firstly we need to convert the denominators to positive numbers.

7/-12 = (7 × -1)/ (-12 × -1) = -7/12

By using the commutativity law, the addition of rational numbers is commutative.

∴ a/b + c/d = c/d + a/b

In order to verify the above property let us consider the given fraction

4/9 and -7/12 as

4/9 + -7/12 and -7/12 + 4/9

The denominators are 9 and 12

By taking LCM for 9 and 12 is 36

We rewrite the given fraction in order to get the same denominator

Now, 4/9 = (4 × 4) / (9 ×4) = 16/36

-7/12 = (-7 ×3) / (12 ×3) = -21/36

Since the denominators are same we can add them directly

16/36 + (-21)/36 = (16 + (-21))/36 = (16-21)/36 = -5/36

-7/12 + 4/9

The denominators are 12 and 9

By taking LCM for 12 and 9 is 36

We rewrite the given fraction in order to get the same denominator

Now, -7/12 = (-7 ×3) / (12 ×3) = -21/36

4/9 = (4 × 4) / (9 ×4) = 16/36

Since the denominators are same we can add them directly

-21/36 + 16/36 = (-21 + 16)/36 = -5/36

∴ 4/9 + -7/12 = -7/12 + 4/9 is satisfied.

(iii) -3/5 and -2/-15

Solution:

By using the commutativity law, the addition of rational numbers is commutative.

∴ a/b + c/d = c/d + a/b

In order to verify the above property let us consider the given fraction

-3/5 and -2/-15 as

-3/5 + -2/-15 and -2/-15 + -3/5

-2/-15 = 2/15

The denominators are 5 and 15

By taking LCM for 5 and 15 is 15

We rewrite the given fraction in order to get the same denominator

Now, -3/5 = (-3 × 3) / (5×3) = -9/15

2/15 = (2 ×1) / (15 ×1) = 2/15

Since the denominators are same we can add them directly

-9/15 + 2/15 = (-9 + 2)/15 = -7/15

-2/-15 + -3/5

-2/-15 = 2/15

The denominators are 15 and 5

By taking LCM for 15 and 5 is 15

We rewrite the given fraction in order to get the same denominator

Now, 2/15 = (2 ×1) / (15 ×1) = 2/15

-3/5 = (-3 × 3) / (5×3) = -9/15

Since the denominators are same we can add them directly

2/15 + -9/15 = (2 + (-9))/15 = (2-9)/15 = -7/15

∴ -3/5 + -2/-15 = -2/-15 + -3/5 is satisfied.

(iv) 2/-7 and 12/-35

Solution: Firstly we need to convert the denominators to positive numbers.

2/-7 = (2 × -1)/ (-7 × -1) = -2/7

12/-35 = (12 × -1)/ (-35 × -1) = -12/35

By using the commutativity law, the addition of rational numbers is commutative.

∴ a/b + c/d = c/d + a/b

In order to verify the above property let us consider the given fraction

-2/7 and -12/35 as

-2/7 + -12/35 and -12/35 + -2/7

The denominators are 7 and 35

By taking LCM for 7 and 35 is 35

We rewrite the given fraction in order to get the same denominator

Now, -2/7 = (-2 × 5) / (7 ×5) = -10/35

-12/35 = (-12 ×1) / (35 ×1) = -12/35

Since the denominators are same we can add them directly

-10/35 + (-12)/35 = (-10 + (-12))/35 = (-10-12)/35 = -22/35

-12/35 + -2/7

The denominators are 35 and 7

By taking LCM for 35 and 7 is 35

We rewrite the given fraction in order to get the same denominator

Now, -12/35 = (-12 ×1) / (35 ×1) = -12/35

-2/7 = (-2 × 5) / (7 ×5) = -10/35

Since the denominators are same we can add them directly

-12/35 + -10/35 = (-12 + (-10))/35 = (-12-10)/35 = -22/35

∴ -2/7 + -12/35 = -12/35 + -2/7 is satisfied.

(v) 4 and -3/5

Solution: By using the commutativity law, the addition of rational numbers is commutative.

∴ a/b + c/d = c/d + a/b

In order to verify the above property let us consider the given fraction

4/1 and -3/5 as

4/1 + -3/5 and -3/5 + 4/1

The denominators are 1 and 5

By taking LCM for 1 and 5 is 5

We rewrite the given fraction in order to get the same denominator

Now, 4/1 = (4 × 5) / (1×5) = 20/5

-3/5 = (-3 ×1) / (5 ×1) = -3/5

Since the denominators are same we can add them directly

20/5 + -3/5 = (20 + (-3))/5 = (20-3)/5 = 17/5

-3/5 + 4/1

The denominators are 5 and 1

By taking LCM for 5 and 1 is 5

We rewrite the given fraction in order to get the same denominator

Now, -3/5 = (-3 ×1) / (5 ×1) = -3/5

4/1 = (4 × 5) / (1×5) = 20/5

Since the denominators are same we can add them directly

-3/5 + 20/5 = (-3 + 20)/5 = 17/5

∴ 4/1 + -3/5 = -3/5 + 4/1 is satisfied.

(vi) -4 and 4/-7

Solution: Firstly we need to convert the denominators to positive numbers.

4/-7 = (4 × -1)/ (-7 × -1) = -4/7

By using the commutativity law, the addition of rational numbers is commutative.

∴ a/b + c/d = c/d + a/b

In order to verify the above property let us consider the given fraction

-4/1 and -4/7 as

-4/1 + -4/7 and -4/7 + -4/1

The denominators are 1 and 7

By taking LCM for 1 and 7 is 7

We rewrite the given fraction in order to get the same denominator

Now, -4/1 = (-4 × 7) / (1×7) = -28/7

-4/7 = (-4 ×1) / (7 ×1) = -4/7

Since the denominators are same we can add them directly

-28/7 + -4/7 = (-28 + (-4))/7 = (-28-4)/7 = -32/7

-4/7 + -4/1

The denominators are 7 and 1

By taking LCM for 7 and 1 is 7

We rewrite the given fraction in order to get the same denominator

Now, -4/7 = (-4 ×1) / (7 ×1) = -4/7

-4/1 = (-4 × 7) / (1×7) = -28/7

Since the denominators are same we can add them directly

-4/7 + -28/7 = (-4 + (-28))/7 = (-4-28)/7 = -32/7

∴ -4/1 + -4/7 = -4/7 + -4/1 is satisfied.

2. Verify associativity of addition of rational numbers i.e., (x + y) + z = x + (y + z), when:

(i) x = ½, y = 2/3, z = -1/5

Solution: As the property states (x + y) + z = x + (y + z)

Use the values as such,

(1/2 + 2/3) + (-1/5) = 1/2 + (2/3 + (-1/5))

Let us consider LHS (1/2 + 2/3) + (-1/5)

Taking LCM for 2 and 3 is 6

(1× 3)/(2×3) + (2×2)/(3×2)

3/6 + 4/6

Since the denominators are same we can add them directly,

3/6 + 4/6 = 7/6

7/6 + (-1/5)

Taking LCM for 6 and 5 is 30

(7×5)/(6×5) + (-1×6)/(5×6)

35/30 + (-6)/30

Since the denominators are same we can add them directly,

(35+(-6))/30 = (35-6)/30 = 29/30

Let us consider RHS 1/2 + (2/3 + (-1/5))

Taking LCM for 3 and 5 is 15

(2/3 + (-1/5)) = (2×5)/(3×5) + (-1×3)/(5×3)

= 10/15 + (-3)/15

Since the denominators are same we can add them directly,

10/15 + (-3)/15 = (10-3)/15 = 7/15

1/2 + 7/15

Taking LCM for 2 and 15 is 30

1/2 + 7/15 = (1×15)/(2×15) + (7×2)/(15×2)

= 15/30 + 14/30

Since the denominators are same we can add them directly,

= (15 + 14)/30 = 29/30

∴ LHS = RHS associativity of addition of rational numbers is verified.

(ii) x = -2/5, y = 4/3, z = -7/10

Solution: As the property states (x + y) + z = x + (y + z)

Use the values as such,

(-2/5 + 4/3) + (-7/10) = -2/5 + (4/3 + (-7/10))

Let us consider LHS (-2/5 + 4/3) + (-7/10)

Taking LCM for 5 and 3 is 15

(-2× 3)/(5×3) + (4×5)/(3×5)

-6/15 + 20/15

Since the denominators are same we can add them directly,

-6/15 + 20/15= (-6+20)/15 = 14/15

14/15 + (-7/10)

Taking LCM for 15 and 10 is 30

(14×2)/(15×2) + (-7×3)/(10×3)

28/30 + (-21)/30

Since the denominators are same we can add them directly,

(28+(-21))/30 = (28-21)/30 = 7/30

Let us consider RHS -2/5 + (4/3 + (-7/10))

Taking LCM for 3 and 10 is 30

(4/3 + (-7/10)) = (4×10)/(3×10) + (-7×3)/(10×3)

= 40/30 + (-21)/30

Since the denominators are same we can add them directly,

40/30 + (-21)/30 = (40-21)/30 = 19/30

-2/5 + 19/30

Taking LCM for 5 and 30 is 30

-2/5 + 19/30 = (-2×6)/(5×6) + (19×1)/(30×1)

= -12/30 + 19/30

Since the denominators are same we can add them directly,

= (-12 + 19)/30 = 7/30

∴ LHS = RHS associativity of addition of rational numbers is verified.

(iii) x = -7/11, y = 2/-5, z = -3/22

Solution: Firstly convert the denominators to positive numbers

2/-5 = (2×-1)/ (-5×-1) = -2/5

As the property states (x + y) + z = x + (y + z)

Use the values as such,

(-7/11 + -2/5) + (-3/22) = -7/11 + (-2/5 + (-3/22))

Let us consider LHS (-7/11 + -2/5) + (-3/22)

Taking LCM for 11 and 5 is 55

(-7×5)/(11×5) + (-2×11)/(5×11)

-35/55 + -22/55

Since the denominators are same we can add them directly,

-35/55 + -22/55 = (-35-22)/55 = -57/55

-57/55 + (-3/22)

Taking LCM for 55 and 22 is 110

(-57×2)/(55×2) + (-3×5)/(22×5)

-114/110 + (-15)/110

Since the denominators are same we can add them directly,

(-114+(-15))/110 = (-114-15)/110 = -129/110

Let us consider RHS -7/11 + (-2/5 + (-3/22))

Taking LCM for 5 and 22 is 110

(-2/5 + (-3/22))= (-2×22)/(5×22) + (-3×5)/(22×5)

= -44/110 + (-15)/110

Since the denominators are same we can add them directly,

-44/110 + (-15)/110 = (-44-15)/110 = -59/110

-7/11 + -59/110

Taking LCM for 11 and 110 is 110

-7/11 + -59/110 = (-7×10)/(11×10) + (-59×1)/(110×1)

= -70/110 + -59/110

Since the denominators are same we can add them directly,

= (-70 -59)/110 = -129/110

∴ LHS = RHS associativity of addition of rational numbers is verified.

(iv) x = -2, y = 3/5, z = -4/3

Solution: As the property states (x + y) + z = x + (y + z)

Use the values as such,

(-2/1 + 3/5) + (-4/3) = -2/1 + (3/5 + (-4/3))

Let us consider LHS (-2/1 + 3/5) + (-4/3)

Taking LCM for 1 and 5 is 5

(-2×5)/(1×5) + (3×1)/(5×1)

-10/5 + 3/5

Since the denominators are same we can add them directly,

-10/5 + 3/5= (-10+3)/5 = -7/5

-7/5 + (-4/3)

Taking LCM for 5 and 3 is 15

(-7×3)/(5×3) + (-4×5)/(3×5)

-21/15 + (-20)/15

Since the denominators are same we can add them directly,

(-21+(-20))/15 = (-21-20)/15 = -41/15

Let us consider RHS -2/1 + (3/5 + (-4/3))

Taking LCM for 5 and 3 is 15

(3/5 + (-4/3)) = (3×3)/(5×3) + (-4×5)/(3×5)

= 9/15 + (-20)/15

Since the denominators are same we can add them directly,

9/15 + (-20)/15 = (9-20)/15 = -11/15

-2/1 + -11/15

Taking LCM for 1 and 15 is 15

-2/1 + -11/15 = (-2×15)/(1×15) + (-11×1)/(15×1)

= -30/15 + -11/15

Since the denominators are same we can add them directly,

= (-30 -11)/15 = -41/15

∴ LHS = RHS associativity of addition of rational numbers is verified.

3. Write the additive of each of the following rational numbers:

(i) -2/17

(ii) 3/-11

(iii) -17/5

(iv) -11/-25

Solution:

(i) The additive inverse of -2/17 is 2/17

(ii) The additive inverse of 3/-11 is 3/11

(iii) The additive inverse of -17/5 is 17/5

(iv) The additive inverse of -11/-25 is -11/25

4. Write the negative(additive) inverse of each of the following:

(i) -2/5

(ii) 7/-9

(iii) -16/13

(iv) -5/1

(v) 0

(vi) 1

(vii) – 1

Solution:

(i) The negative (additive) inverse of -2/5 is 2/5

(ii) The negative (additive) inverse of 7/-9 is 7/9

(iii) The negative (additive) inverse of -16/13 is 16/13

(iv) The negative (additive) inverse of -5/1 is 5

(v) The negative (additive) inverse of 0 is 0

(vi) The negative (additive) inverse of 1 is -1

(vii) The negative (additive) inverse of -1 is 1

5. Using commutativity and associativity of addition of rational numbers, express each of the following as a rational number:

(i) 2/5 + 7/3 + -4/5 + -1/3

Solution: Firstly group the rational numbers with same denominators

2/5 + -4/5 + 7/3 + -1/3

Now the denominators which are same can be added directly.

(2+(-4))/5 + (7+(-1))/3

(2-4)/5 + (7-1)/3

-2/5 + 6/3

By taking LCM for 5 and 3 we get, 15

(-2×3)/(5×3) + (6×5)/(3×5)

-6/15 + 30/15

Since the denominators are same can be added directly

(-6+30)/15 = 24/15

Further can be divided by 3 we get,

24/15 = 8/5

(ii) 3/7 + -4/9 + -11/7 + 7/9

Solution: Firstly group the rational numbers with same denominators

3/7 + -11/7 + -4/9 + 7/9

Now the denominators which are same can be added directly.

(3+ (-11))/7 + (-4+ 7)/9

(3-11)/7 + (-4+7)/9

-8/7 + 3/9

-8/7 + 1/3

By taking LCM for 7 and 3 we get, 21

(-8×3)/ (7×3) + (1×7)/ (3×7)

-24/21 + 7/21

Since the denominators are same can be added directly

(-24+7)/21 = -17/21

(iii) 2/5 + 8/3 + -11/15 + 4/5 + -2/3

Solution: Firstly group the rational numbers with same denominators

2/5 + 4/5 + 8/3 + -2/3 + -11/15

Now the denominators which are same can be added directly.

(2 + 4)/5 + (8 + (-2))/3 + -11/15

6/5 + (8-2)/3 + -11/15

6/5 + 6/3 + -11/15

6/5 + 2/1 + -11/15

By taking LCM for 5, 1 and 15 we get, 15

(6×3)/ (5×3) + (2×15)/ (1×15) + (-11×1)/ (15×1)

18/15 + 30/15 + -11/15

Since the denominators are same can be added directly

(18+30+ (-11))/15 = (18+30-11)/15 = 37/15

(iv) 4/7 + 0 + -8/9 + -13/7 + 17/21

Solution: Firstly group the rational numbers with same denominators

4/7 + -13/7 + -8/9 + 17/21

Now the denominators which are same can be added directly.

(4 + (-13))/7 + -8/9 + 17/21

(4-13)/7 + -8/9 + 17/21

-9/7 + -8/9 + 17/21

By taking LCM for 7, 9 and 21 we get, 63

(-9×9)/ (7×9) + (-8×7)/ (9×7) + (17×3)/ (21×3)

-81/63 + -56/63 + 51/63

Since the denominators are same can be added directly

(-81+(-56)+ 51)/63 = (-81-56+51)/63 = -86/63

6. Re-arrange suitably and find the sum in each of the following:

(i) 11/12 + -17/3 + 11/2 + -25/2

Solution: Firstly group the rational numbers with same denominators

11/12 + -17/3 + (11-25)/2

11/12 + -17/3 + -14/2

By taking LCM for 12, 3 and 2 we get, 12

(11×1)/(12×1) + (-17×4)/(3×4) + (-14×6)/(2×6)

11/12 + -68/12 + -84/12

Since the denominators are same can be added directly

(11-68-84)/12 = -141/12

(ii)-6/7 + -5/6 + -4/9 + -15/7

Solution: Firstly group the rational numbers with same denominators

-6/7 + -15/7 + -5/6 + -4/9

(-6 -15)/7 + -5/6 + -4/9

-21/7 + -5/6 + -4/9

-3/1 + -5/6 + -4/9

By taking LCM for 1, 6 and 9 we get, 18

(-3×18)/(1×18) + (-5×3)/(6×3) + (-4×2)/(9×2)

-54/18 + -15/18 + -8/18

Since the denominators are same can be added directly

(-54-15-8)/18 = -77/18

(iii) 3/5 + 7/3 + 9/ 5+ -13/15 + -7/3

Solution: Firstly group the rational numbers with same denominators

3/5 + 9/5 + 7/3 + -7/3 + -13/15

(3+9)/5 + -13/15

12/5 + -13/15

By taking LCM for 5 and 15 we get, 15

(12×3)/(5×3) + (-13×1)/(15×1)

36/15 + -13/15

Since the denominators are same can be added directly

(36-13)/15 = 23/15

(iv) 4/13 + -5/8 + -8/13 + 9/13

Solution: Firstly group the rational numbers with same denominators

4/13 + -8/13 + 9/13 + -5/8

(4-8+9)/13 + -5/8

5/13 + -5/8

By taking LCM for 13 and 8 we get, 104

(5×8)/(13×8) + (-5×13)/(8×13)

40/104 + -65/104

Since the denominators are same can be added directly

(40-65)/104 = -25/104

(v) 2/3 + -4/5 + 1/3 + 2/5

Solution: Firstly group the rational numbers with same denominators

2/3 + 1/3 + -4/5 + 2/5

(2+1)/3 + (-4+2)/5

3/3 + -2/5

1/1 + -2/5

By taking LCM for 1 and 5 we get, 5

(1×5)/(1×5) + (-2×1)/(5×1)

5/5 + -2/5

Since the denominators are same can be added directly

(5-2)/5 = 3/5

(vi) 1/8 + 5/12 + 2/7 + 7/12 + 9/7 + -5/16

Solution: Firstly group the rational numbers with same denominators

1/8 + 5/12 + 7/12 + 2/7 + 9/7 + -5/16

1/8 + (5+7)/12 + (2+9)/7 + -5/16

1/8 + 12/12 + 11/7 + -5/16

1/8 + 1/1 + 11/7 + -5/16

By taking LCM for 8, 1, 7 and 16 we get, 112

(1×14)/(8×14) + (1×112)/(1×112) + (11×16)/(7×16) + (-5×7)/(16×7)

14/112 + 112/112 + 176/112 + -35/112

Since the denominators are same can be added directly

(14+112+176-35)/112 = 267/112


EXERCISE 1.3 PAGE NO: 1.18

1. Subtract the first rational number from the second in each of the following:

(i) 3/8, 5/8

(ii) -7/9, 4/9

(iii) -2/11, -9/11

(iv) 11/13, -4/13

(v) ¼, -3/8

(vi) -2/3, 5/6

(vii) -6/7, -13/14

(viii) -8/33, -7/22

Solution:

(i) let us subtract

5/8 – 3/8

Since the denominators are same we can subtract directly

(5-3)/8 = 2/8

Further we can divide by 2 we get,

2/8 = 1/4

(ii) let us subtract

4/9 – -7/9

Since the denominators are same we can subtract directly

(4+7)/9 = 11/9

(iii) let us subtract

-9/11 – -2/11

Since the denominators are same we can subtract directly

(-9+2)/11 = -7/11

(iv) let us subtract

-4/13 – 11/13

Since the denominators are same we can subtract directly

(-4-11)/13 = -15/13

(v) let us subtract

-3/8 – 1/4

By taking LCM for 8 and 4 which is 8

-3/8 – 1/4 = (-3×1)/(8×1) – (1×2)/(4×2) = -3/8 – 2/8

Since the denominators are same we can subtract directly

(-3-2)/8 = -5/8

(vi) let us subtract

5/6 – -2/3

By taking LCM for 6 and 3 which is 6

5/6 – -2/3 = (5×1)/(6×1) – (-2×2)/(3×2) = 5/6 – -4/6

Since the denominators are same we can subtract directly

(5+4)/6 = 9/6

Further we can divide by 3 we get,

9/6 = 3/2

(vii) let us subtract

-13/14 – -6/7

By taking LCM for 14 and 7 which is 14

-13/14 – -6/7 = (-13×1)/(14×1) – (-6×2)/(7×2) = -13/14 – -12/14

Since the denominators are same we can subtract directly

(-13+12)/14 = -1/14

(viii) let us subtract

-7/22 – -8/33

By taking LCM for 22 and 33 which is 66

-7/22 – -8/33 = (-7×3)/(22×3) – (-8×2)/(33×2) = -21/66 – -16/66

Since the denominators are same we can subtract directly

(-21+16)/66 = -5/66

2. Evaluate each of the following:

(i) 2/3 – 3/5

Solution: By taking LCM for 3 and 5 which is 15

2/3 – 3/5 = (2×5 – 3×3)/15

= 1/15

(ii) -4/7 – 2/-3

Solution: convert the denominator to positive number by multiplying by -1

2/-3 = -2/3

-4/7 – -2/3

By taking LCM for 7 and 3 which is 21

-4/7 – -2/3 = (-4×3 – -2×7)/21

= (-12+14)/21

= 2/21

(iii) 4/7 – -5/-7

Solution: convert the denominator to positive number by multiplying by -1

-5/-7 = 5/7

4/7 – 5/7

Since the denominators are same we can subtract directly

(4-5)/7 = -1/7

(iv) -2 – 5/9

Solution: By taking LCM for 1 and 9 which is 9

-2/1 – 5/9 = (-2×9 – 5×1)/9

= (-18 – 5)/9

= -23/9

(v) -3/-8 – -2/7

Solution: convert the denominator to positive number by multiplying by -1

-3/-8 = 3/8

3/8 – -2/7

By taking LCM for 8 and 7 which is 56

3/8 – -2/7 = (3×7 – -2×8)/56

= (21 + 16)/56

= 37/56

(vi) -4/13 – -5/26

Solution: By taking LCM for 13 and 26 which is 26

-4/13 – -5/26 = (-4×2 – -5×1)/26

= (-8 + 5)/26

= -3/26

(vii) -5/14 – -2/7

Solution: By taking LCM for 14 and 7 which is 14

-5/14 – -2/7 = (-5×1 – -2×2)/14

= (-5 + 4)/14

= -1/14

(viii) 13/15 – 12/25

Solution: By taking LCM for 15 and 25 which is 75

13/15 – 12/25 = (13×5 – 12×3)/75

= (65 – 36)/75

= 29/75

(ix) -6/13 – -7/13

Solution: Since the denominators are same we can subtract directly

-6/13 – -7/13 = (-6+7)/13

= 1/13

(x) 7/24 – 19/36

Solution: By taking LCM for 24 and 36 which is 72

7/24 – 19/36 = (7×3 – 19×2)/72

= (21 – 38)/72

= -17/72

(xi) 5/63 – -8/21

Solution: By taking LCM for 63 and 21 which is 63

5/63 – -8/21 = (5×1 – -8×3)/63

= (5 + 24)/63

= 29/63

3. The sum of the two numbers is 5/9. If one of the numbers is 1/3, find the other.

Solution: Let us note down the given details

Sum of two numbers = 5/9

One of the number = 1/3

By using the formula,

Other number = sum of number – given number

= 5/9 – 1/3

By taking LCM for 9 and 3 which is 9

5/9 – 1/3 = (5×1 – 1×3)/9

= (5 – 3)/9

= 2/9

∴ the other number is 2/9

4. The sum of the two numbers is -1/3. If one of the numbers is -12/3, find the other.

Solution: Let us note down the given details

Sum of two numbers = -1/3

One of the number = -12/3

By using the formula,

Other number = sum of number – given number

= -1/3 – -12/3

Since the denominators are same we can subtract directly

= (-1+12)/3 = 11/3

∴ the other number is 11/3

5. The sum of the two numbers is -4/3. If one of the numbers is -5, find the other.

Solution: Let us note down the given details

Sum of two numbers = -4/3

One of the number = -5/1

By using the formula,

Other number = sum of number – given number

= -4/3 – -5/1

By taking LCM for 3 and 1 which is 3

-4/3 – -5/1 = (-4×1 – -5×3)/3

= (-4 + 15)/3

= 11/3

∴ the other number is 11/3

6. The sum of the two rational numbers is -8. If one of the numbers is -15/7, find the other.

Solution: Let us note down the given details

Sum of two rational numbers = -8/1

One of the number = -15/7

Let us consider the other number as x

x + -15/7 = -8

(7x -15)/7 = -8

7x -15 = -8×7

7x – 15 = -56

7x = -56+15

x = -41/7

∴ the other number is -41/7

7. What should be added to -7/8 so as to get 5/9?

Solution: Let us consider a number as x to be added to -7/8 to get 5/9

So, -7/8 + x = 5/9

(-7 + 8x)/8 = 5/9

(-7 + 8x) × 9 = 5 × 8

-63 + 72x = 40

72x = 40 + 63

x = 103/72

∴ the required number is 103/72

8. What number should be added to -5/11 so as to get 26/33?

Solution: Let us consider a number as x to be added to -5/11 to get 26/33

So, -5/11 + x = 26/33

x = 26/33 + 5/11

let us take LCM for 33 and 11 which is 33

x = (26×1 + 5×3)/33

= (26 + 15)/33

= 41/33

∴ the required number is 41/33

9. What number should be added to -5/7 to get -2/3?

Solution: Let us consider a number as x to be added to -5/7 to get -2/3

So, -5/7 + x = -2/3

x = -2/3 + 5/7

let us take LCM for 3 and 7 which is 21

x = (-2×7 + 5×3)/21

= (-14 + 15)/21

= 1/21

∴ the required number is 1/21

10. What number should be subtracted from -5/3 to get 5/6?

Solution: Let us consider a number as x to be subtracted from -5/3 to get 5/6

So, -5/3 – x = 5/6

x = -5/3 – 5/6

let us take LCM for 3 and 6 which is 6

x = (-5×2 – 5×1)/6

= (-10 – 5)/6

= -15/6

Further we can divide by 3 we get,

-15/6 = -5/2

∴ the required number is -5/2

11. What number should be subtracted from 3/7 to get 5/4?

Solution: Let us consider a number as x to be subtracted from 3/7 to get 5/4

So, 3/7 – x = 5/4

x = 3/7 – 5/4

let us take LCM for 7 and 4 which is 28

x = (3×4 – 5×7)/28

= (12 – 35)/28

= -23/28

∴ the required number is -23/28

12. What should be added to (2/3 + 3/5) to get -2/15?

Solution: Let us consider a number as x to be added to (2/3 + 3/5) to get -2/15

x + (2/3 + 3/5) = -2/15

By taking LCM of 3 and 5 which is 15 we get,

(15x + 2×5 + 3×3)15 = -2/15

15x + 10 + 9 = -2

15x = -2-19

x = -21/15

Further we can divide by 3 we get,

-21/15 = -7/5

∴ the required number is -7/5

13. What should be added to (1/2 + 1/3 + 1/5) to get 3?

Solution: Let us consider a number as x to be added to (1/2 + 1/3 + 1/5) to get 3

x + (1/2 + 1/3 + 1/5) = 3

By taking LCM of 2, 3 and 5 which is 30 we get,

(30x + 1×15 + 1×10 + 1×6 )30 = 3

30x + 15 + 10 + 6 = 3 × 30

30x + 31 = 90

30x = 90-31

x = 59/30

∴ the required number is 59/30

14. What number should be subtracted from (3/4 – 2/3) to get -1/6?

Solution: Let us consider a number as x to be subtracted from (3/4 – 2/3) to get -1/6

So, (3/4 – 2/3) – x = -1/6

x = 3/4 – 2/3 + 1/6

Let us take LCM for 4 and 3 which is 12

x = (3×3 – 2×4)/12 + 1/6

= (9 – 8)/12 + 1/6

= 1/12 + 1/6

Let us take LCM for 12 and 6 which is 12

= (1×1 + 1×2)/12

= 3/12

Further we can divide by 3 we get,

3/12 = 1/4 ∴ the required number is ¼

15. Fill in the blanks:

(i) -4/13 – -3/26 = ….

Solution:

-4/13 – -3/26

Let us take LCM for 13 and 26 which is 26

(-4×2 + 3×1)/26

(-8+3)/26 = -5/26

(ii) -9/14 + …. = -1

Solution:

Let us consider the number to be added as x

-9/14 + x = -1

x = -1 + 9/14

By taking LCM as 14 we get,

x = (-1×14 + 9)/14

= (-14+9)/14

= -5/14

(iii) -7/9 + …. =3

Solution:

Let us consider the number to be added as x

-7/9 + x = 3

x = 3 + 7/9

By taking LCM as 9 we get,

x = (3×9 + 7)/9

= (27 + 7)/9

= 34/9

(iv) … + 15/23 = 4

Solution:

Let us consider the number to be added as x

x + 15/23 = 4

x = 4 – 15/23

By taking LCM as 23 we get,

x = (4×23 – 15)/23

= (92 – 15)/23

= 77/23


EXERCISE 1.4 PAGE NO: 1.22

1. Simplify each of the following and write as a rational number of the form p/q:

(i) 3/4 + 5/6 + -7/8

Solution:

3/4 + 5/6 -7/8

By taking LCM for 4, 6 and 8 which is 24

((3×6) + (5×4) – (7×3))/24

(18 + 20 – 21)/24

(38-21)/24

17/24

(ii) 2/3 + -5/6 + -7/9

Solution:

2/3 + -5/6 + -7/9

By taking LCM for 3, 6 and 9 which is 18

((2×6) + (-5×3) + (-7×2))/18

(12 – 15 – 14)/18

-17/18

(iii) -11/2 + 7/6 + -5/8

Solution:

-11/2 + 7/6 + -5/8

By taking LCM for 2, 6 and 8 which is 24

((-11×12) + (7×4) + (-5×3))/24

(-132 + 28 – 15)/24

-119/24

(iv) -4/5 + -7/10 + -8/15

Solution:

-4/5 + -7/10 + -8/15

By taking LCM for 5, 10 and 15 which is 30

((-4×6) + (-7×3) + (-8×2))/30

(-24 – 21 – 16)/30

-61/30

(v) -9/10 + 22/15 + 13/-20

Solution:

-9/10 + 22/15 + 13/-20

By taking LCM for 10, 15 and 20 which is 60

((-9×6) + (22×4) + (-13×3))/60

(-54 + 88 – 39)/60

-5/60 = -1/12

(vi) 5/3 + 3/-2 + -7/3 +3

Solution:

5/3 + 3/-2 + -7/3 +3

By taking LCM for 3, 2, 3 and 1 which is 6

((5×2) + (-3×3) + (-7×2) + (3×6))/6

(10 – 9 – 14 + 18)/6

5/6

2. Express each of the following as a rational number of the form p/q:

(i) -8/3 + -1/4 + -11/6 + 3/8 – 3

Solution:

-8/3 + -1/4 + -11/6 + 3/8 – 3

By taking LCM for 3, 4, 6, 8 and 1 which is 24

((-8×8) + (-1×6) + (-11×4) + (3×3) – (3×24))/24

(-64 – 6 – 44 + 9 – 72)/24

-177/24

Further divide by 3 we get,

-177/24 = -59/8

(ii) 6/7 + 1 + -7/9 + 19/21 + -12/7

Solution:

6/7 + 1 + -7/9 + 19/21 + -12/7

By taking LCM for 7, 1, 9, 21 and 7 which is 63

((6×9) + (1×63) + (-7×7) + (19×3) + (-12×9))/63

(54 + 63 – 49 + 57 – 108)/63

17/63

(iii) 15/2 + 9/8 + -11/3 + 6 + -7/6

Solution:

15/2 + 9/8 + -11/3 + 6 + -7/6

By taking LCM for 2, 8, 3, 1 and 6 which is 24

((15×12) + (9×3) + (-11×8) + (6×24) + (-7×4))/24

(180 + 27 – 88 + 144 – 28)/24

235/24

(iv) -7/4 +0 + -9/5 + 19/10 + 11/14

Solution:

-7/4 +0 + -9/5 + 19/10 + 11/14

By taking LCM for 4, 5, 10 and 14 which is 140

((-7×35) + (-9×28) + (19×14) + (11×10))/140

(-245 – 252 + 266 + 110)/140

-121/140

(v) -7/4 +5/3 + -1/2 + -5/6 + 2

Solution:

-7/4 +5/3 + -1/2 + -5/6 + 2

By taking LCM for 4, 3, 2, 6 and 1 which is 12

((-7×3) + (5×4) + (-1×6) + (-5×2) + (2×12))/12

(-21 + 20 – 6 – 10 + 24)/12

7/12

3. Simplify:

(i) -3/2 + 5/4 – 7/4

Solution:

-3/2 + 5/4 – 7/4

By taking LCM for 2 and 4 which is 4

((-3×2) + (5×1) – (7×1))/4

(-6 + 5 – 7)/4

-8/4

Further divide by 2 we get,

-8/2 = -2

(ii) 5/3 – 7/6 + -2/3

Solution:

5/3 – 7/6 + -2/3

By taking LCM for 3 and 6 which is 6

((5×2) – (7×1) + (-2×2))/6

(10 – 7 – 4)/6

-1/6

(iii) 5/4 – 7/6 – -2/3

Solution:

5/4 – 7/6 – -2/3

By taking LCM for 4, 6 and 3 which is 12

((5×3) – (7×2) – (-2×4))/12

(15 – 14 + 8)/12

9/12

Further can divide by 3 we get,

9/12 = 3/4

(iv) -2/5 – -3/10 – -4/7

Solution:

-2/5 – -3/10 – -4/7

By taking LCM for 5, 10 and 7 which is 70

((-2×14) – (-3×7) – (-4×10))/70

(-28 + 21 + 40)/70

33/70

(v) 5/6 + -2/5 – -2/15

Solution:

5/6 + -2/5 – -2/15

By taking LCM for 6, 5 and 15 which is 30

((5×5) + (-2×6) – (-2×2))/30

(25 – 12 + 4)/30

17/30

(vi) 3/8 – -2/9 + -5/36

Solution:

3/8 – -2/9 + -5/36

By taking LCM for 8, 9 and 36 which is 72

((3×9) – (-2×8) + (-5×2))/72

(27 + 16 – 10)/72

33/72

Further can divide by 3 we get,

33/72 = 11/24


EXERCISE 1.5 PAGE NO: 1.25

1. Multiply:

(i) 7/11 by 5/4

Solution:

7/11 by 5/4

(7/11) × (5/4) = (7×5)/(11×4)

= 35/44

(ii) 5/7 by -3/4

Solution:

5/7 by -3/4

(5/7) × (-3/4) = (5×-3)/(7×4)

= -15/28

(iii) -2/9 by 5/11

Solution:

-2/9 by 5/11

(-2/9) × (5/11) = (-2×5)/(9×11)

= -10/99

(iv) -3/17 by -5/-4

Solution:

-3/17 by -5/-4

(-3/17) × (-5/-4) = (-3×-5)/(17×-4)

= 15/-68

= -15/68

(v) 9/-7 by 36/-11

Solution:

9/-7 by 36/-11

(9/-7) × (36/-11) = (9×36)/(-7×-11)

= 324/77

(vi) -11/13 by -21/7

Solution:

-11/13 by -21/7

(-11/13) × (-21/7) = (-11×-21)/(13×7)

= 231/91 = 33/13

(vii) -3/5 by -4/7

Solution:

-3/5 by -4/7

(-3/5) × (-4/7) = (-3×-4)/(5×7)

= 12/35

(viii) -15/11 by 7

Solution:

-15/11 by 7

(-15/11) × 7 = (-15×7)/11

= -105/11

2. Multiply:

(i) -5/17 by 51/-60

Solution:

-5/17 by 51/-60

(-5/17) × (51/-60) = (-5×51)/(17×-60)

= -255/-1020

Further can divide by 255 we get,

-255/-1020 = 1/4

(ii) -6/11 by -55/36

Solution:

-6/11 by -55/36

(-6/11) × (-55/36) = (-6×-55)/(11×36)

= 330/396

Further can divide by 66 we get,

330/396 = 5/6

(iii) -8/25 by -5/16

Solution:

-8/25 by -5/16

(-8/25) × (-5/16) = (-8×-5)/(25×16)

= 40/400

Further can divide by 40 we get,

40/400 = 1/10

(iv) 6/7 by -49/36

Solution:

6/7 by -49/36

(6/7) × (-49/36) = (6×-49)/(7×36)

= 294/252

Further can divide by 42 we get,

294/252 = -7/6

(v) 8/-9 by -7/-16

Solution:

8/-9 by -7/-16

(8/-9) × (-7/-16) = (8×-7)/(-9×-16)

= -56/144

Further can divide by 8 we get,

-56/144 = -7/18

(vi) -8/9 by 3/64

Solution:

-8/9 by 3/64

(-8/9) × (3/64) = (-8×3)/(9×64)

= -24/576

Further can divide by 24 we get,

-24/576 = -1/24

3. Simplify each of the following and express the result as a rational number in standard form:

(i) (-16/21) × (14/5)

Solution:

(-16/21) × (14/5) = (-16/3) × (2/5) (divisible by 7)

= (-16×2)/(3×5)

= -32/15

(ii) (7/6) × (-3/28)

Solution:

(7/6) × (-3/28) = (1/2) × (-1/4) (divisible by 7 and 3)

= -1/8

(iii) (-19/36) × 16

Solution:

-19/36 × 16 = (-19/9) × 4 (divisible by 4)

= (-19×4)/9 = -76/9

(iv) (-13/9) × (27/-26)

Solution:

(-13/9) × (27/-26) = (-1/1) × (3/-2) (divisible by 13 and 9)

= -3/-2 = 3/2

(v) (-9/16) × (-64/-27)

Solution:

(-9/16) × (-64/-27) = (-1/1) × (-4/-3) (divisible by 9 and 16)

= 4/-3 = -4/3

(vi) (-50/7) × (14/3)

Solution:

(-50/7) × (14/3) = (-50/1) × (2/3) (divisible by 7)

= (-50×2)/(1×3)

= -100/3

(vii) (-11/9) × (-81/-88)

Solution:

(-11/9) × (-81/-88) = (-1/1) × (-9/-8) (divisible by 11 and 9)

= (-1×-9)/(1×-8)

= 9/-8 = -9/8

(viii) (-5/9) × (72/-25)

Solution:

(-5/9) × (72/-25) = (-1/1) × (8/-5) (divisible by 5 and 9)

= (-1×8)/(1×-5)

= -8/-5 = 8/5

4. Simplify:

(i) ((25/8) × (2/5)) – ((3/5) × (-10/9))

Solution:

((25/8) × (2/5)) – ((3/5) × (-10/9)) = (25×2)/(8×5) – (3×-10)/(5×9)

= 50/40 – -30/45

= 5/4 + 2/3 (divisible by 5 and 3)

By taking LCM for 4 and 3 which is 12

= ((5×3) + (2×4))/12

= (15+8)/12

= 23/12

(ii) ((1/2) × (1/4)) + ((1/2) × 6)

Solution:

((1/2) × (1/4)) + ((1/2) × 6) = (1×1)/(2×4) + (1×3) (divisible by 2)

= 1/8 +3

By taking LCM for 8 and 1 which is 8

= ((1×1) + (3×8))/8

= (1+24)/8

= 25/8

(iii) (-5 × (2/15)) – (-6 × (2/9))

Solution:

(-5 × (2/15)) – (-6 × (2/9)) = (-1 × (2/3)) – (-2 × (2/3)) (divisible by 5 and 3)

= (-2/3) + (4/3)

Since the denominators are same we can add directly

= (-2+4)/3

= 2/3

(iv) ((-9/4) × (5/3)) + ((13/2) × (5/6))

Solution:

((-9/4) × (5/3)) + ((13/2) × (5/6)) = (-9×5)/(4×3) + (13×5)/(2×6)

= -45/12 + 65/12

Since the denominators are same we can add directly

= (-45+65)/12

= 20/12 (divisible by 2)

= 10/6 (divisible by 2)

= 5/3

(v) ((-4/3) × (12/-5)) + ((3/7) × (21/15))

Solution:

((-4/3) × (12/-5)) + ((3/7) × (21/15)) = ((-4/1) × (4/-5)) + ((1/1) × (3/5)) (divisible by 3, 7)

= (-4×4)/(1×-5) + (1×3)/(1×5)

= -16/-5 + 3/5

Since the denominators are same we can add directly

= (16+3)/5

= 19/5

(vi) ((13/5) × (8/3)) – ((-5/2) × (11/3))

Solution:

((13/5) × (8/3)) – ((-5/2) × (11/3)) = (13×8)/(5×3) – (-5×11)/(2×3)

= 104/15 + 55/6

By taking LCM for 15 and 6 which is 30

= ((104×2) + (55×5))/30

= (208+275)/30

= 483/30

(vii) ((13/7) × (11/26)) – ((-4/3) × (5/6))

Solution:

((13/7) × (11/26)) – ((-4/3) × (5/6)) = ((1/7) × (11/2)) – ((-2/3) × (5/3)) (divisible by 13, 2)

= (1×11)/(7×2) – (-2×5)/(3×3)

= 11/14 + 10/9

By taking LCM for 14 and 9 which is 126

= ((11×9) + (10×14))/126

= (99+140)/126

= 239/126

(viii) ((8/5) × (-3/2)) + ((-3/10) × (11/16))

Solution:

((8/5) × (-3/2)) + ((-3/10) × (11/16)) = ((4/5) × (-3/1)) + ((-3/10) × (11/16)) (divisible by 2)

= (4×-3)/(5×1) + (-3×11)/(10×16)

= -12/5 – 33/160

By taking LCM for 5 and 160 which is 160

= ((-12×32) – (33×1))/160

= (-384 – 33)/160

= -417/160

5. Simplify:

(i) ((3/2) × (1/6)) + ((5/3) × (7/2) – (13/8) × (4/3))

Solution:

((3/2) × (1/6)) + ((5/3) × (7/2) – (13/8) × (4/3)) =

((1/2) × (1/2)) + ((5/3) × (7/2) – (13/2) × (1/3))

(1×1)/(2×2) + (5×7)/(3×2) – (13×1)/(2×3)

1/4 + 35/6 – 13/6

By taking LCM for 4 and 6 which is 24

((1×6) + (35×4) – (13×4))/24

(6 + 140 – 52)/24

94/24

Further divide by 2 we get, 94/24 = 47/12

(ii) ((1/4) × (2/7)) – ((5/14) × (-2/3) + (3/7) × (9/2))

Solution:

((1/4) × (2/7)) – ((5/14) × (-2/3) + (3/7) × (9/2)) =

((1/2) × (1/7)) – ((5/7) × (-1/3) + (3/7) × (9/2))

(1×1)/(2×7) – (5×-1)/(7×3) + (3×9)/(7×2)

1/14 + 5/21 + 27/14

By taking LCM for 14 and 21 which is 42

((1×3) + (5×2) + (27×3))/42

(3 + 10 + 81)/42

94/42

Further divide by 2 we get, 94/42 = 47/21

(iii) ((13/9) × (-15/2)) + ((7/3) × (8/5) + (3/5) × (1/2))

Solution:

((13/3) × (-5/2)) + ((7/3) × (8/5) + (3/5) × (1/2)) =

(13×-5)/(3×2) + (7×8)/(3×5) + (3×1)/(5×2)

-65/6 + 56/15 + 3/10

By taking LCM for 6, 15 and 10 which is 30

((-65×5) + (56×2) + (3×3))/30

(-325 + 112 + 9)/30

-204/30

Further divide by 2 we get, -204/30 = -102/15

(iv) ((3/11) × (5/6)) – ((9/12) × (4/3) + (5/13) × (6/15))

Solution:

((3/11) × (5/6)) – ((9/12) × (4/3) + (5/13) × (6/15)) =

((1/11) × (5/2)) – ((1/1) × (1/1) + (1/13) × (2/1))

(1×5)/(11×2) – 1/1 + (1×2)/(13×1)

5/22 – 1/1 + 2/13

By taking LCM for 22, 1 and 13 which is 286

((5×13) – (1×286) + (2×22))/286

(65 – 286 + 44)/286

-177/286


EXERCISE 1.6 PAGE NO: 1.31

1. Verify the property: x × y = y × x by taking:

(i) x = -1/3, y = 2/7

Solution:

By using the property

x × y = y × x

-1/3 × 2/7 = 2/7 × -1/3

(-1×2)/(3×7) = (2×-1)/(7×3)

-2/21 = -2/21

Hence, the property is satisfied.

(ii) x = -3/5, y = -11/13

Solution:

By using the property

x × y = y × x

-3/5 × -11/13 = -11/13 × -3/5

(-3×-11)/(5×13) = (-11×-3)/(13×5)

33/65 = 33/65

Hence, the property is satisfied.

(iii) x = 2, y = 7/-8

Solution:

By using the property

x × y = y × x

2 × 7/-8 = 7/-8 × 2

(2×7)/-8 = (7×2)/-8

14/-8 = 14/-8

-14/8 = -14/8

Hence, the property is satisfied.

(iv) x = 0, y = -15/8

Solution:

By using the property

x × y = y × x

0 × -15/8 = -15/8 × 0

0 = 0

Hence, the property is satisfied.

2. Verify the property: x × (y × z) = (x × y) × z by taking:

(i) x = -7/3, y = 12/5, z = 4/9

Solution:

By using the property

x × (y × z) = (x × y) × z

-7/3 × (12/5 × 4/9) = (-7/3 × 12/5) × 4/9

(-7×12×4)/(3×5×9) = (-7×12×4)/(3×5×9)

-336/135 = -336/135

Hence, the property is satisfied.

(ii) x = 0, y = -3/5, z = -9/4

Solution:

By using the property

x × (y × z) = (x × y) × z

0 × (-3/5 × -9/4) = (0 × -3/5) × -9/4

0 = 0

Hence, the property is satisfied.

(iii) x = 1/2, y = 5/-4, z = -7/5

Solution:

By using the property

x × (y × z) = (x × y) × z

1/2 × (5/-4 × -7/5) = (1/2 × 5/-4) × -7/5

(1×5×-7)/(2×-4×5) = (1×5×-7)/(2×-4×5)

-35/-40 = -35/-40

35/40 = 35/40

Hence, the property is satisfied.

(iv) x = 5/7, y = -12/13, z = -7/18

Solution:

By using the property

x × (y × z) = (x × y) × z

5/7 × (-12/13 × -7/18) = (5/7 × -12/13) × -7/18

(5×-12×-7)/(7×13×18) = (5×-12×-7)/(7×13×18)

420/1638 = 420/1638

Hence, the property is satisfied.

3. Verify the property: x × (y + z) = x × y + x × z by taking:

(i) x = -3/7, y = 12/13, z = -5/6

Solution:

By using the property

x × (y + z) = x × y + x × z

-3/7 × (12/13 + -5/6) = -3/7 × 12/13 + -3/7 × -5/6

-3/7 × ((12×6) + (-5×13))/78 = (-3×12)/(7×13) + (-3×-5)/(7×6)

-3/7 × (72-65)/78 = -36/91 + 15/42

-3/7 × 7/78 = (-36×6 + 15×13)/546

-1/26 = (196-216)/546

= -21/546

= -1/26

Hence, the property is verified.

(ii) x = -12/5, y = -15/4, z = 8/3

Solution:

By using the property

x × (y + z) = x × y + x × z

-12/5 × (-15/4 + 8/3) = -12/5 × -15/4 + -12/5 × 8/3

-12/5 × ((-15×3) + (8×4))/12 = (-12×-15)/(5×4) + (-12×8)/(5×3)

-12/5 × (-45+32)/12 = 180/20 – 96/15

-12/5 × -13/12 = 9 – 32/5

13/5 = (9×5 – 32×1)/5

= (45-32)/5

= 13/5

Hence, the property is verified.

(iii) x = -8/3, y = 5/6, z = -13/12

Solution:

By using the property

x × (y + z) = x × y + x × z

-8/3 × (5/6 + -13/12) = -8/3 × 5/6 + -8/3 × -13/12

-8/3 × ((5×2) – (13×1))/12 = (-8×5)/(3×6) + (-8×-13)/(3×12)

-8/3 × (10-13)/12 = -40/18 + 104/36

-8/3 × -3/12 = (-40×2 + 104×1)/36

2/3 = (-80+104)/36

= 24/36

= 2/3

Hence, the property is verified.

(iv) x = -3/4, y = -5/2, z = 7/6

Solution:

By using the property

x × (y + z) = x × y + x × z

-3/4 × (-5/2 + 7/6) = -3/4 × -5/2 + -3/4 × 7/6

-3/4 × ((-5×3) + (7×1))/6 = (-3×-5)/(4×2) + (-3×7)/(4×6)

-3/4 × (-15+7)/6 = 15/8 – 21/24

-3/4 × -8/6 = (15×3 – 21×1)/24

-3/4 × -4/3 = (45-21)/24

1 = 24/24

= 1

Hence, the property is verified.

4. Use the distributivity of multiplication of rational numbers over their addition to simplify:

(i) 3/5 × ((35/24) + (10/1))

Solution:

3/5 × 35/24 + 3/5 × 10

1/1 × 7/8 + 6/1

By taking LCM for 8 and 1 which is 8

7/8 + 6 = (7×1 + 6×8)/8

= (7+48)/8

= 55/8

(ii) -5/4 × ((8/5) + (16/5))

Solution:

-5/4 × 8/5 + -5/4 × 16/5

-1/1 × 2/1 + -1/1 × 4/1

-2 + -4

-2 – 4

-6

(iii) 2/7 × ((7/16) – (21/4))

Solution:

2/7 × 7/16 – 2/7 × 21/4

1/1 × 1/8 – 1/1 × 3/2

1/8 – 3/2

By taking LCM for 8 and 2 which is 8

1/8 – 3/2 = (1×1 – 3×4)/8

= (1 – 12)/8

= -11/8

(iv) 3/4 × ((8/9) – 40)

Solution:

3/4 × 8/9 – 3/4 × 40

1/1 × 2/3 – 3/1 × 10

2/3 – 30/1

By taking LCM for 3 and 1 which is 3

2/3 – 30/1 = (2×1 – 30×3)/3

= (2 – 90)/3

= -88/3

5. Find the multiplicative inverse (reciprocal) of each of the following rational numbers:

(i) 9

(ii) -7

(iii) 12/5

(iv) -7/9

(v) -3/-5

(vi) 2/3 × 9/4

(vii) -5/8 × 16/15

(viii) -2 × -3/5

(ix) -1

(x) 0/3

(xi) 1

Solution:

(i) The reciprocal of 9 is 1/9

(ii) The reciprocal of -7 is -1/7

(iii) The reciprocal of 12/5 is 5/12

(iv) The reciprocal of -7/9 is 9/-7

(v) The reciprocal of -3/-5 is 5/3

(vi) The reciprocal of 2/3 × 9/4 is

Firstly solve for 2/3 × 9/4 = 1/1 × 3/2 = 3/2

∴ The reciprocal of 3/2 is 2/3

(vii) The reciprocal of -5/8 × 16/15

Firstly solve for -5/8 × 16/15 = -1/1 × 2/3 = -2/3

∴ The reciprocal of -2/3 is 3/-2

(viii) The reciprocal of -2 × -3/5

Firstly solve for -2 × -3/5 = 6/5

∴ The reciprocal of 6/5 is 5/6

(ix) The reciprocal of -1 is -1

(x) The reciprocal of 0/3 does not exist

(xi) The reciprocal of 1 is 1

6. Name the property of multiplication of rational numbers illustrated by the following statements:

(i) -5/16 × 8/15 = 8/15 × -5/16

(ii) -17/5 ×9 = 9 × -17/5

(iii) 7/4 × (-8/3 + -13/12) = 7/4 × -8/3 + 7/4 × -13/12

(iv) -5/9 × (4/15 × -9/8) = (-5/9 × 4/15) × -9/8

(v) 13/-17 × 1 = 13/-17 = 1 × 13/-17

(vi) -11/16 × 16/-11 = 1

(vii) 2/13 × 0 = 0 = 0 × 2/13

(viii) -3/2 × 5/4 + -3/2 × -7/6 = -3/2 × (5/4 + -7/6)

Solution:

(i) -5/16 × 8/15 = 8/15 × -5/16

According to commutative law, a/b × c/d = c/d × a/b

The above rational number satisfies commutative property.

(ii) -17/5 ×9 = 9 × -17/5

According to commutative law, a/b × c/d = c/d × a/b

The above rational number satisfies commutative property.

(iii) 7/4 × (-8/3 + -13/12) = 7/4 × -8/3 + 7/4 × -13/12

According to given rational number, a/b × (c/d + e/f) = (a/b × c/d) + (a/b × e/f)

Distributivity of multiplication over addition satisfies.

(iv) -5/9 × (4/15 × -9/8) = (-5/9 × 4/15) × -9/8

According to associative law, a/b × (c/d × e/f ) = (a/b × c/d) × e/f

The above rational number satisfies associativity of multiplication.

(v) 13/-17 × 1 = 13/-17 = 1 × 13/-17

Existence of identity for multiplication satisfies for the given rational number.

(vi) -11/16 × 16/-11 = 1

Existence of multiplication inverse satisfies for the given rational number.

(vii) 2/13 × 0 = 0 = 0 × 2/13

By using a/b × 0 = 0 × a/b

Multiplication of zero satisfies for the given rational number.

(viii) -3/2 × 5/4 + -3/2 × -7/6 = -3/2 × (5/4 + -7/6)

According to distributive law, (a/b × c/d) + (a/b × e/f ) = a/b × (c/d + e/f)

The above rational number satisfies distributive law.

7. Fill in the blanks:

(i) The product of two positive rational numbers is always…

(ii) The product of a positive rational number and a negative rational number is always….

(iii) The product of two negative rational numbers is always…

(iv) The reciprocal of a positive rational numbers is…

(v) The reciprocal of a negative rational numbers is…

(vi) Zero has …. Reciprocal.

(vii) The product of a rational number and its reciprocal is…

(viii) The numbers … and … are their own reciprocals.

(ix) If a is reciprocal of b, then the reciprocal of b is.

(x) The number 0 is … the reciprocal of any number.

(xi) reciprocal of 1/a, a ≠ 0 is …

(xii) (17×12)-1 = 17-1 × …

Solution:

(i) The product of two positive rational numbers is always positive.

(ii) The product of a positive rational number and a negative rational number is always negative.

(iii) The product of two negative rational numbers is always positive.

(iv) The reciprocal of a positive rational numbers is positive.

(v) The reciprocal of a negative rational numbers is negative.

(vi) Zero has no Reciprocal.

(vii) The product of a rational number and its reciprocal is 1.

(viii) The numbers 1 and -1 are their own reciprocals.

(ix) If a is reciprocal of b, then the reciprocal of b is a.

(x) The number 0 is not the reciprocal of any number.

(xi) reciprocal of 1/a, a ≠ 0 is a.

(xii) (17×12)-1 = 17-1 × 12-1

8. Fill in the blanks:

(i) -4 × 7/9 = 79 × …

Solution:

-4 × 7/9 = 79 × -4

By using commutative property.

(ii) 5/11 × -3/8 = -3/8 × …

Solution:

5/11 × -3/8 = -3/8 × 5/11

By using commutative property.

(iii) 1/2 × (3/4 + -5/12) = 1/2 × … + … × -5/12

Solution:

1/2 × (3/4 + -5/12) = 1/2 × 3/4 + 1/2 × -5/12

By using distributive property.

(iv) -4/5 × (5/7 + -8/9) = (-4/5 × …) + -4/5 × -8/9

Solution:

-4/5 × (5/7 + -8/9) = (-4/5 × 5/7) + -4/5 × -8/9

By using distributive property.


EXERCISE 1.7 PAGE NO: 1.35

1. Divide:

(i) 1 by 1/2

Solution:

1/1/2 = 1 × 2/1 = 2

(ii) 5 by -5/7

Solution:

5/-5/7 = 5 × 7/-5 = -7

(iii) -3/4 by 9/-16

Solution:

(-3/4) / (9/-16)

(-3/4) × -16/9 = 4/3

(iv) -7/8 by -21/16

Solution:

(-7/8) / (-21/16)

(-7/8) × 16/-21 = 2/3

(v) 7/-4 by 63/64

Solution:

(7/-4) / (63/64)

(7/-4) × 64/63 = -16/9

(vi) 0 by -7/5

Solution:

0 / (7/5) = 0

(vii) -3/4 by -6

Solution:

(-3/4) / -6

(-3/4) × 1/-6 = 1/8

(viii) 2/3 by -7/12

Solution:

(2/3) / (-7/12)

(2/3) × 12/-7 = -8/7

(ix) -4 by -3/5

Solution:

-4 / (-3/5)

-4 × 5/-3 = 20/3

(x) -3/13 by -4/65

Solution:

(-3/13) / (-4/65)

(-3/13) × (65/-4) = 15/4

2. Find the value and express as a rational number in standard form:

(i) 2/5 ÷ 26/15

Solution:

(2/5) / (26/15)

(2/5) × (15/26)

(2/1) × (3/26) = (2×3)/ (1×26) = 6/26 = 3/13

(ii) 10/3 ÷ -35/12

Solution:

(10/3) / (-35/12)

(10/3) × (12/-35)

(10/1) × (4/-35) = (10×4)/ (1×-35) = -40/35 = -8/7

(iii) -6 ÷ -8/17

Solution:

-6 / (-8/17)

-6 × (17/-8)

-3 × (17/-4) = (-3×17)/ (1×-4) = 51/4

(iv) -40/99 ÷ -20

Solution:

(-40/99) / -20

(-40/99) × (1/-20)

(-2/99) × (1/-1) = (-2×1)/ (99×-1) = 2/99

(v) -22/27 ÷ -110/18

Solution:

(-22/27) / (-110/18)

(-22/27) × (18/-110)

(-1/9) × (6/-5)

(-1/3) × (2/-5) = (-1×2) / (3×-5) = 2/15

(vi) -36/125 ÷ -3/75

Solution:

(-36/125) / (-3/75)

(-36/125) × (75/-3)

(-12/25) × (15/-1)

(-12/5) × (3/-1) = (-12×3) / (5×-1) = 36/5

3. The product of two rational numbers is 15. If one of the numbers is -10, find the other.

Solution:

We know that the product of two rational numbers = 15

One of the number = -10

∴ other number can be obtained by dividing the product by the given number.

Other number = 15/-10

= -3/2

4. The product of two rational numbers is -8/9. If one of the numbers is -4/15, find the other.

Solution:

We know that the product of two rational numbers = -8/9

One of the number = -4/15

∴ other number is obtained by dividing the product by the given number.

Other number = (-8/9)/(-4/15)

= (-8/9) × (15/-4)

= (-2/3) × (5/-1)

= (-2×5) /(3×-1)

= -10/-3

= 10/3

5. By what number should we multiply -1/6 so that the product may be -23/9?

Solution:

Let us consider a number = x

So, x × -1/6 = -23/9

x = (-23/9)/(-1/6)

x = (-23/9) × (6/-1)

= (-23/3) × (2×-1)

= (-23×-2)/(3×1)

= 46/3

6. By what number should we multiply -15/28 so that the product may be -5/7?

Solution:

Let us consider a number = x

So, x × -15/28 = -5/7

x = (-5/7)/(-15/28)

x = (-5/7) × (28/-15)

= (-1/1) × (4×-3)

= 4/3

7. By what number should we multiply -8/13 so that the product may be 24?

Solution:

Let us consider a number = x

So, x × -8/13 = 24

x = (24)/(-8/13)

x = (24) × (13/-8)

= (3) × (13×-1)

= -39

8. By what number should -3/4 be multiplied in order to produce 2/3?

Solution:

Let us consider a number = x

So, x × -3/4 = 2/3

x = (2/3)/(-3/4)

x = (2/3) × (4/-3)

= -8/9

9. Find (x+y) ÷ (x-y), if

(i) x= 2/3, y= 3/2

Solution:

(x+y) ÷ (x-y)

(2/3 + 3/2) / (2/3 – 3/2)

((2×2 + 3×3)/6) / ((2×2 – 3×3)/6)

((4+9)/6) / ((4-9)/6)

(13/6) / (-5/6)

(13/6) × (6/-5)

-13/5

(ii) x= 2/5, y= 1/2

Solution:

(x+y) ÷ (x-y)

(2/5 + 1/2) / (2/5 – 1/2)

((2×2 + 1×5)/10) / ((2×2 – 1×5)/10)

((4+5)/10) / ((4-5)/10)

(9/10) / (-1/10)

(9/10) × (10/-1)

-9

(iii) x= 5/4, y= -1/3

Solution:

(x+y) ÷ (x-y)

(5/4 – 1/3) / (5/4 + 1/3)

((5×3 – 1×4)/12) / ((5×3 + 1×4)/12)

((15-4)/12) / ((15+4)/12)

(11/12) / (19/12)

(11/12) × (12/19)

11/19

(iv) x= 2/7, y= 4/3

Solution:

(x+y) ÷ (x-y)

(2/7 + 4/3) / (2/7 – 4/3)

((2×3 + 4×7)/21) / ((2×3 – 4×7)/21)

((6+28)/21) / ((6-28)/21)

(34/21) / (-22/21)

(34/21) × (21/-22)

-34/22

-17/11

(v) x= 1/4, y= 3/2

Solution:

(x+y) ÷ (x-y)

(1/4 + 3/2) / (1/4 – 3/2)

((1×1 + 3×2)/4) / ((1×1 – 3×2)/4)

((1+6)/4) / ((1-6)/4)

(7/4) / (-5/4)

(7/4) × (4/-5) = -7/5

10. The cost of (7frac{2}{3}) meters of rope is Rs 12 ¾. Find the cost per meter.

Solution:

We know that 23/3 meters of rope = Rs 51/4

Let us consider a number = x

So, x × 23/3 = 51/4

x = (51/4)/(23/3)

x = (51/4) × (3/23)

= (51×3) / (4×23)

= 153/92

= (1frac{61}{92})

∴ cost per meter is Rs (1frac{61}{92})

11. The cost of (2frac{1}{3}) meters of cloth is Rs 75 ¼. Find the cost of cloth per meter.

Solution:

We know that 7/3 meters of cloth = Rs 301/4

Let us consider a number = x

So, x × 7/3 = 301/4

x = (301/4)/(7/3)

x = (301/4) × (3/7)

= (301×3) / (4×7)

= (43×3) / (4×1)

= 129/4

= 32.25

∴ cost of cloth per meter is Rs 32.25

12. By what number should -33/16 be divided to get -11/4?

Solution:

Let us consider a number = x

So, (-33/16)/x = -11/4

-33/16 = x × -11/4

x = (-33/16) / (-11/4)

= (-33/16) × (4/-11)

= (-33×4)/(16×-11)

= (-3×1)/(4×-1)

= ¾

13. Divide the sum of -13/5 and 12/7 by the product of -31/7 and -1/2.

Solution:

sum of -13/5 and 12/7

-13/5 + 12/7

((-13×7) + (12×5))/35

(-91+60)/35

-31/35

Product of -31/7 and -1/2

-31/7 × -1/2

(-31×-1)/(7×2)

31/14

∴ by dividing the sum and the product we get,

(-31/35) / (31/14)

(-31/35) × (14/31)

(-31×14)/(35×31)

-14/35

-2/5

14. Divide the sum of 65/12 and 12/7 by their difference.

Solution:

The sum is 65/12 + 12/7

The difference is 65/12 – 12/7

When we divide, (65/12 + 12/7) / (65/12 – 12/7)

((65×7 + 12×12)/84) / ((65×7 – 12×12)/84)

((455+144)/84) / ((455 – 144)/84)

(599/84) / (311/84)

599/84 × 84/311

599/311

15. If 24 trousers of equal size can be prepared in 54 meters of cloth, what length of cloth is required for each trouser?

Solution:

We know that total number trousers = 24

Total length of the cloth = 54

Length of the cloth required for each trouser = total length of the cloth/number of trousers

= 54/24

= 9/4

∴ 9/4 meters is required for each trouser.


EXERCISE 1.8 PAGE NO: 1.43

1. Find a rational number between -3 and 1.

Solution:

Let us consider two rational numbers x and y

We know that between two rational numbers x and y where x < y there is a rational number (x+y)/2

x < (x+y)/2 < y

(-3+1)/2 = -2/2 = -1

So, the rational number between -3 and 1 is -1

∴ -3 < -1 < 1

2. Find any five rational numbers less than 2.

Solution:

Five rational numbers less than 2 are 0, 1/5, 2/5, 3/5, 4/5

3. Find two rational numbers between -2/9 and 5/9

Solution:

The rational numbers between -2/9 and 5/9 is

(-2/9 + 5/9)/2

(1/3)/2

1/6

The rational numbers between -2/9 and 1/6 is

(-2/9 + 1/6)/2

((-2×2 + 1×3)/18)/2

(-4+3)/36

-1/36

∴ the rational numbers between -2/9 and 5/9 are -1/36, 1/6

4. Find two rational numbers between 1/5 and 1/2

Solution:

The rational numbers between 1/5 and 1/2 is

(1/5 + 1/2)/2

((1×2 + 1×5)/10)/2

(2+5)/20 = 7/20

The rational numbers between 1/5 and 7/20 is

(1/5 + 7/20)/2

((1×4 + 7×1)/20)/2

(4+7)/40

11/40

∴ the rational numbers between 1/5 and 1/2 are 7/20, 11/40

5. Find ten rational numbers between 1/4 and 1/2.

Solution:

Firstly convert the given rational numbers into equivalent rational numbers with same denominators.

The LCM for 4 and 2 is 4.

1/4 = 1/4

1/2 = (1×2)/4 = 2/4

1/4 = (1×20 / 4×20) = 20/80

1/2 = (2×20 / 4×20) = 40/80

So, we now know that 21, 22, 23,…39 are integers between numerators 20 and 40.

∴ the rational numbers between 1/4 and 1/2 are 21/80, 22/80, 23/80, …., 39/80

6. Find ten rational numbers between -2/5 and 1/2.

Solution:

Firstly convert the given rational numbers into equivalent rational numbers with same denominators.

The LCM for 5 and 2 is 10.

-2/5 = (-2×2)/10 = -4/10

1/2 = (1×5)/10 = 5/10

-2/5 = (-4×2 / 10×2) = -8/20

1/2 = (5×2 / 10×2) = 10/20

So, we now know that -7, -6, -5,…10 are integers between numerators -8 and 10.

∴ the rational numbers between -2/5 and 1/2 are -7/20, -6/20, -5/20, …., 9/20

7. Find ten rational numbers between 3/5 and 3/4.

Solution:

Firstly convert the given rational numbers into equivalent rational numbers with same denominators.

The LCM for 5 and 4 is 20.

3/5 = 3× 20 / 5×20 = 60/100

3/4 = 3×25 / 4×25 = 75/100

So, we now know that 61, 62, 63,..74 are integers between numerators 60 and 75.

∴ the rational numbers between 3/5 and 3/4 are 61/100, 62/100, 63/100, …., 74/100


RD Sharma Solutions for Class 8 Maths Chapter 1 – Rational Numbers

Here students will be acquainted with detailed concepts discussed in this Chapter as listed below.

  • Introduction to rational numbers.
  • Review about rational numbers.
  • Addition of rational numbers and their properties.
  • Subtraction of rational numbers and their properties.
  • Simplification of expressions involving addition and subtraction.
  • Properties of multiplication of rational numbers.
  • Division of rational numbers.
  • Representation of rational numbers on the number line.

All Chapter RD Sharma Solutions For Class 8 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

In this chapter, we provide RD Sharma Solutions for Class 8 Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 8 Maths pdf, free RD Sharma Solutions for Class 8 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Solutions for Class 8 Maths Chapter 1- Rational Numbers are available here. Our expert faculty team has prepared solutions in order to help you with your exam preparation to obtain good marks in Maths. If you wish to secure an excellent score, solving RD Sharma Class 8 Solutions is an utmost necessity. This chapter mainly deals with problems based on rational numbers, whole numbers, natural numbers, representation of rational numbers on the number line. In order to help you understand and solve the problems, we, at BYJU’S, have prepared the RD Sharma Class 8 Chapter 1 where solutions are solved in detail. Download pdf of Class 8 Chapter 1, in their respective links.

Chapter 1 Rational Numbers

RD Sharma Solutions for Class 8 Maths Chapter 1 – Rational Numbers

Chapter 1- Rational Numbers contains 8 exercises and the RD Sharma Solutions present on this page provide the solutions for the questions present in each exercise.

Access Answers to Maths RD Sharma Solutions for Class 8 Chapter 1 Rational Numbers

EXERCISE 1.1 PAGE NO: 1.5

1. Add the following rational numbers:

(i) -5/7 and 3/7

(ii) -15/4 and 7/4

(iii) -8/11 and -4/11

(iv) 6/13 and -9/13

Solution:

Since the denominators are of same positive numbers we can add them directly

(i) -5/7 + 3/7 = (-5+3)/7 = -2/7

(ii) -15/4 + 7/4 = (-15+7)/4 = -8/4

Further dividing by 4 we get,

-8/4 = -2

(iii) -8/11 + -4/11 = (-8 + (-4))/11 = (-8-4)/11 = -12/11

(iv) 6/13 + -9/13 = (6 + (-9))/13 = (6-9)/13 = -3/13

2. Add the following rational numbers:

(i) 3/4 and -5/8

Solution: The denominators are 4 and 8

By taking LCM for 4 and 8 is 8

We rewrite the given fraction in order to get the same denominator

3/4 = (3×2) / (4×2) = 6/8 and

-5/8 = (-5×1) / (8×1) = -5/8

Since the denominators are same we can add them directly

6/8 + -5/8 = (6 + (-5))/8 = (6-5)/8 = 1/8

(ii) 5/-9 and 7/3

Solution: Firstly we need to convert the denominators to positive numbers.

5/-9 = (5 × -1)/ (-9 × -1) = -5/9

The denominators are 9 and 3

By taking LCM for 9 and 3 is 9

We rewrite the given fraction in order to get the same denominator

-5/9 = (-5×1) / (9×1) = -5/9 and

7/3 = (7×3) / (3×3) = 21/9

Since the denominators are same we can add them directly

-5/9 + 21/9 = (-5+21)/9 = 16/9

(iii) -3 and 3/5

Solution: The denominators are 1 and 5

By taking LCM for 1 and 5 is 5

We rewrite the given fraction in order to get the same denominator

-3/1 = (-3×5) / (1×5) = -15/5 and

3/5 = (3×1) / (5×1) = 3/5

Now, the denominators are same we can add them directly

-15/5 + 3/5 = (-15+3)/5 = -12/5

(iv) -7/27 and 11/18

Solution: The denominators are 27 and 18

By taking LCM for 27 and 18 is 54

We rewrite the given fraction in order to get the same denominator

-7/27 = (-7×2) / (27×2) = -14/54 and

11/18 = (11×3) / (18×3) = 33/54

Now, the denominators are same we can add them directly

-14/54 + 33/54 = (-14+33)/54 = 19/54

(v) 31/-4 and -5/8

Solution: Firstly we need to convert the denominators to positive numbers.

31/-4 = (31 × -1)/ (-4 × -1) = -31/4

The denominators are 4 and 8

By taking LCM for 4 and 8 is 8

We rewrite the given fraction in order to get the same denominator

-31/4 = (-31×2) / (4×2) = -62/8 and

-5/8 = (-5×1) / (8×1) = -5/8

Since the denominators are same we can add them directly

-62/8 + (-5)/8 = (-62 + (-5))/8 = (-62-5)/8 = -67/8

(vi) 5/36 and -7/12

Solution: The denominators are 36 and 12

By taking LCM for 36 and 12 is 36

We rewrite the given fraction in order to get the same denominator

5/36 = (5×1) / (36×1) = 5/36 and

-7/12 = (-7×3) / (12×3) = -21/36

Now, the denominators are same we can add them directly

5/36 + -21/36 = (5 + (-21))/36 = 5-21/36 = -16/36 = -4/9

(vii) -5/16 and 7/24

Solution: The denominators are 16 and 24

By taking LCM for 16 and 24 is 48

We rewrite the given fraction in order to get the same denominator

-5/16 = (-5×3) / (16×3) = -15/48 and

7/24 = (7×2) / (24×2) = 14/48

Now, the denominators are same we can add them directly

-15/48 + 14/48 = (-15 + 14)/48 = -1/48

(viii) 7/-18 and 8/27

Solution: Firstly we need to convert the denominators to positive numbers.

7/-18 = (7 × -1)/ (-18 × -1) = -7/18

The denominators are 18 and 27

By taking LCM for 18 and 27 is 54

We rewrite the given fraction in order to get the same denominator

-7/18 = (-7×3) / (18×3) = -21/54 and

8/27 = (8×2) / (27×2) = 16/54

Since the denominators are same we can add them directly

-21/54 + 16/54 = (-21 + 16)/54 = -5/54

3.Simplify:

(i) 8/9 + -11/6

Solution: let us take the LCM for 9 and 6 which is 18

(8×2)/(9×2) + (-11×3)/(6×3)

16/18 + -33/18

Since the denominators are same we can add them directly

(16-33)/18 = -17/18

(ii) 3 + 5/-7

Solution: Firstly convert the denominator to positive number

5/-7 = (5×-1)/(-7×-1) = -5/7

3/1 + -5/7

Now let us take the LCM for 1 and 7 which is 7

(3×7)/(1×7) + (-5×1)/(7×1)

21/7 + -5/7

Since the denominators are same we can add them directly

(21-5)/7 = 16/7

(iii) 1/-12 + 2/-15

Solution: Firstly convert the denominator to positive number

1/-12 = (1×-1)/(-12×-1) = -1/12

2/-15 = (2×-1)/(-15×-1) = -2/15

-1/12 + -2/15

Now let us take the LCM for 12 and 15 which is 60

(-1×5)/(12×5) + (-2×4)/(15×4)

-5/60 + -8/60

Since the denominators are same we can add them directly

(-5-8)/60 = -13/60

(iv) -8/19 + -4/57

Solution: let us take the LCM for 19 and 57 which is 57

(-8×3)/(19×3) + (-4×1)/(57×1)

-24/57 + -4/57

Since the denominators are same we can add them directly

(-24-4)/57 = -28/57

(v) 7/9 + 3/-4

Solution: Firstly convert the denominator to positive number

3/-4 = (3×-1)/(-4×-1) = -3/4

7/9 + -3/4

Now let us take the LCM for 9 and 4 which is 36

(7×4)/(9×4) + (-3×9)/(4×9)

28/36 + -27/36

Since the denominators are same we can add them directly

(28-27)/36 = 1/36

(vi) 5/26 + 11/-39

Solution: Firstly convert the denominator to positive number

11/-39 = (11×-1)/(-39×-1) = -11/39

5/26 + -11/39

Now let us take the LCM for 26 and 39 which is 78

(5×3)/(26×3) + (-11×2)/(39×2)

15/78 + -22/78

Since the denominators are same we can add them directly

(15-22)/78 = -7/78

(vii) -16/9 + -5/12

Solution: let us take the LCM for 9 and 12 which is 108

(-16×12)/(9×12) + (-5×9)/(12×9)

-192/108 + -45/108

Since the denominators are same we can add them directly

(-192-45)/108 = -237/108

Further divide the fraction by 3 we get,

-237/108 = -79/36

(viii) -13/8 + 5/36

Solution: let us take the LCM for 8 and 36 which is 72

(-13×9)/(8×9) + (5×2)/(36×2)

-117/72 + 10/72

Since the denominators are same we can add them directly

(-117+10)/72 = -107/72

(ix) 0 + -3/5

Solution: We know that anything added to 0 results in the same.

0 + -3/5 = -3/5

(x) 1 + -4/5

Solution: let us take the LCM for 1 and 5 which is 5

(1×5)/(1×5) + (-4×1)/(5×1)

5/5 + -4/5

Since the denominators are same we can add them directly

(5-4)/5 = 1/5

4. Add and express the sum as a mixed fraction:

(i) -12/5 and 43/10

Solution: let us add the given fraction

-12/5 + 43/10

let us take the LCM for 5 and 10 which is 10

(-12×2)/(5×2) + (43×1)/(10×1)

-24/10 + 43/10

Since the denominators are same we can add them directly

(-24+43)/10 = 19/10

19/10 can be written as (1frac{9}{10}) in mixed fraction.

(ii) 24/7 and -11/4

Solution: let us add the given fraction

24/7 + -11/4

let us take the LCM for 7 and 4 which is 28

(24×4)/(7×4) + (-11×7)/(4×7)

96/28 + -77/28

Since the denominators are same we can add them directly

(96-77)/28 = 19/28

(iii) -31/6 and -27/8

Solution: let us add the given fraction

-31/6 + -27/8

let us take the LCM for 6 and 8 which is 24

(-31×4)/(6×4) + (-27×3)/(8×3)

-124/24 + -81/24

Since the denominators are same we can add them directly

(-124-81)/24 = -205/24

-205/24 can be written as (-8frac{13}{24}) in mixed fraction.

(iv) 101/6 and 7/8

Solution: let us add the given fraction

101/6 + 7/8

let us take the LCM for 6 and 8 which is 24

(101×4)/(6×4) + (7×3)/(8×3)

404/24 + 21/24

Since the denominators are same we can add them directly

(404+21)/24 = 425/24

425/24 can be written as (17frac{17}{24}) in mixed fraction.


EXERCISE 1.2 PAGE NO: 1.14

1. Verify commutativity of addition of rational numbers for each of the following pairs of rational numbers:

(i) -11/5 and 4/7

Solution: By using the commutativity law, the addition of rational numbers is commutative ∴ a/b + c/d = c/d + a/b

In order to verify the above property let us consider the given fraction

-11/5 and 4/7 as

-11/5 + 4/7 and 4/7 + -11/5

The denominators are 5 and 7

By taking LCM for 5 and 7 is 35

We rewrite the given fraction in order to get the same denominator

Now, -11/5 = (-11 × 7) / (5 ×7) = -77/35

4/7 = (4 ×5) / (7 ×5) = 20/35

Since the denominators are same we can add them directly

-77/35 + 20/35 = (-77+20)/35 = -57/35

4/7 + -11/5

The denominators are 7 and 5

By taking LCM for 7 and 5 is 35

We rewrite the given fraction in order to get the same denominator

Now, 4/7 = (4 × 5) / (7 ×5) = 20/35

-11/5 = (-11 ×7) / (5 ×7) = -77/35

Since the denominators are same we can add them directly

20/35 + -77/35 = (20 + (-77))/35 = (20-77)/35 = -57/35

∴ -11/5 + 4/7 = 4/7 + -11/5 is satisfied.

(ii) 4/9 and 7/-12

Solution: Firstly we need to convert the denominators to positive numbers.

7/-12 = (7 × -1)/ (-12 × -1) = -7/12

By using the commutativity law, the addition of rational numbers is commutative.

∴ a/b + c/d = c/d + a/b

In order to verify the above property let us consider the given fraction

4/9 and -7/12 as

4/9 + -7/12 and -7/12 + 4/9

The denominators are 9 and 12

By taking LCM for 9 and 12 is 36

We rewrite the given fraction in order to get the same denominator

Now, 4/9 = (4 × 4) / (9 ×4) = 16/36

-7/12 = (-7 ×3) / (12 ×3) = -21/36

Since the denominators are same we can add them directly

16/36 + (-21)/36 = (16 + (-21))/36 = (16-21)/36 = -5/36

-7/12 + 4/9

The denominators are 12 and 9

By taking LCM for 12 and 9 is 36

We rewrite the given fraction in order to get the same denominator

Now, -7/12 = (-7 ×3) / (12 ×3) = -21/36

4/9 = (4 × 4) / (9 ×4) = 16/36

Since the denominators are same we can add them directly

-21/36 + 16/36 = (-21 + 16)/36 = -5/36

∴ 4/9 + -7/12 = -7/12 + 4/9 is satisfied.

(iii) -3/5 and -2/-15

Solution:

By using the commutativity law, the addition of rational numbers is commutative.

∴ a/b + c/d = c/d + a/b

In order to verify the above property let us consider the given fraction

-3/5 and -2/-15 as

-3/5 + -2/-15 and -2/-15 + -3/5

-2/-15 = 2/15

The denominators are 5 and 15

By taking LCM for 5 and 15 is 15

We rewrite the given fraction in order to get the same denominator

Now, -3/5 = (-3 × 3) / (5×3) = -9/15

2/15 = (2 ×1) / (15 ×1) = 2/15

Since the denominators are same we can add them directly

-9/15 + 2/15 = (-9 + 2)/15 = -7/15

-2/-15 + -3/5

-2/-15 = 2/15

The denominators are 15 and 5

By taking LCM for 15 and 5 is 15

We rewrite the given fraction in order to get the same denominator

Now, 2/15 = (2 ×1) / (15 ×1) = 2/15

-3/5 = (-3 × 3) / (5×3) = -9/15

Since the denominators are same we can add them directly

2/15 + -9/15 = (2 + (-9))/15 = (2-9)/15 = -7/15

∴ -3/5 + -2/-15 = -2/-15 + -3/5 is satisfied.

(iv) 2/-7 and 12/-35

Solution: Firstly we need to convert the denominators to positive numbers.

2/-7 = (2 × -1)/ (-7 × -1) = -2/7

12/-35 = (12 × -1)/ (-35 × -1) = -12/35

By using the commutativity law, the addition of rational numbers is commutative.

∴ a/b + c/d = c/d + a/b

In order to verify the above property let us consider the given fraction

-2/7 and -12/35 as

-2/7 + -12/35 and -12/35 + -2/7

The denominators are 7 and 35

By taking LCM for 7 and 35 is 35

We rewrite the given fraction in order to get the same denominator

Now, -2/7 = (-2 × 5) / (7 ×5) = -10/35

-12/35 = (-12 ×1) / (35 ×1) = -12/35

Since the denominators are same we can add them directly

-10/35 + (-12)/35 = (-10 + (-12))/35 = (-10-12)/35 = -22/35

-12/35 + -2/7

The denominators are 35 and 7

By taking LCM for 35 and 7 is 35

We rewrite the given fraction in order to get the same denominator

Now, -12/35 = (-12 ×1) / (35 ×1) = -12/35

-2/7 = (-2 × 5) / (7 ×5) = -10/35

Since the denominators are same we can add them directly

-12/35 + -10/35 = (-12 + (-10))/35 = (-12-10)/35 = -22/35

∴ -2/7 + -12/35 = -12/35 + -2/7 is satisfied.

(v) 4 and -3/5

Solution: By using the commutativity law, the addition of rational numbers is commutative.

∴ a/b + c/d = c/d + a/b

In order to verify the above property let us consider the given fraction

4/1 and -3/5 as

4/1 + -3/5 and -3/5 + 4/1

The denominators are 1 and 5

By taking LCM for 1 and 5 is 5

We rewrite the given fraction in order to get the same denominator

Now, 4/1 = (4 × 5) / (1×5) = 20/5

-3/5 = (-3 ×1) / (5 ×1) = -3/5

Since the denominators are same we can add them directly

20/5 + -3/5 = (20 + (-3))/5 = (20-3)/5 = 17/5

-3/5 + 4/1

The denominators are 5 and 1

By taking LCM for 5 and 1 is 5

We rewrite the given fraction in order to get the same denominator

Now, -3/5 = (-3 ×1) / (5 ×1) = -3/5

4/1 = (4 × 5) / (1×5) = 20/5

Since the denominators are same we can add them directly

-3/5 + 20/5 = (-3 + 20)/5 = 17/5

∴ 4/1 + -3/5 = -3/5 + 4/1 is satisfied.

(vi) -4 and 4/-7

Solution: Firstly we need to convert the denominators to positive numbers.

4/-7 = (4 × -1)/ (-7 × -1) = -4/7

By using the commutativity law, the addition of rational numbers is commutative.

∴ a/b + c/d = c/d + a/b

In order to verify the above property let us consider the given fraction

-4/1 and -4/7 as

-4/1 + -4/7 and -4/7 + -4/1

The denominators are 1 and 7

By taking LCM for 1 and 7 is 7

We rewrite the given fraction in order to get the same denominator

Now, -4/1 = (-4 × 7) / (1×7) = -28/7

-4/7 = (-4 ×1) / (7 ×1) = -4/7

Since the denominators are same we can add them directly

-28/7 + -4/7 = (-28 + (-4))/7 = (-28-4)/7 = -32/7

-4/7 + -4/1

The denominators are 7 and 1

By taking LCM for 7 and 1 is 7

We rewrite the given fraction in order to get the same denominator

Now, -4/7 = (-4 ×1) / (7 ×1) = -4/7

-4/1 = (-4 × 7) / (1×7) = -28/7

Since the denominators are same we can add them directly

-4/7 + -28/7 = (-4 + (-28))/7 = (-4-28)/7 = -32/7

∴ -4/1 + -4/7 = -4/7 + -4/1 is satisfied.

2. Verify associativity of addition of rational numbers i.e., (x + y) + z = x + (y + z), when:

(i) x = ½, y = 2/3, z = -1/5

Solution: As the property states (x + y) + z = x + (y + z)

Use the values as such,

(1/2 + 2/3) + (-1/5) = 1/2 + (2/3 + (-1/5))

Let us consider LHS (1/2 + 2/3) + (-1/5)

Taking LCM for 2 and 3 is 6

(1× 3)/(2×3) + (2×2)/(3×2)

3/6 + 4/6

Since the denominators are same we can add them directly,

3/6 + 4/6 = 7/6

7/6 + (-1/5)

Taking LCM for 6 and 5 is 30

(7×5)/(6×5) + (-1×6)/(5×6)

35/30 + (-6)/30

Since the denominators are same we can add them directly,

(35+(-6))/30 = (35-6)/30 = 29/30

Let us consider RHS 1/2 + (2/3 + (-1/5))

Taking LCM for 3 and 5 is 15

(2/3 + (-1/5)) = (2×5)/(3×5) + (-1×3)/(5×3)

= 10/15 + (-3)/15

Since the denominators are same we can add them directly,

10/15 + (-3)/15 = (10-3)/15 = 7/15

1/2 + 7/15

Taking LCM for 2 and 15 is 30

1/2 + 7/15 = (1×15)/(2×15) + (7×2)/(15×2)

= 15/30 + 14/30

Since the denominators are same we can add them directly,

= (15 + 14)/30 = 29/30

∴ LHS = RHS associativity of addition of rational numbers is verified.

(ii) x = -2/5, y = 4/3, z = -7/10

Solution: As the property states (x + y) + z = x + (y + z)

Use the values as such,

(-2/5 + 4/3) + (-7/10) = -2/5 + (4/3 + (-7/10))

Let us consider LHS (-2/5 + 4/3) + (-7/10)

Taking LCM for 5 and 3 is 15

(-2× 3)/(5×3) + (4×5)/(3×5)

-6/15 + 20/15

Since the denominators are same we can add them directly,

-6/15 + 20/15= (-6+20)/15 = 14/15

14/15 + (-7/10)

Taking LCM for 15 and 10 is 30

(14×2)/(15×2) + (-7×3)/(10×3)

28/30 + (-21)/30

Since the denominators are same we can add them directly,

(28+(-21))/30 = (28-21)/30 = 7/30

Let us consider RHS -2/5 + (4/3 + (-7/10))

Taking LCM for 3 and 10 is 30

(4/3 + (-7/10)) = (4×10)/(3×10) + (-7×3)/(10×3)

= 40/30 + (-21)/30

Since the denominators are same we can add them directly,

40/30 + (-21)/30 = (40-21)/30 = 19/30

-2/5 + 19/30

Taking LCM for 5 and 30 is 30

-2/5 + 19/30 = (-2×6)/(5×6) + (19×1)/(30×1)

= -12/30 + 19/30

Since the denominators are same we can add them directly,

= (-12 + 19)/30 = 7/30

∴ LHS = RHS associativity of addition of rational numbers is verified.

(iii) x = -7/11, y = 2/-5, z = -3/22

Solution: Firstly convert the denominators to positive numbers

2/-5 = (2×-1)/ (-5×-1) = -2/5

As the property states (x + y) + z = x + (y + z)

Use the values as such,

(-7/11 + -2/5) + (-3/22) = -7/11 + (-2/5 + (-3/22))

Let us consider LHS (-7/11 + -2/5) + (-3/22)

Taking LCM for 11 and 5 is 55

(-7×5)/(11×5) + (-2×11)/(5×11)

-35/55 + -22/55

Since the denominators are same we can add them directly,

-35/55 + -22/55 = (-35-22)/55 = -57/55

-57/55 + (-3/22)

Taking LCM for 55 and 22 is 110

(-57×2)/(55×2) + (-3×5)/(22×5)

-114/110 + (-15)/110

Since the denominators are same we can add them directly,

(-114+(-15))/110 = (-114-15)/110 = -129/110

Let us consider RHS -7/11 + (-2/5 + (-3/22))

Taking LCM for 5 and 22 is 110

(-2/5 + (-3/22))= (-2×22)/(5×22) + (-3×5)/(22×5)

= -44/110 + (-15)/110

Since the denominators are same we can add them directly,

-44/110 + (-15)/110 = (-44-15)/110 = -59/110

-7/11 + -59/110

Taking LCM for 11 and 110 is 110

-7/11 + -59/110 = (-7×10)/(11×10) + (-59×1)/(110×1)

= -70/110 + -59/110

Since the denominators are same we can add them directly,

= (-70 -59)/110 = -129/110

∴ LHS = RHS associativity of addition of rational numbers is verified.

(iv) x = -2, y = 3/5, z = -4/3

Solution: As the property states (x + y) + z = x + (y + z)

Use the values as such,

(-2/1 + 3/5) + (-4/3) = -2/1 + (3/5 + (-4/3))

Let us consider LHS (-2/1 + 3/5) + (-4/3)

Taking LCM for 1 and 5 is 5

(-2×5)/(1×5) + (3×1)/(5×1)

-10/5 + 3/5

Since the denominators are same we can add them directly,

-10/5 + 3/5= (-10+3)/5 = -7/5

-7/5 + (-4/3)

Taking LCM for 5 and 3 is 15

(-7×3)/(5×3) + (-4×5)/(3×5)

-21/15 + (-20)/15

Since the denominators are same we can add them directly,

(-21+(-20))/15 = (-21-20)/15 = -41/15

Let us consider RHS -2/1 + (3/5 + (-4/3))

Taking LCM for 5 and 3 is 15

(3/5 + (-4/3)) = (3×3)/(5×3) + (-4×5)/(3×5)

= 9/15 + (-20)/15

Since the denominators are same we can add them directly,

9/15 + (-20)/15 = (9-20)/15 = -11/15

-2/1 + -11/15

Taking LCM for 1 and 15 is 15

-2/1 + -11/15 = (-2×15)/(1×15) + (-11×1)/(15×1)

= -30/15 + -11/15

Since the denominators are same we can add them directly,

= (-30 -11)/15 = -41/15

∴ LHS = RHS associativity of addition of rational numbers is verified.

3. Write the additive of each of the following rational numbers:

(i) -2/17

(ii) 3/-11

(iii) -17/5

(iv) -11/-25

Solution:

(i) The additive inverse of -2/17 is 2/17

(ii) The additive inverse of 3/-11 is 3/11

(iii) The additive inverse of -17/5 is 17/5

(iv) The additive inverse of -11/-25 is -11/25

4. Write the negative(additive) inverse of each of the following:

(i) -2/5

(ii) 7/-9

(iii) -16/13

(iv) -5/1

(v) 0

(vi) 1

(vii) – 1

Solution:

(i) The negative (additive) inverse of -2/5 is 2/5

(ii) The negative (additive) inverse of 7/-9 is 7/9

(iii) The negative (additive) inverse of -16/13 is 16/13

(iv) The negative (additive) inverse of -5/1 is 5

(v) The negative (additive) inverse of 0 is 0

(vi) The negative (additive) inverse of 1 is -1

(vii) The negative (additive) inverse of -1 is 1

5. Using commutativity and associativity of addition of rational numbers, express each of the following as a rational number:

(i) 2/5 + 7/3 + -4/5 + -1/3

Solution: Firstly group the rational numbers with same denominators

2/5 + -4/5 + 7/3 + -1/3

Now the denominators which are same can be added directly.

(2+(-4))/5 + (7+(-1))/3

(2-4)/5 + (7-1)/3

-2/5 + 6/3

By taking LCM for 5 and 3 we get, 15

(-2×3)/(5×3) + (6×5)/(3×5)

-6/15 + 30/15

Since the denominators are same can be added directly

(-6+30)/15 = 24/15

Further can be divided by 3 we get,

24/15 = 8/5

(ii) 3/7 + -4/9 + -11/7 + 7/9

Solution: Firstly group the rational numbers with same denominators

3/7 + -11/7 + -4/9 + 7/9

Now the denominators which are same can be added directly.

(3+ (-11))/7 + (-4+ 7)/9

(3-11)/7 + (-4+7)/9

-8/7 + 3/9

-8/7 + 1/3

By taking LCM for 7 and 3 we get, 21

(-8×3)/ (7×3) + (1×7)/ (3×7)

-24/21 + 7/21

Since the denominators are same can be added directly

(-24+7)/21 = -17/21

(iii) 2/5 + 8/3 + -11/15 + 4/5 + -2/3

Solution: Firstly group the rational numbers with same denominators

2/5 + 4/5 + 8/3 + -2/3 + -11/15

Now the denominators which are same can be added directly.

(2 + 4)/5 + (8 + (-2))/3 + -11/15

6/5 + (8-2)/3 + -11/15

6/5 + 6/3 + -11/15

6/5 + 2/1 + -11/15

By taking LCM for 5, 1 and 15 we get, 15

(6×3)/ (5×3) + (2×15)/ (1×15) + (-11×1)/ (15×1)

18/15 + 30/15 + -11/15

Since the denominators are same can be added directly

(18+30+ (-11))/15 = (18+30-11)/15 = 37/15

(iv) 4/7 + 0 + -8/9 + -13/7 + 17/21

Solution: Firstly group the rational numbers with same denominators

4/7 + -13/7 + -8/9 + 17/21

Now the denominators which are same can be added directly.

(4 + (-13))/7 + -8/9 + 17/21

(4-13)/7 + -8/9 + 17/21

-9/7 + -8/9 + 17/21

By taking LCM for 7, 9 and 21 we get, 63

(-9×9)/ (7×9) + (-8×7)/ (9×7) + (17×3)/ (21×3)

-81/63 + -56/63 + 51/63

Since the denominators are same can be added directly

(-81+(-56)+ 51)/63 = (-81-56+51)/63 = -86/63

6. Re-arrange suitably and find the sum in each of the following:

(i) 11/12 + -17/3 + 11/2 + -25/2

Solution: Firstly group the rational numbers with same denominators

11/12 + -17/3 + (11-25)/2

11/12 + -17/3 + -14/2

By taking LCM for 12, 3 and 2 we get, 12

(11×1)/(12×1) + (-17×4)/(3×4) + (-14×6)/(2×6)

11/12 + -68/12 + -84/12

Since the denominators are same can be added directly

(11-68-84)/12 = -141/12

(ii)-6/7 + -5/6 + -4/9 + -15/7

Solution: Firstly group the rational numbers with same denominators

-6/7 + -15/7 + -5/6 + -4/9

(-6 -15)/7 + -5/6 + -4/9

-21/7 + -5/6 + -4/9

-3/1 + -5/6 + -4/9

By taking LCM for 1, 6 and 9 we get, 18

(-3×18)/(1×18) + (-5×3)/(6×3) + (-4×2)/(9×2)

-54/18 + -15/18 + -8/18

Since the denominators are same can be added directly

(-54-15-8)/18 = -77/18

(iii) 3/5 + 7/3 + 9/ 5+ -13/15 + -7/3

Solution: Firstly group the rational numbers with same denominators

3/5 + 9/5 + 7/3 + -7/3 + -13/15

(3+9)/5 + -13/15

12/5 + -13/15

By taking LCM for 5 and 15 we get, 15

(12×3)/(5×3) + (-13×1)/(15×1)

36/15 + -13/15

Since the denominators are same can be added directly

(36-13)/15 = 23/15

(iv) 4/13 + -5/8 + -8/13 + 9/13

Solution: Firstly group the rational numbers with same denominators

4/13 + -8/13 + 9/13 + -5/8

(4-8+9)/13 + -5/8

5/13 + -5/8

By taking LCM for 13 and 8 we get, 104

(5×8)/(13×8) + (-5×13)/(8×13)

40/104 + -65/104

Since the denominators are same can be added directly

(40-65)/104 = -25/104

(v) 2/3 + -4/5 + 1/3 + 2/5

Solution: Firstly group the rational numbers with same denominators

2/3 + 1/3 + -4/5 + 2/5

(2+1)/3 + (-4+2)/5

3/3 + -2/5

1/1 + -2/5

By taking LCM for 1 and 5 we get, 5

(1×5)/(1×5) + (-2×1)/(5×1)

5/5 + -2/5

Since the denominators are same can be added directly

(5-2)/5 = 3/5

(vi) 1/8 + 5/12 + 2/7 + 7/12 + 9/7 + -5/16

Solution: Firstly group the rational numbers with same denominators

1/8 + 5/12 + 7/12 + 2/7 + 9/7 + -5/16

1/8 + (5+7)/12 + (2+9)/7 + -5/16

1/8 + 12/12 + 11/7 + -5/16

1/8 + 1/1 + 11/7 + -5/16

By taking LCM for 8, 1, 7 and 16 we get, 112

(1×14)/(8×14) + (1×112)/(1×112) + (11×16)/(7×16) + (-5×7)/(16×7)

14/112 + 112/112 + 176/112 + -35/112

Since the denominators are same can be added directly

(14+112+176-35)/112 = 267/112


EXERCISE 1.3 PAGE NO: 1.18

1. Subtract the first rational number from the second in each of the following:

(i) 3/8, 5/8

(ii) -7/9, 4/9

(iii) -2/11, -9/11

(iv) 11/13, -4/13

(v) ¼, -3/8

(vi) -2/3, 5/6

(vii) -6/7, -13/14

(viii) -8/33, -7/22

Solution:

(i) let us subtract

5/8 – 3/8

Since the denominators are same we can subtract directly

(5-3)/8 = 2/8

Further we can divide by 2 we get,

2/8 = 1/4

(ii) let us subtract

4/9 – -7/9

Since the denominators are same we can subtract directly

(4+7)/9 = 11/9

(iii) let us subtract

-9/11 – -2/11

Since the denominators are same we can subtract directly

(-9+2)/11 = -7/11

(iv) let us subtract

-4/13 – 11/13

Since the denominators are same we can subtract directly

(-4-11)/13 = -15/13

(v) let us subtract

-3/8 – 1/4

By taking LCM for 8 and 4 which is 8

-3/8 – 1/4 = (-3×1)/(8×1) – (1×2)/(4×2) = -3/8 – 2/8

Since the denominators are same we can subtract directly

(-3-2)/8 = -5/8

(vi) let us subtract

5/6 – -2/3

By taking LCM for 6 and 3 which is 6

5/6 – -2/3 = (5×1)/(6×1) – (-2×2)/(3×2) = 5/6 – -4/6

Since the denominators are same we can subtract directly

(5+4)/6 = 9/6

Further we can divide by 3 we get,

9/6 = 3/2

(vii) let us subtract

-13/14 – -6/7

By taking LCM for 14 and 7 which is 14

-13/14 – -6/7 = (-13×1)/(14×1) – (-6×2)/(7×2) = -13/14 – -12/14

Since the denominators are same we can subtract directly

(-13+12)/14 = -1/14

(viii) let us subtract

-7/22 – -8/33

By taking LCM for 22 and 33 which is 66

-7/22 – -8/33 = (-7×3)/(22×3) – (-8×2)/(33×2) = -21/66 – -16/66

Since the denominators are same we can subtract directly

(-21+16)/66 = -5/66

2. Evaluate each of the following:

(i) 2/3 – 3/5

Solution: By taking LCM for 3 and 5 which is 15

2/3 – 3/5 = (2×5 – 3×3)/15

= 1/15

(ii) -4/7 – 2/-3

Solution: convert the denominator to positive number by multiplying by -1

2/-3 = -2/3

-4/7 – -2/3

By taking LCM for 7 and 3 which is 21

-4/7 – -2/3 = (-4×3 – -2×7)/21

= (-12+14)/21

= 2/21

(iii) 4/7 – -5/-7

Solution: convert the denominator to positive number by multiplying by -1

-5/-7 = 5/7

4/7 – 5/7

Since the denominators are same we can subtract directly

(4-5)/7 = -1/7

(iv) -2 – 5/9

Solution: By taking LCM for 1 and 9 which is 9

-2/1 – 5/9 = (-2×9 – 5×1)/9

= (-18 – 5)/9

= -23/9

(v) -3/-8 – -2/7

Solution: convert the denominator to positive number by multiplying by -1

-3/-8 = 3/8

3/8 – -2/7

By taking LCM for 8 and 7 which is 56

3/8 – -2/7 = (3×7 – -2×8)/56

= (21 + 16)/56

= 37/56

(vi) -4/13 – -5/26

Solution: By taking LCM for 13 and 26 which is 26

-4/13 – -5/26 = (-4×2 – -5×1)/26

= (-8 + 5)/26

= -3/26

(vii) -5/14 – -2/7

Solution: By taking LCM for 14 and 7 which is 14

-5/14 – -2/7 = (-5×1 – -2×2)/14

= (-5 + 4)/14

= -1/14

(viii) 13/15 – 12/25

Solution: By taking LCM for 15 and 25 which is 75

13/15 – 12/25 = (13×5 – 12×3)/75

= (65 – 36)/75

= 29/75

(ix) -6/13 – -7/13

Solution: Since the denominators are same we can subtract directly

-6/13 – -7/13 = (-6+7)/13

= 1/13

(x) 7/24 – 19/36

Solution: By taking LCM for 24 and 36 which is 72

7/24 – 19/36 = (7×3 – 19×2)/72

= (21 – 38)/72

= -17/72

(xi) 5/63 – -8/21

Solution: By taking LCM for 63 and 21 which is 63

5/63 – -8/21 = (5×1 – -8×3)/63

= (5 + 24)/63

= 29/63

3. The sum of the two numbers is 5/9. If one of the numbers is 1/3, find the other.

Solution: Let us note down the given details

Sum of two numbers = 5/9

One of the number = 1/3

By using the formula,

Other number = sum of number – given number

= 5/9 – 1/3

By taking LCM for 9 and 3 which is 9

5/9 – 1/3 = (5×1 – 1×3)/9

= (5 – 3)/9

= 2/9

∴ the other number is 2/9

4. The sum of the two numbers is -1/3. If one of the numbers is -12/3, find the other.

Solution: Let us note down the given details

Sum of two numbers = -1/3

One of the number = -12/3

By using the formula,

Other number = sum of number – given number

= -1/3 – -12/3

Since the denominators are same we can subtract directly

= (-1+12)/3 = 11/3

∴ the other number is 11/3

5. The sum of the two numbers is -4/3. If one of the numbers is -5, find the other.

Solution: Let us note down the given details

Sum of two numbers = -4/3

One of the number = -5/1

By using the formula,

Other number = sum of number – given number

= -4/3 – -5/1

By taking LCM for 3 and 1 which is 3

-4/3 – -5/1 = (-4×1 – -5×3)/3

= (-4 + 15)/3

= 11/3

∴ the other number is 11/3

6. The sum of the two rational numbers is -8. If one of the numbers is -15/7, find the other.

Solution: Let us note down the given details

Sum of two rational numbers = -8/1

One of the number = -15/7

Let us consider the other number as x

x + -15/7 = -8

(7x -15)/7 = -8

7x -15 = -8×7

7x – 15 = -56

7x = -56+15

x = -41/7

∴ the other number is -41/7

7. What should be added to -7/8 so as to get 5/9?

Solution: Let us consider a number as x to be added to -7/8 to get 5/9

So, -7/8 + x = 5/9

(-7 + 8x)/8 = 5/9

(-7 + 8x) × 9 = 5 × 8

-63 + 72x = 40

72x = 40 + 63

x = 103/72

∴ the required number is 103/72

8. What number should be added to -5/11 so as to get 26/33?

Solution: Let us consider a number as x to be added to -5/11 to get 26/33

So, -5/11 + x = 26/33

x = 26/33 + 5/11

let us take LCM for 33 and 11 which is 33

x = (26×1 + 5×3)/33

= (26 + 15)/33

= 41/33

∴ the required number is 41/33

9. What number should be added to -5/7 to get -2/3?

Solution: Let us consider a number as x to be added to -5/7 to get -2/3

So, -5/7 + x = -2/3

x = -2/3 + 5/7

let us take LCM for 3 and 7 which is 21

x = (-2×7 + 5×3)/21

= (-14 + 15)/21

= 1/21

∴ the required number is 1/21

10. What number should be subtracted from -5/3 to get 5/6?

Solution: Let us consider a number as x to be subtracted from -5/3 to get 5/6

So, -5/3 – x = 5/6

x = -5/3 – 5/6

let us take LCM for 3 and 6 which is 6

x = (-5×2 – 5×1)/6

= (-10 – 5)/6

= -15/6

Further we can divide by 3 we get,

-15/6 = -5/2

∴ the required number is -5/2

11. What number should be subtracted from 3/7 to get 5/4?

Solution: Let us consider a number as x to be subtracted from 3/7 to get 5/4

So, 3/7 – x = 5/4

x = 3/7 – 5/4

let us take LCM for 7 and 4 which is 28

x = (3×4 – 5×7)/28

= (12 – 35)/28

= -23/28

∴ the required number is -23/28

12. What should be added to (2/3 + 3/5) to get -2/15?

Solution: Let us consider a number as x to be added to (2/3 + 3/5) to get -2/15

x + (2/3 + 3/5) = -2/15

By taking LCM of 3 and 5 which is 15 we get,

(15x + 2×5 + 3×3)15 = -2/15

15x + 10 + 9 = -2

15x = -2-19

x = -21/15

Further we can divide by 3 we get,

-21/15 = -7/5

∴ the required number is -7/5

13. What should be added to (1/2 + 1/3 + 1/5) to get 3?

Solution: Let us consider a number as x to be added to (1/2 + 1/3 + 1/5) to get 3

x + (1/2 + 1/3 + 1/5) = 3

By taking LCM of 2, 3 and 5 which is 30 we get,

(30x + 1×15 + 1×10 + 1×6 )30 = 3

30x + 15 + 10 + 6 = 3 × 30

30x + 31 = 90

30x = 90-31

x = 59/30

∴ the required number is 59/30

14. What number should be subtracted from (3/4 – 2/3) to get -1/6?

Solution: Let us consider a number as x to be subtracted from (3/4 – 2/3) to get -1/6

So, (3/4 – 2/3) – x = -1/6

x = 3/4 – 2/3 + 1/6

Let us take LCM for 4 and 3 which is 12

x = (3×3 – 2×4)/12 + 1/6

= (9 – 8)/12 + 1/6

= 1/12 + 1/6

Let us take LCM for 12 and 6 which is 12

= (1×1 + 1×2)/12

= 3/12

Further we can divide by 3 we get,

3/12 = 1/4 ∴ the required number is ¼

15. Fill in the blanks:

(i) -4/13 – -3/26 = ….

Solution:

-4/13 – -3/26

Let us take LCM for 13 and 26 which is 26

(-4×2 + 3×1)/26

(-8+3)/26 = -5/26

(ii) -9/14 + …. = -1

Solution:

Let us consider the number to be added as x

-9/14 + x = -1

x = -1 + 9/14

By taking LCM as 14 we get,

x = (-1×14 + 9)/14

= (-14+9)/14

= -5/14

(iii) -7/9 + …. =3

Solution:

Let us consider the number to be added as x

-7/9 + x = 3

x = 3 + 7/9

By taking LCM as 9 we get,

x = (3×9 + 7)/9

= (27 + 7)/9

= 34/9

(iv) … + 15/23 = 4

Solution:

Let us consider the number to be added as x

x + 15/23 = 4

x = 4 – 15/23

By taking LCM as 23 we get,

x = (4×23 – 15)/23

= (92 – 15)/23

= 77/23


EXERCISE 1.4 PAGE NO: 1.22

1. Simplify each of the following and write as a rational number of the form p/q:

(i) 3/4 + 5/6 + -7/8

Solution:

3/4 + 5/6 -7/8

By taking LCM for 4, 6 and 8 which is 24

((3×6) + (5×4) – (7×3))/24

(18 + 20 – 21)/24

(38-21)/24

17/24

(ii) 2/3 + -5/6 + -7/9

Solution:

2/3 + -5/6 + -7/9

By taking LCM for 3, 6 and 9 which is 18

((2×6) + (-5×3) + (-7×2))/18

(12 – 15 – 14)/18

-17/18

(iii) -11/2 + 7/6 + -5/8

Solution:

-11/2 + 7/6 + -5/8

By taking LCM for 2, 6 and 8 which is 24

((-11×12) + (7×4) + (-5×3))/24

(-132 + 28 – 15)/24

-119/24

(iv) -4/5 + -7/10 + -8/15

Solution:

-4/5 + -7/10 + -8/15

By taking LCM for 5, 10 and 15 which is 30

((-4×6) + (-7×3) + (-8×2))/30

(-24 – 21 – 16)/30

-61/30

(v) -9/10 + 22/15 + 13/-20

Solution:

-9/10 + 22/15 + 13/-20

By taking LCM for 10, 15 and 20 which is 60

((-9×6) + (22×4) + (-13×3))/60

(-54 + 88 – 39)/60

-5/60 = -1/12

(vi) 5/3 + 3/-2 + -7/3 +3

Solution:

5/3 + 3/-2 + -7/3 +3

By taking LCM for 3, 2, 3 and 1 which is 6

((5×2) + (-3×3) + (-7×2) + (3×6))/6

(10 – 9 – 14 + 18)/6

5/6

2. Express each of the following as a rational number of the form p/q:

(i) -8/3 + -1/4 + -11/6 + 3/8 – 3

Solution:

-8/3 + -1/4 + -11/6 + 3/8 – 3

By taking LCM for 3, 4, 6, 8 and 1 which is 24

((-8×8) + (-1×6) + (-11×4) + (3×3) – (3×24))/24

(-64 – 6 – 44 + 9 – 72)/24

-177/24

Further divide by 3 we get,

-177/24 = -59/8

(ii) 6/7 + 1 + -7/9 + 19/21 + -12/7

Solution:

6/7 + 1 + -7/9 + 19/21 + -12/7

By taking LCM for 7, 1, 9, 21 and 7 which is 63

((6×9) + (1×63) + (-7×7) + (19×3) + (-12×9))/63

(54 + 63 – 49 + 57 – 108)/63

17/63

(iii) 15/2 + 9/8 + -11/3 + 6 + -7/6

Solution:

15/2 + 9/8 + -11/3 + 6 + -7/6

By taking LCM for 2, 8, 3, 1 and 6 which is 24

((15×12) + (9×3) + (-11×8) + (6×24) + (-7×4))/24

(180 + 27 – 88 + 144 – 28)/24

235/24

(iv) -7/4 +0 + -9/5 + 19/10 + 11/14

Solution:

-7/4 +0 + -9/5 + 19/10 + 11/14

By taking LCM for 4, 5, 10 and 14 which is 140

((-7×35) + (-9×28) + (19×14) + (11×10))/140

(-245 – 252 + 266 + 110)/140

-121/140

(v) -7/4 +5/3 + -1/2 + -5/6 + 2

Solution:

-7/4 +5/3 + -1/2 + -5/6 + 2

By taking LCM for 4, 3, 2, 6 and 1 which is 12

((-7×3) + (5×4) + (-1×6) + (-5×2) + (2×12))/12

(-21 + 20 – 6 – 10 + 24)/12

7/12

3. Simplify:

(i) -3/2 + 5/4 – 7/4

Solution:

-3/2 + 5/4 – 7/4

By taking LCM for 2 and 4 which is 4

((-3×2) + (5×1) – (7×1))/4

(-6 + 5 – 7)/4

-8/4

Further divide by 2 we get,

-8/2 = -2

(ii) 5/3 – 7/6 + -2/3

Solution:

5/3 – 7/6 + -2/3

By taking LCM for 3 and 6 which is 6

((5×2) – (7×1) + (-2×2))/6

(10 – 7 – 4)/6

-1/6

(iii) 5/4 – 7/6 – -2/3

Solution:

5/4 – 7/6 – -2/3

By taking LCM for 4, 6 and 3 which is 12

((5×3) – (7×2) – (-2×4))/12

(15 – 14 + 8)/12

9/12

Further can divide by 3 we get,

9/12 = 3/4

(iv) -2/5 – -3/10 – -4/7

Solution:

-2/5 – -3/10 – -4/7

By taking LCM for 5, 10 and 7 which is 70

((-2×14) – (-3×7) – (-4×10))/70

(-28 + 21 + 40)/70

33/70

(v) 5/6 + -2/5 – -2/15

Solution:

5/6 + -2/5 – -2/15

By taking LCM for 6, 5 and 15 which is 30

((5×5) + (-2×6) – (-2×2))/30

(25 – 12 + 4)/30

17/30

(vi) 3/8 – -2/9 + -5/36

Solution:

3/8 – -2/9 + -5/36

By taking LCM for 8, 9 and 36 which is 72

((3×9) – (-2×8) + (-5×2))/72

(27 + 16 – 10)/72

33/72

Further can divide by 3 we get,

33/72 = 11/24


EXERCISE 1.5 PAGE NO: 1.25

1. Multiply:

(i) 7/11 by 5/4

Solution:

7/11 by 5/4

(7/11) × (5/4) = (7×5)/(11×4)

= 35/44

(ii) 5/7 by -3/4

Solution:

5/7 by -3/4

(5/7) × (-3/4) = (5×-3)/(7×4)

= -15/28

(iii) -2/9 by 5/11

Solution:

-2/9 by 5/11

(-2/9) × (5/11) = (-2×5)/(9×11)

= -10/99

(iv) -3/17 by -5/-4

Solution:

-3/17 by -5/-4

(-3/17) × (-5/-4) = (-3×-5)/(17×-4)

= 15/-68

= -15/68

(v) 9/-7 by 36/-11

Solution:

9/-7 by 36/-11

(9/-7) × (36/-11) = (9×36)/(-7×-11)

= 324/77

(vi) -11/13 by -21/7

Solution:

-11/13 by -21/7

(-11/13) × (-21/7) = (-11×-21)/(13×7)

= 231/91 = 33/13

(vii) -3/5 by -4/7

Solution:

-3/5 by -4/7

(-3/5) × (-4/7) = (-3×-4)/(5×7)

= 12/35

(viii) -15/11 by 7

Solution:

-15/11 by 7

(-15/11) × 7 = (-15×7)/11

= -105/11

2. Multiply:

(i) -5/17 by 51/-60

Solution:

-5/17 by 51/-60

(-5/17) × (51/-60) = (-5×51)/(17×-60)

= -255/-1020

Further can divide by 255 we get,

-255/-1020 = 1/4

(ii) -6/11 by -55/36

Solution:

-6/11 by -55/36

(-6/11) × (-55/36) = (-6×-55)/(11×36)

= 330/396

Further can divide by 66 we get,

330/396 = 5/6

(iii) -8/25 by -5/16

Solution:

-8/25 by -5/16

(-8/25) × (-5/16) = (-8×-5)/(25×16)

= 40/400

Further can divide by 40 we get,

40/400 = 1/10

(iv) 6/7 by -49/36

Solution:

6/7 by -49/36

(6/7) × (-49/36) = (6×-49)/(7×36)

= 294/252

Further can divide by 42 we get,

294/252 = -7/6

(v) 8/-9 by -7/-16

Solution:

8/-9 by -7/-16

(8/-9) × (-7/-16) = (8×-7)/(-9×-16)

= -56/144

Further can divide by 8 we get,

-56/144 = -7/18

(vi) -8/9 by 3/64

Solution:

-8/9 by 3/64

(-8/9) × (3/64) = (-8×3)/(9×64)

= -24/576

Further can divide by 24 we get,

-24/576 = -1/24

3. Simplify each of the following and express the result as a rational number in standard form:

(i) (-16/21) × (14/5)

Solution:

(-16/21) × (14/5) = (-16/3) × (2/5) (divisible by 7)

= (-16×2)/(3×5)

= -32/15

(ii) (7/6) × (-3/28)

Solution:

(7/6) × (-3/28) = (1/2) × (-1/4) (divisible by 7 and 3)

= -1/8

(iii) (-19/36) × 16

Solution:

-19/36 × 16 = (-19/9) × 4 (divisible by 4)

= (-19×4)/9 = -76/9

(iv) (-13/9) × (27/-26)

Solution:

(-13/9) × (27/-26) = (-1/1) × (3/-2) (divisible by 13 and 9)

= -3/-2 = 3/2

(v) (-9/16) × (-64/-27)

Solution:

(-9/16) × (-64/-27) = (-1/1) × (-4/-3) (divisible by 9 and 16)

= 4/-3 = -4/3

(vi) (-50/7) × (14/3)

Solution:

(-50/7) × (14/3) = (-50/1) × (2/3) (divisible by 7)

= (-50×2)/(1×3)

= -100/3

(vii) (-11/9) × (-81/-88)

Solution:

(-11/9) × (-81/-88) = (-1/1) × (-9/-8) (divisible by 11 and 9)

= (-1×-9)/(1×-8)

= 9/-8 = -9/8

(viii) (-5/9) × (72/-25)

Solution:

(-5/9) × (72/-25) = (-1/1) × (8/-5) (divisible by 5 and 9)

= (-1×8)/(1×-5)

= -8/-5 = 8/5

4. Simplify:

(i) ((25/8) × (2/5)) – ((3/5) × (-10/9))

Solution:

((25/8) × (2/5)) – ((3/5) × (-10/9)) = (25×2)/(8×5) – (3×-10)/(5×9)

= 50/40 – -30/45

= 5/4 + 2/3 (divisible by 5 and 3)

By taking LCM for 4 and 3 which is 12

= ((5×3) + (2×4))/12

= (15+8)/12

= 23/12

(ii) ((1/2) × (1/4)) + ((1/2) × 6)

Solution:

((1/2) × (1/4)) + ((1/2) × 6) = (1×1)/(2×4) + (1×3) (divisible by 2)

= 1/8 +3

By taking LCM for 8 and 1 which is 8

= ((1×1) + (3×8))/8

= (1+24)/8

= 25/8

(iii) (-5 × (2/15)) – (-6 × (2/9))

Solution:

(-5 × (2/15)) – (-6 × (2/9)) = (-1 × (2/3)) – (-2 × (2/3)) (divisible by 5 and 3)

= (-2/3) + (4/3)

Since the denominators are same we can add directly

= (-2+4)/3

= 2/3

(iv) ((-9/4) × (5/3)) + ((13/2) × (5/6))

Solution:

((-9/4) × (5/3)) + ((13/2) × (5/6)) = (-9×5)/(4×3) + (13×5)/(2×6)

= -45/12 + 65/12

Since the denominators are same we can add directly

= (-45+65)/12

= 20/12 (divisible by 2)

= 10/6 (divisible by 2)

= 5/3

(v) ((-4/3) × (12/-5)) + ((3/7) × (21/15))

Solution:

((-4/3) × (12/-5)) + ((3/7) × (21/15)) = ((-4/1) × (4/-5)) + ((1/1) × (3/5)) (divisible by 3, 7)

= (-4×4)/(1×-5) + (1×3)/(1×5)

= -16/-5 + 3/5

Since the denominators are same we can add directly

= (16+3)/5

= 19/5

(vi) ((13/5) × (8/3)) – ((-5/2) × (11/3))

Solution:

((13/5) × (8/3)) – ((-5/2) × (11/3)) = (13×8)/(5×3) – (-5×11)/(2×3)

= 104/15 + 55/6

By taking LCM for 15 and 6 which is 30

= ((104×2) + (55×5))/30

= (208+275)/30

= 483/30

(vii) ((13/7) × (11/26)) – ((-4/3) × (5/6))

Solution:

((13/7) × (11/26)) – ((-4/3) × (5/6)) = ((1/7) × (11/2)) – ((-2/3) × (5/3)) (divisible by 13, 2)

= (1×11)/(7×2) – (-2×5)/(3×3)

= 11/14 + 10/9

By taking LCM for 14 and 9 which is 126

= ((11×9) + (10×14))/126

= (99+140)/126

= 239/126

(viii) ((8/5) × (-3/2)) + ((-3/10) × (11/16))

Solution:

((8/5) × (-3/2)) + ((-3/10) × (11/16)) = ((4/5) × (-3/1)) + ((-3/10) × (11/16)) (divisible by 2)

= (4×-3)/(5×1) + (-3×11)/(10×16)

= -12/5 – 33/160

By taking LCM for 5 and 160 which is 160

= ((-12×32) – (33×1))/160

= (-384 – 33)/160

= -417/160

5. Simplify:

(i) ((3/2) × (1/6)) + ((5/3) × (7/2) – (13/8) × (4/3))

Solution:

((3/2) × (1/6)) + ((5/3) × (7/2) – (13/8) × (4/3)) =

((1/2) × (1/2)) + ((5/3) × (7/2) – (13/2) × (1/3))

(1×1)/(2×2) + (5×7)/(3×2) – (13×1)/(2×3)

1/4 + 35/6 – 13/6

By taking LCM for 4 and 6 which is 24

((1×6) + (35×4) – (13×4))/24

(6 + 140 – 52)/24

94/24

Further divide by 2 we get, 94/24 = 47/12

(ii) ((1/4) × (2/7)) – ((5/14) × (-2/3) + (3/7) × (9/2))

Solution:

((1/4) × (2/7)) – ((5/14) × (-2/3) + (3/7) × (9/2)) =

((1/2) × (1/7)) – ((5/7) × (-1/3) + (3/7) × (9/2))

(1×1)/(2×7) – (5×-1)/(7×3) + (3×9)/(7×2)

1/14 + 5/21 + 27/14

By taking LCM for 14 and 21 which is 42

((1×3) + (5×2) + (27×3))/42

(3 + 10 + 81)/42

94/42

Further divide by 2 we get, 94/42 = 47/21

(iii) ((13/9) × (-15/2)) + ((7/3) × (8/5) + (3/5) × (1/2))

Solution:

((13/3) × (-5/2)) + ((7/3) × (8/5) + (3/5) × (1/2)) =

(13×-5)/(3×2) + (7×8)/(3×5) + (3×1)/(5×2)

-65/6 + 56/15 + 3/10

By taking LCM for 6, 15 and 10 which is 30

((-65×5) + (56×2) + (3×3))/30

(-325 + 112 + 9)/30

-204/30

Further divide by 2 we get, -204/30 = -102/15

(iv) ((3/11) × (5/6)) – ((9/12) × (4/3) + (5/13) × (6/15))

Solution:

((3/11) × (5/6)) – ((9/12) × (4/3) + (5/13) × (6/15)) =

((1/11) × (5/2)) – ((1/1) × (1/1) + (1/13) × (2/1))

(1×5)/(11×2) – 1/1 + (1×2)/(13×1)

5/22 – 1/1 + 2/13

By taking LCM for 22, 1 and 13 which is 286

((5×13) – (1×286) + (2×22))/286

(65 – 286 + 44)/286

-177/286


EXERCISE 1.6 PAGE NO: 1.31

1. Verify the property: x × y = y × x by taking:

(i) x = -1/3, y = 2/7

Solution:

By using the property

x × y = y × x

-1/3 × 2/7 = 2/7 × -1/3

(-1×2)/(3×7) = (2×-1)/(7×3)

-2/21 = -2/21

Hence, the property is satisfied.

(ii) x = -3/5, y = -11/13

Solution:

By using the property

x × y = y × x

-3/5 × -11/13 = -11/13 × -3/5

(-3×-11)/(5×13) = (-11×-3)/(13×5)

33/65 = 33/65

Hence, the property is satisfied.

(iii) x = 2, y = 7/-8

Solution:

By using the property

x × y = y × x

2 × 7/-8 = 7/-8 × 2

(2×7)/-8 = (7×2)/-8

14/-8 = 14/-8

-14/8 = -14/8

Hence, the property is satisfied.

(iv) x = 0, y = -15/8

Solution:

By using the property

x × y = y × x

0 × -15/8 = -15/8 × 0

0 = 0

Hence, the property is satisfied.

2. Verify the property: x × (y × z) = (x × y) × z by taking:

(i) x = -7/3, y = 12/5, z = 4/9

Solution:

By using the property

x × (y × z) = (x × y) × z

-7/3 × (12/5 × 4/9) = (-7/3 × 12/5) × 4/9

(-7×12×4)/(3×5×9) = (-7×12×4)/(3×5×9)

-336/135 = -336/135

Hence, the property is satisfied.

(ii) x = 0, y = -3/5, z = -9/4

Solution:

By using the property

x × (y × z) = (x × y) × z

0 × (-3/5 × -9/4) = (0 × -3/5) × -9/4

0 = 0

Hence, the property is satisfied.

(iii) x = 1/2, y = 5/-4, z = -7/5

Solution:

By using the property

x × (y × z) = (x × y) × z

1/2 × (5/-4 × -7/5) = (1/2 × 5/-4) × -7/5

(1×5×-7)/(2×-4×5) = (1×5×-7)/(2×-4×5)

-35/-40 = -35/-40

35/40 = 35/40

Hence, the property is satisfied.

(iv) x = 5/7, y = -12/13, z = -7/18

Solution:

By using the property

x × (y × z) = (x × y) × z

5/7 × (-12/13 × -7/18) = (5/7 × -12/13) × -7/18

(5×-12×-7)/(7×13×18) = (5×-12×-7)/(7×13×18)

420/1638 = 420/1638

Hence, the property is satisfied.

3. Verify the property: x × (y + z) = x × y + x × z by taking:

(i) x = -3/7, y = 12/13, z = -5/6

Solution:

By using the property

x × (y + z) = x × y + x × z

-3/7 × (12/13 + -5/6) = -3/7 × 12/13 + -3/7 × -5/6

-3/7 × ((12×6) + (-5×13))/78 = (-3×12)/(7×13) + (-3×-5)/(7×6)

-3/7 × (72-65)/78 = -36/91 + 15/42

-3/7 × 7/78 = (-36×6 + 15×13)/546

-1/26 = (196-216)/546

= -21/546

= -1/26

Hence, the property is verified.

(ii) x = -12/5, y = -15/4, z = 8/3

Solution:

By using the property

x × (y + z) = x × y + x × z

-12/5 × (-15/4 + 8/3) = -12/5 × -15/4 + -12/5 × 8/3

-12/5 × ((-15×3) + (8×4))/12 = (-12×-15)/(5×4) + (-12×8)/(5×3)

-12/5 × (-45+32)/12 = 180/20 – 96/15

-12/5 × -13/12 = 9 – 32/5

13/5 = (9×5 – 32×1)/5

= (45-32)/5

= 13/5

Hence, the property is verified.

(iii) x = -8/3, y = 5/6, z = -13/12

Solution:

By using the property

x × (y + z) = x × y + x × z

-8/3 × (5/6 + -13/12) = -8/3 × 5/6 + -8/3 × -13/12

-8/3 × ((5×2) – (13×1))/12 = (-8×5)/(3×6) + (-8×-13)/(3×12)

-8/3 × (10-13)/12 = -40/18 + 104/36

-8/3 × -3/12 = (-40×2 + 104×1)/36

2/3 = (-80+104)/36

= 24/36

= 2/3

Hence, the property is verified.

(iv) x = -3/4, y = -5/2, z = 7/6

Solution:

By using the property

x × (y + z) = x × y + x × z

-3/4 × (-5/2 + 7/6) = -3/4 × -5/2 + -3/4 × 7/6

-3/4 × ((-5×3) + (7×1))/6 = (-3×-5)/(4×2) + (-3×7)/(4×6)

-3/4 × (-15+7)/6 = 15/8 – 21/24

-3/4 × -8/6 = (15×3 – 21×1)/24

-3/4 × -4/3 = (45-21)/24

1 = 24/24

= 1

Hence, the property is verified.

4. Use the distributivity of multiplication of rational numbers over their addition to simplify:

(i) 3/5 × ((35/24) + (10/1))

Solution:

3/5 × 35/24 + 3/5 × 10

1/1 × 7/8 + 6/1

By taking LCM for 8 and 1 which is 8

7/8 + 6 = (7×1 + 6×8)/8

= (7+48)/8

= 55/8

(ii) -5/4 × ((8/5) + (16/5))

Solution:

-5/4 × 8/5 + -5/4 × 16/5

-1/1 × 2/1 + -1/1 × 4/1

-2 + -4

-2 – 4

-6

(iii) 2/7 × ((7/16) – (21/4))

Solution:

2/7 × 7/16 – 2/7 × 21/4

1/1 × 1/8 – 1/1 × 3/2

1/8 – 3/2

By taking LCM for 8 and 2 which is 8

1/8 – 3/2 = (1×1 – 3×4)/8

= (1 – 12)/8

= -11/8

(iv) 3/4 × ((8/9) – 40)

Solution:

3/4 × 8/9 – 3/4 × 40

1/1 × 2/3 – 3/1 × 10

2/3 – 30/1

By taking LCM for 3 and 1 which is 3

2/3 – 30/1 = (2×1 – 30×3)/3

= (2 – 90)/3

= -88/3

5. Find the multiplicative inverse (reciprocal) of each of the following rational numbers:

(i) 9

(ii) -7

(iii) 12/5

(iv) -7/9

(v) -3/-5

(vi) 2/3 × 9/4

(vii) -5/8 × 16/15

(viii) -2 × -3/5

(ix) -1

(x) 0/3

(xi) 1

Solution:

(i) The reciprocal of 9 is 1/9

(ii) The reciprocal of -7 is -1/7

(iii) The reciprocal of 12/5 is 5/12

(iv) The reciprocal of -7/9 is 9/-7

(v) The reciprocal of -3/-5 is 5/3

(vi) The reciprocal of 2/3 × 9/4 is

Firstly solve for 2/3 × 9/4 = 1/1 × 3/2 = 3/2

∴ The reciprocal of 3/2 is 2/3

(vii) The reciprocal of -5/8 × 16/15

Firstly solve for -5/8 × 16/15 = -1/1 × 2/3 = -2/3

∴ The reciprocal of -2/3 is 3/-2

(viii) The reciprocal of -2 × -3/5

Firstly solve for -2 × -3/5 = 6/5

∴ The reciprocal of 6/5 is 5/6

(ix) The reciprocal of -1 is -1

(x) The reciprocal of 0/3 does not exist

(xi) The reciprocal of 1 is 1

6. Name the property of multiplication of rational numbers illustrated by the following statements:

(i) -5/16 × 8/15 = 8/15 × -5/16

(ii) -17/5 ×9 = 9 × -17/5

(iii) 7/4 × (-8/3 + -13/12) = 7/4 × -8/3 + 7/4 × -13/12

(iv) -5/9 × (4/15 × -9/8) = (-5/9 × 4/15) × -9/8

(v) 13/-17 × 1 = 13/-17 = 1 × 13/-17

(vi) -11/16 × 16/-11 = 1

(vii) 2/13 × 0 = 0 = 0 × 2/13

(viii) -3/2 × 5/4 + -3/2 × -7/6 = -3/2 × (5/4 + -7/6)

Solution:

(i) -5/16 × 8/15 = 8/15 × -5/16

According to commutative law, a/b × c/d = c/d × a/b

The above rational number satisfies commutative property.

(ii) -17/5 ×9 = 9 × -17/5

According to commutative law, a/b × c/d = c/d × a/b

The above rational number satisfies commutative property.

(iii) 7/4 × (-8/3 + -13/12) = 7/4 × -8/3 + 7/4 × -13/12

According to given rational number, a/b × (c/d + e/f) = (a/b × c/d) + (a/b × e/f)

Distributivity of multiplication over addition satisfies.

(iv) -5/9 × (4/15 × -9/8) = (-5/9 × 4/15) × -9/8

According to associative law, a/b × (c/d × e/f ) = (a/b × c/d) × e/f

The above rational number satisfies associativity of multiplication.

(v) 13/-17 × 1 = 13/-17 = 1 × 13/-17

Existence of identity for multiplication satisfies for the given rational number.

(vi) -11/16 × 16/-11 = 1

Existence of multiplication inverse satisfies for the given rational number.

(vii) 2/13 × 0 = 0 = 0 × 2/13

By using a/b × 0 = 0 × a/b

Multiplication of zero satisfies for the given rational number.

(viii) -3/2 × 5/4 + -3/2 × -7/6 = -3/2 × (5/4 + -7/6)

According to distributive law, (a/b × c/d) + (a/b × e/f ) = a/b × (c/d + e/f)

The above rational number satisfies distributive law.

7. Fill in the blanks:

(i) The product of two positive rational numbers is always…

(ii) The product of a positive rational number and a negative rational number is always….

(iii) The product of two negative rational numbers is always…

(iv) The reciprocal of a positive rational numbers is…

(v) The reciprocal of a negative rational numbers is…

(vi) Zero has …. Reciprocal.

(vii) The product of a rational number and its reciprocal is…

(viii) The numbers … and … are their own reciprocals.

(ix) If a is reciprocal of b, then the reciprocal of b is.

(x) The number 0 is … the reciprocal of any number.

(xi) reciprocal of 1/a, a ≠ 0 is …

(xii) (17×12)-1 = 17-1 × …

Solution:

(i) The product of two positive rational numbers is always positive.

(ii) The product of a positive rational number and a negative rational number is always negative.

(iii) The product of two negative rational numbers is always positive.

(iv) The reciprocal of a positive rational numbers is positive.

(v) The reciprocal of a negative rational numbers is negative.

(vi) Zero has no Reciprocal.

(vii) The product of a rational number and its reciprocal is 1.

(viii) The numbers 1 and -1 are their own reciprocals.

(ix) If a is reciprocal of b, then the reciprocal of b is a.

(x) The number 0 is not the reciprocal of any number.

(xi) reciprocal of 1/a, a ≠ 0 is a.

(xii) (17×12)-1 = 17-1 × 12-1

8. Fill in the blanks:

(i) -4 × 7/9 = 79 × …

Solution:

-4 × 7/9 = 79 × -4

By using commutative property.

(ii) 5/11 × -3/8 = -3/8 × …

Solution:

5/11 × -3/8 = -3/8 × 5/11

By using commutative property.

(iii) 1/2 × (3/4 + -5/12) = 1/2 × … + … × -5/12

Solution:

1/2 × (3/4 + -5/12) = 1/2 × 3/4 + 1/2 × -5/12

By using distributive property.

(iv) -4/5 × (5/7 + -8/9) = (-4/5 × …) + -4/5 × -8/9

Solution:

-4/5 × (5/7 + -8/9) = (-4/5 × 5/7) + -4/5 × -8/9

By using distributive property.


EXERCISE 1.7 PAGE NO: 1.35

1. Divide:

(i) 1 by 1/2

Solution:

1/1/2 = 1 × 2/1 = 2

(ii) 5 by -5/7

Solution:

5/-5/7 = 5 × 7/-5 = -7

(iii) -3/4 by 9/-16

Solution:

(-3/4) / (9/-16)

(-3/4) × -16/9 = 4/3

(iv) -7/8 by -21/16

Solution:

(-7/8) / (-21/16)

(-7/8) × 16/-21 = 2/3

(v) 7/-4 by 63/64

Solution:

(7/-4) / (63/64)

(7/-4) × 64/63 = -16/9

(vi) 0 by -7/5

Solution:

0 / (7/5) = 0

(vii) -3/4 by -6

Solution:

(-3/4) / -6

(-3/4) × 1/-6 = 1/8

(viii) 2/3 by -7/12

Solution:

(2/3) / (-7/12)

(2/3) × 12/-7 = -8/7

(ix) -4 by -3/5

Solution:

-4 / (-3/5)

-4 × 5/-3 = 20/3

(x) -3/13 by -4/65

Solution:

(-3/13) / (-4/65)

(-3/13) × (65/-4) = 15/4

2. Find the value and express as a rational number in standard form:

(i) 2/5 ÷ 26/15

Solution:

(2/5) / (26/15)

(2/5) × (15/26)

(2/1) × (3/26) = (2×3)/ (1×26) = 6/26 = 3/13

(ii) 10/3 ÷ -35/12

Solution:

(10/3) / (-35/12)

(10/3) × (12/-35)

(10/1) × (4/-35) = (10×4)/ (1×-35) = -40/35 = -8/7

(iii) -6 ÷ -8/17

Solution:

-6 / (-8/17)

-6 × (17/-8)

-3 × (17/-4) = (-3×17)/ (1×-4) = 51/4

(iv) -40/99 ÷ -20

Solution:

(-40/99) / -20

(-40/99) × (1/-20)

(-2/99) × (1/-1) = (-2×1)/ (99×-1) = 2/99

(v) -22/27 ÷ -110/18

Solution:

(-22/27) / (-110/18)

(-22/27) × (18/-110)

(-1/9) × (6/-5)

(-1/3) × (2/-5) = (-1×2) / (3×-5) = 2/15

(vi) -36/125 ÷ -3/75

Solution:

(-36/125) / (-3/75)

(-36/125) × (75/-3)

(-12/25) × (15/-1)

(-12/5) × (3/-1) = (-12×3) / (5×-1) = 36/5

3. The product of two rational numbers is 15. If one of the numbers is -10, find the other.

Solution:

We know that the product of two rational numbers = 15

One of the number = -10

∴ other number can be obtained by dividing the product by the given number.

Other number = 15/-10

= -3/2

4. The product of two rational numbers is -8/9. If one of the numbers is -4/15, find the other.

Solution:

We know that the product of two rational numbers = -8/9

One of the number = -4/15

∴ other number is obtained by dividing the product by the given number.

Other number = (-8/9)/(-4/15)

= (-8/9) × (15/-4)

= (-2/3) × (5/-1)

= (-2×5) /(3×-1)

= -10/-3

= 10/3

5. By what number should we multiply -1/6 so that the product may be -23/9?

Solution:

Let us consider a number = x

So, x × -1/6 = -23/9

x = (-23/9)/(-1/6)

x = (-23/9) × (6/-1)

= (-23/3) × (2×-1)

= (-23×-2)/(3×1)

= 46/3

6. By what number should we multiply -15/28 so that the product may be -5/7?

Solution:

Let us consider a number = x

So, x × -15/28 = -5/7

x = (-5/7)/(-15/28)

x = (-5/7) × (28/-15)

= (-1/1) × (4×-3)

= 4/3

7. By what number should we multiply -8/13 so that the product may be 24?

Solution:

Let us consider a number = x

So, x × -8/13 = 24

x = (24)/(-8/13)

x = (24) × (13/-8)

= (3) × (13×-1)

= -39

8. By what number should -3/4 be multiplied in order to produce 2/3?

Solution:

Let us consider a number = x

So, x × -3/4 = 2/3

x = (2/3)/(-3/4)

x = (2/3) × (4/-3)

= -8/9

9. Find (x+y) ÷ (x-y), if

(i) x= 2/3, y= 3/2

Solution:

(x+y) ÷ (x-y)

(2/3 + 3/2) / (2/3 – 3/2)

((2×2 + 3×3)/6) / ((2×2 – 3×3)/6)

((4+9)/6) / ((4-9)/6)

(13/6) / (-5/6)

(13/6) × (6/-5)

-13/5

(ii) x= 2/5, y= 1/2

Solution:

(x+y) ÷ (x-y)

(2/5 + 1/2) / (2/5 – 1/2)

((2×2 + 1×5)/10) / ((2×2 – 1×5)/10)

((4+5)/10) / ((4-5)/10)

(9/10) / (-1/10)

(9/10) × (10/-1)

-9

(iii) x= 5/4, y= -1/3

Solution:

(x+y) ÷ (x-y)

(5/4 – 1/3) / (5/4 + 1/3)

((5×3 – 1×4)/12) / ((5×3 + 1×4)/12)

((15-4)/12) / ((15+4)/12)

(11/12) / (19/12)

(11/12) × (12/19)

11/19

(iv) x= 2/7, y= 4/3

Solution:

(x+y) ÷ (x-y)

(2/7 + 4/3) / (2/7 – 4/3)

((2×3 + 4×7)/21) / ((2×3 – 4×7)/21)

((6+28)/21) / ((6-28)/21)

(34/21) / (-22/21)

(34/21) × (21/-22)

-34/22

-17/11

(v) x= 1/4, y= 3/2

Solution:

(x+y) ÷ (x-y)

(1/4 + 3/2) / (1/4 – 3/2)

((1×1 + 3×2)/4) / ((1×1 – 3×2)/4)

((1+6)/4) / ((1-6)/4)

(7/4) / (-5/4)

(7/4) × (4/-5) = -7/5

10. The cost of (7frac{2}{3}) meters of rope is Rs 12 ¾. Find the cost per meter.

Solution:

We know that 23/3 meters of rope = Rs 51/4

Let us consider a number = x

So, x × 23/3 = 51/4

x = (51/4)/(23/3)

x = (51/4) × (3/23)

= (51×3) / (4×23)

= 153/92

= (1frac{61}{92})

∴ cost per meter is Rs (1frac{61}{92})

11. The cost of (2frac{1}{3}) meters of cloth is Rs 75 ¼. Find the cost of cloth per meter.

Solution:

We know that 7/3 meters of cloth = Rs 301/4

Let us consider a number = x

So, x × 7/3 = 301/4

x = (301/4)/(7/3)

x = (301/4) × (3/7)

= (301×3) / (4×7)

= (43×3) / (4×1)

= 129/4

= 32.25

∴ cost of cloth per meter is Rs 32.25

12. By what number should -33/16 be divided to get -11/4?

Solution:

Let us consider a number = x

So, (-33/16)/x = -11/4

-33/16 = x × -11/4

x = (-33/16) / (-11/4)

= (-33/16) × (4/-11)

= (-33×4)/(16×-11)

= (-3×1)/(4×-1)

= ¾

13. Divide the sum of -13/5 and 12/7 by the product of -31/7 and -1/2.

Solution:

sum of -13/5 and 12/7

-13/5 + 12/7

((-13×7) + (12×5))/35

(-91+60)/35

-31/35

Product of -31/7 and -1/2

-31/7 × -1/2

(-31×-1)/(7×2)

31/14

∴ by dividing the sum and the product we get,

(-31/35) / (31/14)

(-31/35) × (14/31)

(-31×14)/(35×31)

-14/35

-2/5

14. Divide the sum of 65/12 and 12/7 by their difference.

Solution:

The sum is 65/12 + 12/7

The difference is 65/12 – 12/7

When we divide, (65/12 + 12/7) / (65/12 – 12/7)

((65×7 + 12×12)/84) / ((65×7 – 12×12)/84)

((455+144)/84) / ((455 – 144)/84)

(599/84) / (311/84)

599/84 × 84/311

599/311

15. If 24 trousers of equal size can be prepared in 54 meters of cloth, what length of cloth is required for each trouser?

Solution:

We know that total number trousers = 24

Total length of the cloth = 54

Length of the cloth required for each trouser = total length of the cloth/number of trousers

= 54/24

= 9/4

∴ 9/4 meters is required for each trouser.


EXERCISE 1.8 PAGE NO: 1.43

1. Find a rational number between -3 and 1.

Solution:

Let us consider two rational numbers x and y

We know that between two rational numbers x and y where x < y there is a rational number (x+y)/2

x < (x+y)/2 < y

(-3+1)/2 = -2/2 = -1

So, the rational number between -3 and 1 is -1

∴ -3 < -1 < 1

2. Find any five rational numbers less than 2.

Solution:

Five rational numbers less than 2 are 0, 1/5, 2/5, 3/5, 4/5

3. Find two rational numbers between -2/9 and 5/9

Solution:

The rational numbers between -2/9 and 5/9 is

(-2/9 + 5/9)/2

(1/3)/2

1/6

The rational numbers between -2/9 and 1/6 is

(-2/9 + 1/6)/2

((-2×2 + 1×3)/18)/2

(-4+3)/36

-1/36

∴ the rational numbers between -2/9 and 5/9 are -1/36, 1/6

4. Find two rational numbers between 1/5 and 1/2

Solution:

The rational numbers between 1/5 and 1/2 is

(1/5 + 1/2)/2

((1×2 + 1×5)/10)/2

(2+5)/20 = 7/20

The rational numbers between 1/5 and 7/20 is

(1/5 + 7/20)/2

((1×4 + 7×1)/20)/2

(4+7)/40

11/40

∴ the rational numbers between 1/5 and 1/2 are 7/20, 11/40

5. Find ten rational numbers between 1/4 and 1/2.

Solution:

Firstly convert the given rational numbers into equivalent rational numbers with same denominators.

The LCM for 4 and 2 is 4.

1/4 = 1/4

1/2 = (1×2)/4 = 2/4

1/4 = (1×20 / 4×20) = 20/80

1/2 = (2×20 / 4×20) = 40/80

So, we now know that 21, 22, 23,…39 are integers between numerators 20 and 40.

∴ the rational numbers between 1/4 and 1/2 are 21/80, 22/80, 23/80, …., 39/80

6. Find ten rational numbers between -2/5 and 1/2.

Solution:

Firstly convert the given rational numbers into equivalent rational numbers with same denominators.

The LCM for 5 and 2 is 10.

-2/5 = (-2×2)/10 = -4/10

1/2 = (1×5)/10 = 5/10

-2/5 = (-4×2 / 10×2) = -8/20

1/2 = (5×2 / 10×2) = 10/20

So, we now know that -7, -6, -5,…10 are integers between numerators -8 and 10.

∴ the rational numbers between -2/5 and 1/2 are -7/20, -6/20, -5/20, …., 9/20

7. Find ten rational numbers between 3/5 and 3/4.

Solution:

Firstly convert the given rational numbers into equivalent rational numbers with same denominators.

The LCM for 5 and 4 is 20.

3/5 = 3× 20 / 5×20 = 60/100

3/4 = 3×25 / 4×25 = 75/100

So, we now know that 61, 62, 63,..74 are integers between numerators 60 and 75.

∴ the rational numbers between 3/5 and 3/4 are 61/100, 62/100, 63/100, …., 74/100


RD Sharma Solutions for Class 8 Maths Chapter 1 – Rational Numbers

Here students will be acquainted with detailed concepts discussed in this Chapter as listed below.

  • Introduction to rational numbers.
  • Review about rational numbers.
  • Addition of rational numbers and their properties.
  • Subtraction of rational numbers and their properties.
  • Simplification of expressions involving addition and subtraction.
  • Properties of multiplication of rational numbers.
  • Division of rational numbers.
  • Representation of rational numbers on the number line.

All Chapter RD Sharma Solutions For Class 8 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

In this chapter, we provide RD Sharma Solutions for Class 9 Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Maths pdf, free RD Sharma Solutions for Class 9 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Class 9th solutions Chapter 16 Circles

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length (frac { 1 }{ 2 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = (frac { 1 }{ 2 }) (AB).
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length (frac { 3 }{ 4 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A1, A2, A3 and A4.
(iii) JoinA4B.
(iv) From A3, draw a line parallel to A4B which meets AB at C.
C is the required point and AC = (frac { 3 }{ 4 }) (AB).
Class 9 RD Sharma Solutions Chapter 16 Circles

RD Sharma Solutions Class 9 Chapter 16 Circles Ex 16.2

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = (frac { 1 }{ 2 }) x 108°
∴ We shall bisect it.
RD Sharma Solutions Class 9 Chapter 16 Circles
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =(frac { 1 }{ 2 }) x 90° = 45°.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ (frac { 1 }{ 2 }) ∠DCA + (frac { 1 }{ 2 }) ∠DCB = 180° x (frac { 1 }{ 2 }) = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
Class 9 RD Sharma Solutions Chapter 16 Circles

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x (frac { 1 }{ 2 }) = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + (frac { 1 }{ 2 }) x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
RD Sharma Class 9 Book Chapter 16 Circles
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
Circles With Solutions PDF RD Sharma Class 9 Solutions

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 (frac { 1 }{ 2 })°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
RD Sharma Class 9 Maths Book Questions Chapter 16 Circles
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
RD Sharma Mathematics Class 9 Solutions Chapter 16 Circles
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
RD Sharma Math Solution Class 9 Chapter 16 Circles
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
RD Sharma Class 9 Questions Chapter 16 Circles
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
Maths RD Sharma Class 9 Chapter 16 Circles
(vi) 22 (frac { 1 }{ 2 })°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 (frac { 1 }{ 2 })°
RD Sharma Class 9 PDF Chapter 16 Circles

RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles Ex 16.3

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
Circles Class 9 RD Sharma Solutions

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
Class 9 RD Sharma Solutions Chapter 16 Circles
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
RD Sharma Class 9 Book Chapter 16 Circles

Solutions of Chapter 16 Circles Ex 16.1 and 16.2 are given below:
Chapter 16 Circles Ex 16.3
Chapter 16 Circles Ex 16.4
Chapter 16 Circles Ex 16.5

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1 1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 1
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 2
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 3
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 4
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 5
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 6
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 7
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 8
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 9
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 10
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 11
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 12
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 13
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 14
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 16
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 17

RD Sharma Class 9 Solutions CirclesClass 9 Maths Sample Papers

All Chapter RD Sharma Solutions For Class 9 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

In this chapter, we provide RD Sharma Solutions for Class 9 Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Maths pdf, free RD Sharma Solutions for Class 9 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Class 9th solutions Chapter 16 Circles

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length (frac { 1 }{ 2 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = (frac { 1 }{ 2 }) (AB).
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length (frac { 3 }{ 4 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A1, A2, A3 and A4.
(iii) JoinA4B.
(iv) From A3, draw a line parallel to A4B which meets AB at C.
C is the required point and AC = (frac { 3 }{ 4 }) (AB).
Class 9 RD Sharma Solutions Chapter 16 Circles

RD Sharma Solutions Class 9 Chapter 16 Circles Ex 16.2

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = (frac { 1 }{ 2 }) x 108°
∴ We shall bisect it.
RD Sharma Solutions Class 9 Chapter 16 Circles
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =(frac { 1 }{ 2 }) x 90° = 45°.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ (frac { 1 }{ 2 }) ∠DCA + (frac { 1 }{ 2 }) ∠DCB = 180° x (frac { 1 }{ 2 }) = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
Class 9 RD Sharma Solutions Chapter 16 Circles

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x (frac { 1 }{ 2 }) = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + (frac { 1 }{ 2 }) x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
RD Sharma Class 9 Book Chapter 16 Circles
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
Circles With Solutions PDF RD Sharma Class 9 Solutions

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 (frac { 1 }{ 2 })°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
RD Sharma Class 9 Maths Book Questions Chapter 16 Circles
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
RD Sharma Mathematics Class 9 Solutions Chapter 16 Circles
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
RD Sharma Math Solution Class 9 Chapter 16 Circles
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
RD Sharma Class 9 Questions Chapter 16 Circles
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
Maths RD Sharma Class 9 Chapter 16 Circles
(vi) 22 (frac { 1 }{ 2 })°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 (frac { 1 }{ 2 })°
RD Sharma Class 9 PDF Chapter 16 Circles

RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles Ex 16.3

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
Circles Class 9 RD Sharma Solutions

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
Class 9 RD Sharma Solutions Chapter 16 Circles
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
RD Sharma Class 9 Book Chapter 16 Circles

Solutions of Chapter 16 Circles Ex 16.1 and 16.2 are given below:
Chapter 16 Circles Ex 16.3
Chapter 16 Circles Ex 16.4
Chapter 16 Circles Ex 16.5

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1 1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 1
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 2
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 3
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 4
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 5
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 6
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 7
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 8
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 9
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 10
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 11
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 12
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 13
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 14
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 16
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 17

RD Sharma Class 9 Solutions CirclesClass 9 Maths Sample Papers

All Chapter RD Sharma Solutions For Class 9 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

In this chapter, we provide RD Sharma Solutions for Class 9 Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Maths pdf, free RD Sharma Solutions for Class 9 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Class 9th solutions Chapter 16 Circles

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length (frac { 1 }{ 2 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = (frac { 1 }{ 2 }) (AB).
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length (frac { 3 }{ 4 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A1, A2, A3 and A4.
(iii) JoinA4B.
(iv) From A3, draw a line parallel to A4B which meets AB at C.
C is the required point and AC = (frac { 3 }{ 4 }) (AB).
Class 9 RD Sharma Solutions Chapter 16 Circles

RD Sharma Solutions Class 9 Chapter 16 Circles Ex 16.2

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = (frac { 1 }{ 2 }) x 108°
∴ We shall bisect it.
RD Sharma Solutions Class 9 Chapter 16 Circles
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =(frac { 1 }{ 2 }) x 90° = 45°.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ (frac { 1 }{ 2 }) ∠DCA + (frac { 1 }{ 2 }) ∠DCB = 180° x (frac { 1 }{ 2 }) = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
Class 9 RD Sharma Solutions Chapter 16 Circles

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x (frac { 1 }{ 2 }) = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + (frac { 1 }{ 2 }) x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
RD Sharma Class 9 Book Chapter 16 Circles
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
Circles With Solutions PDF RD Sharma Class 9 Solutions

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 (frac { 1 }{ 2 })°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
RD Sharma Class 9 Maths Book Questions Chapter 16 Circles
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
RD Sharma Mathematics Class 9 Solutions Chapter 16 Circles
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
RD Sharma Math Solution Class 9 Chapter 16 Circles
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
RD Sharma Class 9 Questions Chapter 16 Circles
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
Maths RD Sharma Class 9 Chapter 16 Circles
(vi) 22 (frac { 1 }{ 2 })°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 (frac { 1 }{ 2 })°
RD Sharma Class 9 PDF Chapter 16 Circles

RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles Ex 16.3

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
Circles Class 9 RD Sharma Solutions

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
Class 9 RD Sharma Solutions Chapter 16 Circles
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
RD Sharma Class 9 Book Chapter 16 Circles

Solutions of Chapter 16 Circles Ex 16.1 and 16.2 are given below:
Chapter 16 Circles Ex 16.3
Chapter 16 Circles Ex 16.4
Chapter 16 Circles Ex 16.5

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1 1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 1
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 2
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 3
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 4
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 5
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 6
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 7
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 8
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 9
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 10
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 11
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 12
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 13
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 14
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 16
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 17

RD Sharma Class 9 Solutions CirclesClass 9 Maths Sample Papers

All Chapter RD Sharma Solutions For Class 9 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

In this chapter, we provide RD Sharma Solutions for Class 9 Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Maths pdf, free RD Sharma Solutions for Class 9 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Class 9th solutions Chapter 16 Circles

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length (frac { 1 }{ 2 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = (frac { 1 }{ 2 }) (AB).
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length (frac { 3 }{ 4 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A1, A2, A3 and A4.
(iii) JoinA4B.
(iv) From A3, draw a line parallel to A4B which meets AB at C.
C is the required point and AC = (frac { 3 }{ 4 }) (AB).
Class 9 RD Sharma Solutions Chapter 16 Circles

RD Sharma Solutions Class 9 Chapter 16 Circles Ex 16.2

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = (frac { 1 }{ 2 }) x 108°
∴ We shall bisect it.
RD Sharma Solutions Class 9 Chapter 16 Circles
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =(frac { 1 }{ 2 }) x 90° = 45°.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ (frac { 1 }{ 2 }) ∠DCA + (frac { 1 }{ 2 }) ∠DCB = 180° x (frac { 1 }{ 2 }) = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
Class 9 RD Sharma Solutions Chapter 16 Circles

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x (frac { 1 }{ 2 }) = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + (frac { 1 }{ 2 }) x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
RD Sharma Class 9 Book Chapter 16 Circles
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
Circles With Solutions PDF RD Sharma Class 9 Solutions

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 (frac { 1 }{ 2 })°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
RD Sharma Class 9 Maths Book Questions Chapter 16 Circles
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
RD Sharma Mathematics Class 9 Solutions Chapter 16 Circles
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
RD Sharma Math Solution Class 9 Chapter 16 Circles
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
RD Sharma Class 9 Questions Chapter 16 Circles
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
Maths RD Sharma Class 9 Chapter 16 Circles
(vi) 22 (frac { 1 }{ 2 })°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 (frac { 1 }{ 2 })°
RD Sharma Class 9 PDF Chapter 16 Circles

RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles Ex 16.3

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
Circles Class 9 RD Sharma Solutions

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
Class 9 RD Sharma Solutions Chapter 16 Circles
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
RD Sharma Class 9 Book Chapter 16 Circles

Solutions of Chapter 16 Circles Ex 16.1 and 16.2 are given below:
Chapter 16 Circles Ex 16.3
Chapter 16 Circles Ex 16.4
Chapter 16 Circles Ex 16.5

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1 1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 1
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 2
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 3
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 4
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 5
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 6
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 7
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 8
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 9
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 10
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 11
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 12
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 13
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 14
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 16
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 17

RD Sharma Class 9 Solutions CirclesClass 9 Maths Sample Papers

All Chapter RD Sharma Solutions For Class 9 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

In this chapter, we provide RD Sharma Solutions for Class 9 Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Maths pdf, free RD Sharma Solutions for Class 9 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Class 9th solutions Chapter 16 Circles

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length (frac { 1 }{ 2 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = (frac { 1 }{ 2 }) (AB).
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length (frac { 3 }{ 4 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A1, A2, A3 and A4.
(iii) JoinA4B.
(iv) From A3, draw a line parallel to A4B which meets AB at C.
C is the required point and AC = (frac { 3 }{ 4 }) (AB).
Class 9 RD Sharma Solutions Chapter 16 Circles

RD Sharma Solutions Class 9 Chapter 16 Circles Ex 16.2

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = (frac { 1 }{ 2 }) x 108°
∴ We shall bisect it.
RD Sharma Solutions Class 9 Chapter 16 Circles
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =(frac { 1 }{ 2 }) x 90° = 45°.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ (frac { 1 }{ 2 }) ∠DCA + (frac { 1 }{ 2 }) ∠DCB = 180° x (frac { 1 }{ 2 }) = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
Class 9 RD Sharma Solutions Chapter 16 Circles

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x (frac { 1 }{ 2 }) = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + (frac { 1 }{ 2 }) x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
RD Sharma Class 9 Book Chapter 16 Circles
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
Circles With Solutions PDF RD Sharma Class 9 Solutions

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 (frac { 1 }{ 2 })°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
RD Sharma Class 9 Maths Book Questions Chapter 16 Circles
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
RD Sharma Mathematics Class 9 Solutions Chapter 16 Circles
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
RD Sharma Math Solution Class 9 Chapter 16 Circles
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
RD Sharma Class 9 Questions Chapter 16 Circles
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
Maths RD Sharma Class 9 Chapter 16 Circles
(vi) 22 (frac { 1 }{ 2 })°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 (frac { 1 }{ 2 })°
RD Sharma Class 9 PDF Chapter 16 Circles

RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles Ex 16.3

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
Circles Class 9 RD Sharma Solutions

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
Class 9 RD Sharma Solutions Chapter 16 Circles
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
RD Sharma Class 9 Book Chapter 16 Circles

Solutions of Chapter 16 Circles Ex 16.1 and 16.2 are given below:
Chapter 16 Circles Ex 16.3
Chapter 16 Circles Ex 16.4
Chapter 16 Circles Ex 16.5

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1 1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 1
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 2
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 3
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 4
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 5
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 6
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 7
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 8
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 9
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 10
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 11
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 12
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 13
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 14
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 16
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 17

RD Sharma Class 9 Solutions CirclesClass 9 Maths Sample Papers

All Chapter RD Sharma Solutions For Class 9 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

In this chapter, we provide RD Sharma Solutions for Class 9 Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Maths pdf, free RD Sharma Solutions for Class 9 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Class 9th solutions Chapter 16 Circles

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length (frac { 1 }{ 2 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = (frac { 1 }{ 2 }) (AB).
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length (frac { 3 }{ 4 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A1, A2, A3 and A4.
(iii) JoinA4B.
(iv) From A3, draw a line parallel to A4B which meets AB at C.
C is the required point and AC = (frac { 3 }{ 4 }) (AB).
Class 9 RD Sharma Solutions Chapter 16 Circles

RD Sharma Solutions Class 9 Chapter 16 Circles Ex 16.2

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = (frac { 1 }{ 2 }) x 108°
∴ We shall bisect it.
RD Sharma Solutions Class 9 Chapter 16 Circles
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =(frac { 1 }{ 2 }) x 90° = 45°.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ (frac { 1 }{ 2 }) ∠DCA + (frac { 1 }{ 2 }) ∠DCB = 180° x (frac { 1 }{ 2 }) = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
Class 9 RD Sharma Solutions Chapter 16 Circles

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x (frac { 1 }{ 2 }) = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + (frac { 1 }{ 2 }) x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
RD Sharma Class 9 Book Chapter 16 Circles
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
Circles With Solutions PDF RD Sharma Class 9 Solutions

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 (frac { 1 }{ 2 })°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
RD Sharma Class 9 Maths Book Questions Chapter 16 Circles
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
RD Sharma Mathematics Class 9 Solutions Chapter 16 Circles
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
RD Sharma Math Solution Class 9 Chapter 16 Circles
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
RD Sharma Class 9 Questions Chapter 16 Circles
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
Maths RD Sharma Class 9 Chapter 16 Circles
(vi) 22 (frac { 1 }{ 2 })°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 (frac { 1 }{ 2 })°
RD Sharma Class 9 PDF Chapter 16 Circles

RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles Ex 16.3

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
Circles Class 9 RD Sharma Solutions

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
Class 9 RD Sharma Solutions Chapter 16 Circles
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
RD Sharma Class 9 Book Chapter 16 Circles

Solutions of Chapter 16 Circles Ex 16.1 and 16.2 are given below:
Chapter 16 Circles Ex 16.3
Chapter 16 Circles Ex 16.4
Chapter 16 Circles Ex 16.5

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1 1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 1
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 2
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 3
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 4
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 5
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 6
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 7
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 8
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 9
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 10
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 11
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 12
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 13
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 14
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 16
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 17

RD Sharma Class 9 Solutions CirclesClass 9 Maths Sample Papers

All Chapter RD Sharma Solutions For Class 9 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

In this chapter, we provide RD Sharma Solutions for Class 9 Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Maths pdf, free RD Sharma Solutions for Class 9 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Class 9th solutions Chapter 16 Circles

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length (frac { 1 }{ 2 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = (frac { 1 }{ 2 }) (AB).
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length (frac { 3 }{ 4 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A1, A2, A3 and A4.
(iii) JoinA4B.
(iv) From A3, draw a line parallel to A4B which meets AB at C.
C is the required point and AC = (frac { 3 }{ 4 }) (AB).
Class 9 RD Sharma Solutions Chapter 16 Circles

RD Sharma Solutions Class 9 Chapter 16 Circles Ex 16.2

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = (frac { 1 }{ 2 }) x 108°
∴ We shall bisect it.
RD Sharma Solutions Class 9 Chapter 16 Circles
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =(frac { 1 }{ 2 }) x 90° = 45°.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ (frac { 1 }{ 2 }) ∠DCA + (frac { 1 }{ 2 }) ∠DCB = 180° x (frac { 1 }{ 2 }) = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
Class 9 RD Sharma Solutions Chapter 16 Circles

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x (frac { 1 }{ 2 }) = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + (frac { 1 }{ 2 }) x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
RD Sharma Class 9 Book Chapter 16 Circles
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
Circles With Solutions PDF RD Sharma Class 9 Solutions

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 (frac { 1 }{ 2 })°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
RD Sharma Class 9 Maths Book Questions Chapter 16 Circles
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
RD Sharma Mathematics Class 9 Solutions Chapter 16 Circles
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
RD Sharma Math Solution Class 9 Chapter 16 Circles
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
RD Sharma Class 9 Questions Chapter 16 Circles
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
Maths RD Sharma Class 9 Chapter 16 Circles
(vi) 22 (frac { 1 }{ 2 })°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 (frac { 1 }{ 2 })°
RD Sharma Class 9 PDF Chapter 16 Circles

RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles Ex 16.3

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
Circles Class 9 RD Sharma Solutions

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
Class 9 RD Sharma Solutions Chapter 16 Circles
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
RD Sharma Class 9 Book Chapter 16 Circles

Solutions of Chapter 16 Circles Ex 16.1 and 16.2 are given below:
Chapter 16 Circles Ex 16.3
Chapter 16 Circles Ex 16.4
Chapter 16 Circles Ex 16.5

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1 1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 1
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 2
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 3
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 4
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 5
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 6
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 7
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 8
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 9
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 10
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 11
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 12
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 13
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 14
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 16
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 17

RD Sharma Class 9 Solutions CirclesClass 9 Maths Sample Papers

All Chapter RD Sharma Solutions For Class 9 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

In this chapter, we provide RD Sharma Solutions for Class 9 Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Maths pdf, free RD Sharma Solutions for Class 9 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Class 9th solutions Chapter 16 Circles

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length (frac { 1 }{ 2 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = (frac { 1 }{ 2 }) (AB).
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length (frac { 3 }{ 4 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A1, A2, A3 and A4.
(iii) JoinA4B.
(iv) From A3, draw a line parallel to A4B which meets AB at C.
C is the required point and AC = (frac { 3 }{ 4 }) (AB).
Class 9 RD Sharma Solutions Chapter 16 Circles

RD Sharma Solutions Class 9 Chapter 16 Circles Ex 16.2

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = (frac { 1 }{ 2 }) x 108°
∴ We shall bisect it.
RD Sharma Solutions Class 9 Chapter 16 Circles
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =(frac { 1 }{ 2 }) x 90° = 45°.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ (frac { 1 }{ 2 }) ∠DCA + (frac { 1 }{ 2 }) ∠DCB = 180° x (frac { 1 }{ 2 }) = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
Class 9 RD Sharma Solutions Chapter 16 Circles

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x (frac { 1 }{ 2 }) = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + (frac { 1 }{ 2 }) x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
RD Sharma Class 9 Book Chapter 16 Circles
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
Circles With Solutions PDF RD Sharma Class 9 Solutions

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 (frac { 1 }{ 2 })°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
RD Sharma Class 9 Maths Book Questions Chapter 16 Circles
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
RD Sharma Mathematics Class 9 Solutions Chapter 16 Circles
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
RD Sharma Math Solution Class 9 Chapter 16 Circles
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
RD Sharma Class 9 Questions Chapter 16 Circles
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
Maths RD Sharma Class 9 Chapter 16 Circles
(vi) 22 (frac { 1 }{ 2 })°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 (frac { 1 }{ 2 })°
RD Sharma Class 9 PDF Chapter 16 Circles

RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles Ex 16.3

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
Circles Class 9 RD Sharma Solutions

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
Class 9 RD Sharma Solutions Chapter 16 Circles
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
RD Sharma Class 9 Book Chapter 16 Circles

Solutions of Chapter 16 Circles Ex 16.1 and 16.2 are given below:
Chapter 16 Circles Ex 16.3
Chapter 16 Circles Ex 16.4
Chapter 16 Circles Ex 16.5

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1 1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 1
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 2
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 3
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 4
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 5
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 6
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 7
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 8
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 9
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 10
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 11
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 12
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 13
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 14
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 16
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 17

RD Sharma Class 9 Solutions CirclesClass 9 Maths Sample Papers

All Chapter RD Sharma Solutions For Class 9 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

In this chapter, we provide RD Sharma Solutions for Class 9 Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Maths pdf, free RD Sharma Solutions for Class 9 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Class 9th solutions Chapter 16 Circles

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length (frac { 1 }{ 2 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = (frac { 1 }{ 2 }) (AB).
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length (frac { 3 }{ 4 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A1, A2, A3 and A4.
(iii) JoinA4B.
(iv) From A3, draw a line parallel to A4B which meets AB at C.
C is the required point and AC = (frac { 3 }{ 4 }) (AB).
Class 9 RD Sharma Solutions Chapter 16 Circles

RD Sharma Solutions Class 9 Chapter 16 Circles Ex 16.2

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = (frac { 1 }{ 2 }) x 108°
∴ We shall bisect it.
RD Sharma Solutions Class 9 Chapter 16 Circles
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =(frac { 1 }{ 2 }) x 90° = 45°.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ (frac { 1 }{ 2 }) ∠DCA + (frac { 1 }{ 2 }) ∠DCB = 180° x (frac { 1 }{ 2 }) = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
Class 9 RD Sharma Solutions Chapter 16 Circles

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x (frac { 1 }{ 2 }) = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + (frac { 1 }{ 2 }) x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
RD Sharma Class 9 Book Chapter 16 Circles
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
Circles With Solutions PDF RD Sharma Class 9 Solutions

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 (frac { 1 }{ 2 })°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
RD Sharma Class 9 Maths Book Questions Chapter 16 Circles
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
RD Sharma Mathematics Class 9 Solutions Chapter 16 Circles
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
RD Sharma Math Solution Class 9 Chapter 16 Circles
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
RD Sharma Class 9 Questions Chapter 16 Circles
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
Maths RD Sharma Class 9 Chapter 16 Circles
(vi) 22 (frac { 1 }{ 2 })°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 (frac { 1 }{ 2 })°
RD Sharma Class 9 PDF Chapter 16 Circles

RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles Ex 16.3

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
Circles Class 9 RD Sharma Solutions

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
Class 9 RD Sharma Solutions Chapter 16 Circles
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
RD Sharma Class 9 Book Chapter 16 Circles

Solutions of Chapter 16 Circles Ex 16.1 and 16.2 are given below:
Chapter 16 Circles Ex 16.3
Chapter 16 Circles Ex 16.4
Chapter 16 Circles Ex 16.5

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1 1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 1
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 2
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 3
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 4
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 5
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 6
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 7
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 8
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 9
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 10
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 11
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 12
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 13
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 14
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 16
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 17

RD Sharma Class 9 Solutions CirclesClass 9 Maths Sample Papers

All Chapter RD Sharma Solutions For Class 9 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

In this chapter, we provide RD Sharma Solutions for Class 9 Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Maths pdf, free RD Sharma Solutions for Class 9 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Class 9th solutions Chapter 16 Circles

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length (frac { 1 }{ 2 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = (frac { 1 }{ 2 }) (AB).
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length (frac { 3 }{ 4 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A1, A2, A3 and A4.
(iii) JoinA4B.
(iv) From A3, draw a line parallel to A4B which meets AB at C.
C is the required point and AC = (frac { 3 }{ 4 }) (AB).
Class 9 RD Sharma Solutions Chapter 16 Circles

RD Sharma Solutions Class 9 Chapter 16 Circles Ex 16.2

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = (frac { 1 }{ 2 }) x 108°
∴ We shall bisect it.
RD Sharma Solutions Class 9 Chapter 16 Circles
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =(frac { 1 }{ 2 }) x 90° = 45°.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ (frac { 1 }{ 2 }) ∠DCA + (frac { 1 }{ 2 }) ∠DCB = 180° x (frac { 1 }{ 2 }) = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
Class 9 RD Sharma Solutions Chapter 16 Circles

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x (frac { 1 }{ 2 }) = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + (frac { 1 }{ 2 }) x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
RD Sharma Class 9 Book Chapter 16 Circles
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
Circles With Solutions PDF RD Sharma Class 9 Solutions

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 (frac { 1 }{ 2 })°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
RD Sharma Class 9 Maths Book Questions Chapter 16 Circles
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
RD Sharma Mathematics Class 9 Solutions Chapter 16 Circles
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
RD Sharma Math Solution Class 9 Chapter 16 Circles
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
RD Sharma Class 9 Questions Chapter 16 Circles
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
Maths RD Sharma Class 9 Chapter 16 Circles
(vi) 22 (frac { 1 }{ 2 })°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 (frac { 1 }{ 2 })°
RD Sharma Class 9 PDF Chapter 16 Circles

RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles Ex 16.3

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
Circles Class 9 RD Sharma Solutions

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
Class 9 RD Sharma Solutions Chapter 16 Circles
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
RD Sharma Class 9 Book Chapter 16 Circles

Solutions of Chapter 16 Circles Ex 16.1 and 16.2 are given below:
Chapter 16 Circles Ex 16.3
Chapter 16 Circles Ex 16.4
Chapter 16 Circles Ex 16.5

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1 1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 1
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 2
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 3
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 4
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 5
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 6
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 7
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 8
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 9
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 10
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 11
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 12
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 13
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 14
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 16
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 17

RD Sharma Class 9 Solutions CirclesClass 9 Maths Sample Papers

All Chapter RD Sharma Solutions For Class 9 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

In this chapter, we provide RD Sharma Solutions for Class 9 Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Maths pdf, free RD Sharma Solutions for Class 9 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Class 9th solutions Chapter 16 Circles

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length (frac { 1 }{ 2 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = (frac { 1 }{ 2 }) (AB).
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length (frac { 3 }{ 4 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A1, A2, A3 and A4.
(iii) JoinA4B.
(iv) From A3, draw a line parallel to A4B which meets AB at C.
C is the required point and AC = (frac { 3 }{ 4 }) (AB).
Class 9 RD Sharma Solutions Chapter 16 Circles

RD Sharma Solutions Class 9 Chapter 16 Circles Ex 16.2

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = (frac { 1 }{ 2 }) x 108°
∴ We shall bisect it.
RD Sharma Solutions Class 9 Chapter 16 Circles
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =(frac { 1 }{ 2 }) x 90° = 45°.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ (frac { 1 }{ 2 }) ∠DCA + (frac { 1 }{ 2 }) ∠DCB = 180° x (frac { 1 }{ 2 }) = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
Class 9 RD Sharma Solutions Chapter 16 Circles

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x (frac { 1 }{ 2 }) = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + (frac { 1 }{ 2 }) x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
RD Sharma Class 9 Book Chapter 16 Circles
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
Circles With Solutions PDF RD Sharma Class 9 Solutions

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 (frac { 1 }{ 2 })°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
RD Sharma Class 9 Maths Book Questions Chapter 16 Circles
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
RD Sharma Mathematics Class 9 Solutions Chapter 16 Circles
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
RD Sharma Math Solution Class 9 Chapter 16 Circles
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
RD Sharma Class 9 Questions Chapter 16 Circles
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
Maths RD Sharma Class 9 Chapter 16 Circles
(vi) 22 (frac { 1 }{ 2 })°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 (frac { 1 }{ 2 })°
RD Sharma Class 9 PDF Chapter 16 Circles

RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles Ex 16.3

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
Circles Class 9 RD Sharma Solutions

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
Class 9 RD Sharma Solutions Chapter 16 Circles
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
RD Sharma Class 9 Book Chapter 16 Circles

Solutions of Chapter 16 Circles Ex 16.1 and 16.2 are given below:
Chapter 16 Circles Ex 16.3
Chapter 16 Circles Ex 16.4
Chapter 16 Circles Ex 16.5

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1 1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 1
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 2
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 3
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 4
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 5
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 6
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 7
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 8
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 9
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 10
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 11
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 12
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 13
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 14
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 16
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 17

RD Sharma Class 9 Solutions CirclesClass 9 Maths Sample Papers

All Chapter RD Sharma Solutions For Class 9 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

In this chapter, we provide RD Sharma Solutions for Class 9 Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Maths pdf, free RD Sharma Solutions for Class 9 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Class 9th solutions Chapter 16 Circles

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length (frac { 1 }{ 2 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = (frac { 1 }{ 2 }) (AB).
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length (frac { 3 }{ 4 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A1, A2, A3 and A4.
(iii) JoinA4B.
(iv) From A3, draw a line parallel to A4B which meets AB at C.
C is the required point and AC = (frac { 3 }{ 4 }) (AB).
Class 9 RD Sharma Solutions Chapter 16 Circles

RD Sharma Solutions Class 9 Chapter 16 Circles Ex 16.2

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = (frac { 1 }{ 2 }) x 108°
∴ We shall bisect it.
RD Sharma Solutions Class 9 Chapter 16 Circles
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =(frac { 1 }{ 2 }) x 90° = 45°.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ (frac { 1 }{ 2 }) ∠DCA + (frac { 1 }{ 2 }) ∠DCB = 180° x (frac { 1 }{ 2 }) = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
Class 9 RD Sharma Solutions Chapter 16 Circles

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x (frac { 1 }{ 2 }) = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + (frac { 1 }{ 2 }) x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
RD Sharma Class 9 Book Chapter 16 Circles
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
Circles With Solutions PDF RD Sharma Class 9 Solutions

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 (frac { 1 }{ 2 })°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
RD Sharma Class 9 Maths Book Questions Chapter 16 Circles
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
RD Sharma Mathematics Class 9 Solutions Chapter 16 Circles
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
RD Sharma Math Solution Class 9 Chapter 16 Circles
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
RD Sharma Class 9 Questions Chapter 16 Circles
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
Maths RD Sharma Class 9 Chapter 16 Circles
(vi) 22 (frac { 1 }{ 2 })°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 (frac { 1 }{ 2 })°
RD Sharma Class 9 PDF Chapter 16 Circles

RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles Ex 16.3

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
Circles Class 9 RD Sharma Solutions

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
Class 9 RD Sharma Solutions Chapter 16 Circles
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
RD Sharma Class 9 Book Chapter 16 Circles

Solutions of Chapter 16 Circles Ex 16.1 and 16.2 are given below:
Chapter 16 Circles Ex 16.3
Chapter 16 Circles Ex 16.4
Chapter 16 Circles Ex 16.5

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1 1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 1
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 2
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 3
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 4
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 5
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 6
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 7
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 8
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 9
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 10
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 11
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 12
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 13
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 14
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 16
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 17

RD Sharma Class 9 Solutions CirclesClass 9 Maths Sample Papers

All Chapter RD Sharma Solutions For Class 9 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

In this chapter, we provide RD Sharma Solutions for Class 9 Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Maths pdf, free RD Sharma Solutions for Class 9 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Class 9th solutions Chapter 16 Circles

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length (frac { 1 }{ 2 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = (frac { 1 }{ 2 }) (AB).
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length (frac { 3 }{ 4 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A1, A2, A3 and A4.
(iii) JoinA4B.
(iv) From A3, draw a line parallel to A4B which meets AB at C.
C is the required point and AC = (frac { 3 }{ 4 }) (AB).
Class 9 RD Sharma Solutions Chapter 16 Circles

RD Sharma Solutions Class 9 Chapter 16 Circles Ex 16.2

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = (frac { 1 }{ 2 }) x 108°
∴ We shall bisect it.
RD Sharma Solutions Class 9 Chapter 16 Circles
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =(frac { 1 }{ 2 }) x 90° = 45°.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ (frac { 1 }{ 2 }) ∠DCA + (frac { 1 }{ 2 }) ∠DCB = 180° x (frac { 1 }{ 2 }) = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
Class 9 RD Sharma Solutions Chapter 16 Circles

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x (frac { 1 }{ 2 }) = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + (frac { 1 }{ 2 }) x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
RD Sharma Class 9 Book Chapter 16 Circles
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
Circles With Solutions PDF RD Sharma Class 9 Solutions

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 (frac { 1 }{ 2 })°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
RD Sharma Class 9 Maths Book Questions Chapter 16 Circles
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
RD Sharma Mathematics Class 9 Solutions Chapter 16 Circles
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
RD Sharma Math Solution Class 9 Chapter 16 Circles
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
RD Sharma Class 9 Questions Chapter 16 Circles
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
Maths RD Sharma Class 9 Chapter 16 Circles
(vi) 22 (frac { 1 }{ 2 })°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 (frac { 1 }{ 2 })°
RD Sharma Class 9 PDF Chapter 16 Circles

RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles Ex 16.3

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
Circles Class 9 RD Sharma Solutions

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
Class 9 RD Sharma Solutions Chapter 16 Circles
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
RD Sharma Class 9 Book Chapter 16 Circles

Solutions of Chapter 16 Circles Ex 16.1 and 16.2 are given below:
Chapter 16 Circles Ex 16.3
Chapter 16 Circles Ex 16.4
Chapter 16 Circles Ex 16.5

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1 1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 1
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 2
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 3
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 4
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 5
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 6
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 7
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 8
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 9
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 10
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 11
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 12
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 13
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 14
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 16
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 17

RD Sharma Class 9 Solutions CirclesClass 9 Maths Sample Papers

All Chapter RD Sharma Solutions For Class 9 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

In this chapter, we provide RD Sharma Solutions for Class 9 Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Maths pdf, free RD Sharma Solutions for Class 9 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Class 9th solutions Chapter 16 Circles

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length (frac { 1 }{ 2 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = (frac { 1 }{ 2 }) (AB).
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length (frac { 3 }{ 4 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A1, A2, A3 and A4.
(iii) JoinA4B.
(iv) From A3, draw a line parallel to A4B which meets AB at C.
C is the required point and AC = (frac { 3 }{ 4 }) (AB).
Class 9 RD Sharma Solutions Chapter 16 Circles

RD Sharma Solutions Class 9 Chapter 16 Circles Ex 16.2

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = (frac { 1 }{ 2 }) x 108°
∴ We shall bisect it.
RD Sharma Solutions Class 9 Chapter 16 Circles
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =(frac { 1 }{ 2 }) x 90° = 45°.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ (frac { 1 }{ 2 }) ∠DCA + (frac { 1 }{ 2 }) ∠DCB = 180° x (frac { 1 }{ 2 }) = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
Class 9 RD Sharma Solutions Chapter 16 Circles

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x (frac { 1 }{ 2 }) = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + (frac { 1 }{ 2 }) x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
RD Sharma Class 9 Book Chapter 16 Circles
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
Circles With Solutions PDF RD Sharma Class 9 Solutions

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 (frac { 1 }{ 2 })°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
RD Sharma Class 9 Maths Book Questions Chapter 16 Circles
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
RD Sharma Mathematics Class 9 Solutions Chapter 16 Circles
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
RD Sharma Math Solution Class 9 Chapter 16 Circles
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
RD Sharma Class 9 Questions Chapter 16 Circles
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
Maths RD Sharma Class 9 Chapter 16 Circles
(vi) 22 (frac { 1 }{ 2 })°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 (frac { 1 }{ 2 })°
RD Sharma Class 9 PDF Chapter 16 Circles

RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles Ex 16.3

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
Circles Class 9 RD Sharma Solutions

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
Class 9 RD Sharma Solutions Chapter 16 Circles
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
RD Sharma Class 9 Book Chapter 16 Circles

Solutions of Chapter 16 Circles Ex 16.1 and 16.2 are given below:
Chapter 16 Circles Ex 16.3
Chapter 16 Circles Ex 16.4
Chapter 16 Circles Ex 16.5

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1 1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 1
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 2
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 3
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 4
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 5
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 6
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 7
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 8
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 9
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 10
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 11
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 12
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 13
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 14
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 16
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 17

RD Sharma Class 9 Solutions CirclesClass 9 Maths Sample Papers

All Chapter RD Sharma Solutions For Class 9 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

In this chapter, we provide RD Sharma Solutions for Class 9 Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Maths pdf, free RD Sharma Solutions for Class 9 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Class 9th solutions Chapter 16 Circles

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length (frac { 1 }{ 2 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = (frac { 1 }{ 2 }) (AB).
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length (frac { 3 }{ 4 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A1, A2, A3 and A4.
(iii) JoinA4B.
(iv) From A3, draw a line parallel to A4B which meets AB at C.
C is the required point and AC = (frac { 3 }{ 4 }) (AB).
Class 9 RD Sharma Solutions Chapter 16 Circles

RD Sharma Solutions Class 9 Chapter 16 Circles Ex 16.2

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = (frac { 1 }{ 2 }) x 108°
∴ We shall bisect it.
RD Sharma Solutions Class 9 Chapter 16 Circles
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =(frac { 1 }{ 2 }) x 90° = 45°.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ (frac { 1 }{ 2 }) ∠DCA + (frac { 1 }{ 2 }) ∠DCB = 180° x (frac { 1 }{ 2 }) = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
Class 9 RD Sharma Solutions Chapter 16 Circles

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x (frac { 1 }{ 2 }) = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + (frac { 1 }{ 2 }) x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
RD Sharma Class 9 Book Chapter 16 Circles
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
Circles With Solutions PDF RD Sharma Class 9 Solutions

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 (frac { 1 }{ 2 })°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
RD Sharma Class 9 Maths Book Questions Chapter 16 Circles
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
RD Sharma Mathematics Class 9 Solutions Chapter 16 Circles
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
RD Sharma Math Solution Class 9 Chapter 16 Circles
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
RD Sharma Class 9 Questions Chapter 16 Circles
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
Maths RD Sharma Class 9 Chapter 16 Circles
(vi) 22 (frac { 1 }{ 2 })°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 (frac { 1 }{ 2 })°
RD Sharma Class 9 PDF Chapter 16 Circles

RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles Ex 16.3

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
Circles Class 9 RD Sharma Solutions

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
Class 9 RD Sharma Solutions Chapter 16 Circles
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
RD Sharma Class 9 Book Chapter 16 Circles

Solutions of Chapter 16 Circles Ex 16.1 and 16.2 are given below:
Chapter 16 Circles Ex 16.3
Chapter 16 Circles Ex 16.4
Chapter 16 Circles Ex 16.5

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1 1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 1
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 2
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 3
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 4
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 5
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 6
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 7
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 8
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 9
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 10
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 11
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 12
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 13
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 14
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 16
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 17

RD Sharma Class 9 Solutions CirclesClass 9 Maths Sample Papers

All Chapter RD Sharma Solutions For Class 9 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

In this chapter, we provide RD Sharma Solutions for Class 9 Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Maths pdf, free RD Sharma Solutions for Class 9 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Class 9th solutions Chapter 16 Circles

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length (frac { 1 }{ 2 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = (frac { 1 }{ 2 }) (AB).
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length (frac { 3 }{ 4 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A1, A2, A3 and A4.
(iii) JoinA4B.
(iv) From A3, draw a line parallel to A4B which meets AB at C.
C is the required point and AC = (frac { 3 }{ 4 }) (AB).
Class 9 RD Sharma Solutions Chapter 16 Circles

RD Sharma Solutions Class 9 Chapter 16 Circles Ex 16.2

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = (frac { 1 }{ 2 }) x 108°
∴ We shall bisect it.
RD Sharma Solutions Class 9 Chapter 16 Circles
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =(frac { 1 }{ 2 }) x 90° = 45°.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ (frac { 1 }{ 2 }) ∠DCA + (frac { 1 }{ 2 }) ∠DCB = 180° x (frac { 1 }{ 2 }) = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
Class 9 RD Sharma Solutions Chapter 16 Circles

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x (frac { 1 }{ 2 }) = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + (frac { 1 }{ 2 }) x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
RD Sharma Class 9 Book Chapter 16 Circles
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
Circles With Solutions PDF RD Sharma Class 9 Solutions

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 (frac { 1 }{ 2 })°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
RD Sharma Class 9 Maths Book Questions Chapter 16 Circles
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
RD Sharma Mathematics Class 9 Solutions Chapter 16 Circles
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
RD Sharma Math Solution Class 9 Chapter 16 Circles
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
RD Sharma Class 9 Questions Chapter 16 Circles
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
Maths RD Sharma Class 9 Chapter 16 Circles
(vi) 22 (frac { 1 }{ 2 })°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 (frac { 1 }{ 2 })°
RD Sharma Class 9 PDF Chapter 16 Circles

RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles Ex 16.3

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
Circles Class 9 RD Sharma Solutions

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
Class 9 RD Sharma Solutions Chapter 16 Circles
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
RD Sharma Class 9 Book Chapter 16 Circles

Solutions of Chapter 16 Circles Ex 16.1 and 16.2 are given below:
Chapter 16 Circles Ex 16.3
Chapter 16 Circles Ex 16.4
Chapter 16 Circles Ex 16.5

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1 1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 1
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 2
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 3
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 4
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 5
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 6
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 7
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 8
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 9
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 10
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 11
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 12
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 13
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 14
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 16
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 17

RD Sharma Class 9 Solutions CirclesClass 9 Maths Sample Papers

All Chapter RD Sharma Solutions For Class 9 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

In this chapter, we provide RD Sharma Solutions for Class 9 Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Maths pdf, free RD Sharma Solutions for Class 9 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Class 9th solutions Chapter 16 Circles

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length (frac { 1 }{ 2 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = (frac { 1 }{ 2 }) (AB).
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length (frac { 3 }{ 4 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A1, A2, A3 and A4.
(iii) JoinA4B.
(iv) From A3, draw a line parallel to A4B which meets AB at C.
C is the required point and AC = (frac { 3 }{ 4 }) (AB).
Class 9 RD Sharma Solutions Chapter 16 Circles

RD Sharma Solutions Class 9 Chapter 16 Circles Ex 16.2

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = (frac { 1 }{ 2 }) x 108°
∴ We shall bisect it.
RD Sharma Solutions Class 9 Chapter 16 Circles
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =(frac { 1 }{ 2 }) x 90° = 45°.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ (frac { 1 }{ 2 }) ∠DCA + (frac { 1 }{ 2 }) ∠DCB = 180° x (frac { 1 }{ 2 }) = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
Class 9 RD Sharma Solutions Chapter 16 Circles

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x (frac { 1 }{ 2 }) = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + (frac { 1 }{ 2 }) x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
RD Sharma Class 9 Book Chapter 16 Circles
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
Circles With Solutions PDF RD Sharma Class 9 Solutions

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 (frac { 1 }{ 2 })°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
RD Sharma Class 9 Maths Book Questions Chapter 16 Circles
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
RD Sharma Mathematics Class 9 Solutions Chapter 16 Circles
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
RD Sharma Math Solution Class 9 Chapter 16 Circles
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
RD Sharma Class 9 Questions Chapter 16 Circles
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
Maths RD Sharma Class 9 Chapter 16 Circles
(vi) 22 (frac { 1 }{ 2 })°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 (frac { 1 }{ 2 })°
RD Sharma Class 9 PDF Chapter 16 Circles

RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles Ex 16.3

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
Circles Class 9 RD Sharma Solutions

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
Class 9 RD Sharma Solutions Chapter 16 Circles
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
RD Sharma Class 9 Book Chapter 16 Circles

Solutions of Chapter 16 Circles Ex 16.1 and 16.2 are given below:
Chapter 16 Circles Ex 16.3
Chapter 16 Circles Ex 16.4
Chapter 16 Circles Ex 16.5

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1 1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 1
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 2
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 3
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 4
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 5
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 6
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 7
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 8
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 9
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 10
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 11
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 12
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 13
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 14
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 16
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 17

RD Sharma Class 9 Solutions CirclesClass 9 Maths Sample Papers

All Chapter RD Sharma Solutions For Class 9 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

In this chapter, we provide RD Sharma Solutions for Class 9 Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Maths pdf, free RD Sharma Solutions for Class 9 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Class 9th solutions Chapter 16 Circles

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length (frac { 1 }{ 2 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = (frac { 1 }{ 2 }) (AB).
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length (frac { 3 }{ 4 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A1, A2, A3 and A4.
(iii) JoinA4B.
(iv) From A3, draw a line parallel to A4B which meets AB at C.
C is the required point and AC = (frac { 3 }{ 4 }) (AB).
Class 9 RD Sharma Solutions Chapter 16 Circles

RD Sharma Solutions Class 9 Chapter 16 Circles Ex 16.2

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = (frac { 1 }{ 2 }) x 108°
∴ We shall bisect it.
RD Sharma Solutions Class 9 Chapter 16 Circles
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =(frac { 1 }{ 2 }) x 90° = 45°.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ (frac { 1 }{ 2 }) ∠DCA + (frac { 1 }{ 2 }) ∠DCB = 180° x (frac { 1 }{ 2 }) = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
Class 9 RD Sharma Solutions Chapter 16 Circles

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x (frac { 1 }{ 2 }) = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + (frac { 1 }{ 2 }) x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
RD Sharma Class 9 Book Chapter 16 Circles
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
Circles With Solutions PDF RD Sharma Class 9 Solutions

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 (frac { 1 }{ 2 })°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
RD Sharma Class 9 Maths Book Questions Chapter 16 Circles
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
RD Sharma Mathematics Class 9 Solutions Chapter 16 Circles
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
RD Sharma Math Solution Class 9 Chapter 16 Circles
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
RD Sharma Class 9 Questions Chapter 16 Circles
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
Maths RD Sharma Class 9 Chapter 16 Circles
(vi) 22 (frac { 1 }{ 2 })°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 (frac { 1 }{ 2 })°
RD Sharma Class 9 PDF Chapter 16 Circles

RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles Ex 16.3

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
Circles Class 9 RD Sharma Solutions

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
Class 9 RD Sharma Solutions Chapter 16 Circles
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
RD Sharma Class 9 Book Chapter 16 Circles

Solutions of Chapter 16 Circles Ex 16.1 and 16.2 are given below:
Chapter 16 Circles Ex 16.3
Chapter 16 Circles Ex 16.4
Chapter 16 Circles Ex 16.5

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1 1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 1
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 2
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 3
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 4
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 5
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 6
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 7
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 8
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 9
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 10
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 11
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 12
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 13
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 14
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 16
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 17

RD Sharma Class 9 Solutions CirclesClass 9 Maths Sample Papers

All Chapter RD Sharma Solutions For Class 9 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share alarity.in to your friends.

In this chapter, we provide RD Sharma Solutions for Class 9 Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Maths pdf, free RD Sharma Solutions for Class 9 Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Class 9th solutions Chapter 16 Circles

RD Sharma Class 9 Chapter 16 Circles Ex 16.1

Question 1.
Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.
Solution:
Steps of construction :
(i) Draw a line segment AB = 8.6 cm.
(ii) With centres A and B, and radius more than half of AB, draw arcs which intersect each other at E and F, on both sides of AB.
(iii) Join EF which intersects AB at D.
D is the required mid point of AB.
Measuring each part, it is 4.3 cm long.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment.
Solution:
Steps of construction :
(i) Draw a line segment AB = 5.8 cm.
(ii) With centres A and B and radius more than half of AB, draw arcs intersecting each other at E and F.
(iii) Join EF which is the required perpendicular bisector of AB.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle?
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 5 cm.
(ii) Draw a chord AB of the circle.
(iii) With centres A and B and radius more than half of AB draw arcs intersecting each other at P and Q.
(iv) Join PQ which intersects AB at D and passes through O, the centre of the circle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Draw a circle with centre at point O. Draw its two chords AB and CD such that AB is not parallel to CD. Draw the perpendicular bisectors of AB and CD. At what point do they intersect?
Solution:
Steps of construction :
(i) Draw a circle with centre O and a suitable radius.
(ii) Draw two chords AB and CD which are not parallel to each other.
(iii) Draw the perpendicular bisectors of AB and CD with the help of ruler and compasses. We see that these two chords intersect each other at the centre O, of the circle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a line segment of length 10 cm and bisect it. Further bisect one of the equal parts and measure its length.
Solution:
Steps of construction :
(i) Draw a line segment AB = 10 cm.
(ii) With the help of ruler and compasses, bisect AB at C.
(iii) Bisect again AC at D.
Measuring AD, it is 2.5 cm long.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a line segment AB and bisect it. Bisect one of the equal parts to obtain a line segment of length (frac { 1 }{ 2 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) With the help of ruler and compasses, draw the bisector of AB which bisects AB at C.
(iii) Similarly, in the same way draw the bisector of AC at D.
We see that AC = (frac { 1 }{ 2 }) (AB).
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Draw a line segment AB and by ruler and compasses, obtain a line segment of length (frac { 3 }{ 4 }) (AB).
Solution:
Steps of construction :
(i) Draw a line segment AB.
(ii) Draw a ray AX making an acute angle with A3 and equal 4 parts at A1, A2, A3 and A4.
(iii) JoinA4B.
(iv) From A3, draw a line parallel to A4B which meets AB at C.
C is the required point and AC = (frac { 3 }{ 4 }) (AB).
Class 9 RD Sharma Solutions Chapter 16 Circles

RD Sharma Solutions Class 9 Chapter 16 Circles Ex 16.2

Question 1.
Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC.
Solution:
Steps of construction :
(i) Draw an angle BAC.
(ii) Draw a line DF.
(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.
(iv) Cut off PQ = LM and join DQ and produce it to E, then
∠EDF = ∠BAC.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.
Solution:
Steps of construction :
(i) Draw an angle ABC which is an obtuse i.e. more than 90°.
(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
ThenBD is die bisector of ∠ABC.
On measuring each part, we find each angle = 53°.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Using your protractor, draw an angle of measure 108°. With this angle as given, draw an angle of 54°.
Solution:
Steps of construction :
(i) With the help of protractor draw an angle ABC = 108°.
As 54° = (frac { 1 }{ 2 }) x 108°
∴ We shall bisect it.
RD Sharma Solutions Class 9 Chapter 16 Circles
(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.
(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G
(iv) Join BG and produce it to D.
Then ∠DBC = 54°.

Question 4.
Using protractor, draw a right angle. Bisect it to get an angle of measure 45°.
Solution:
Steps of construction :
(i) Using protractor, draw a right angle ∆ABC.
i. e. ∠ABC = 90°.
(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.
(iii) With centre E and F, draw arcs intersecting each other at G.
(iv) Join BG and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC =(frac { 1 }{ 2 }) x 90° = 45°.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.
Solution:
Steps of construction :
(i) Draw a linear pair ∠DCA and ∠DCB.
(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.
Verification : ∵∠DCA + ∠DCB = 180°
⇒ (frac { 1 }{ 2 }) ∠DCA + (frac { 1 }{ 2 }) ∠DCB = 180° x (frac { 1 }{ 2 }) = 90°
∴ ∠ECF = 90°
i.e. EC and FC are perpendicular to each other.
Circles Class 9 RD Sharma Solutions

Question 6.
Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line
Solution:
Steps of construction :
(i) Draw two lines AB and CD intersecting each other at O.
(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Using ruler and compasses only, draw a right angle.
Solution:
Steps of construction :
(i) Draw a line segment BC.
(ii) With centre B and a suitable radius, draw an arc meeting BC at E.
(iii) With centre E and same radius, cut off arcs EF and FG.
(iv) Bisect arc FG at H.
(v) Join BH and produce it to A.
Then ∠ABC = 90°.
Class 9 RD Sharma Solutions Chapter 16 Circles

Question 8.
Using ruler and compasses only, draw an angle of measure 135°.
Solution:
Steps of construction :
(i) Draw a line DC and take a point B on it.
(ii) With centre B and a suitable radius draw an arc meeting BC at P.
(iii) With centre P, cut off arcs PQ, QR and RS.
(iv) Bisect as QR at T and join BT and produce it to E.
(v) Now bisect the arc KS at RL.
(vi) Join BL and produce it to A.
Now ∠ABC = 135°.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using a protractor, draw an angle of measure 72°. With this angle as given, draw angles of measure 36° and 54°.
Solution:
Steps of construction :
(i) Draw an angle ABC = 12° with the help of protractor.
(ii) With centre B and a suitable radius, draw an arc EF.
(iii) With centre E and F, draw arcs intersecting
each other at G and produce it to D.
Then BD is the bisector of ∠ABC.
∴ ∠DBC = 72° x (frac { 1 }{ 2 }) = 36°.
(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.
∴ ∠PBC = 36° + (frac { 1 }{ 2 }) x 36°
= 36° + 18° = 54°
Hence ∠PBC = 54°
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct the following angles at the initial point of a given ray and justify the construction:
(i) 45°
(ii) 90°
Solution:
(i) 45°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) With centre E, cut off equal arcs EF and FG.
(d) Bisect FG at H.
(e) Join BH and produce to X so that ∠XBC = 90°.
(f) Bisect ∠XBC so that ∠ABC = 45°.
RD Sharma Class 9 Book Chapter 16 Circles
(ii) 90°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG
(c) Now bisect the arc EG at H.
(d) Join BH and produce it to A.
∴ ∠ABC = 90°.
Circles With Solutions PDF RD Sharma Class 9 Solutions

Question 11.
Construct the angles of the following measurements:
(i) 30°
(ii) 75°
(iii) 105°
(iv) 135°
(v) 15°
(vi) 22 (frac { 1 }{ 2 })°
Solution:
(i) 30°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) Cut off arcs EF and bisect it at G.
So that ∠ABC = 30°.
RD Sharma Class 9 Maths Book Questions Chapter 16 Circles
(ii) 75°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF = FG from E.
(d) Bisect the arc FG at K and join BK so that ∠KBC = 90°.
(e) Now bisect arc HF at L and join BL and produce it to A so that ∠ABC = 75°.
RD Sharma Mathematics Class 9 Solutions Chapter 16 Circles
(iii) 105°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius draw an arc meeting BC at E.
(c) From E, cut off arc EF = FG and divide FG at H.
(d) Join BH meeting the arc at K.
(e) Now bisect the arc KG at L.
Join BL and produce it to A.
Then ∠ABC = 105°.
RD Sharma Math Solution Class 9 Chapter 16 Circles
(iv) 135°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) From E, cut off EF = FG = GH.
(d) Bisect arc FG at K, and join them.
(e) Bisect arc KH at L.
(f) Join BL and produce it to A, then ∠ABC = 135°.
RD Sharma Class 9 Questions Chapter 16 Circles
(v) 15°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc meeting BC at E.
(c) Cut off arc EF from E and bisect it at G. Then ∠GBC = 30°.
(d) Again bisect the arc EJ at H.
(e) Join BH and produce it to A.
Then ∠ABC = 15°.
Maths RD Sharma Class 9 Chapter 16 Circles
(vi) 22 (frac { 1 }{ 2 })°
Steps of construction :
(a) Draw a line segment BC.
(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG
(c) Bisect FG at H so that ∠HBC = 90°.
(d) Now bisect ∠HBC at K. So that ∠YBC = 45°.
(e) Again bisect ∠YBC at J. So thttt ∠ABC = 22 (frac { 1 }{ 2 })°
RD Sharma Class 9 PDF Chapter 16 Circles

RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles Ex 16.3

Question 1.
Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.6 cm.
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.
(iii) Join EC.
(iv) Draw perpendicular bisector of CE which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Chapter 16 Circles

Question 2.
Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 4.5 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.
(iii) Draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 Solutions Chapter 16 Circles

Question 3.
Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 3.4 cm.
(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Solutions Class 9 Chapter 16 Circles

Question 4.
Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm!
(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.
(iii) Join EC.
(iv) Draw the perpendicular bisector of EC which intersects BE at A.
(v) Join AC.
∆ABC is the required triangle.
RD Sharma Class 9 PDF Chapter 16 Circles

Question 5.
Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 10 cm.
(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.
(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,
(v) Join AB and AC.
∆ABC is the required triangle.
Circles Class 9 RD Sharma Solutions

Question 6.
Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm and bisect it at D.
(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.
(iii) Join AD and AB and AC.
Then ∆ABC is the required triangle.
RD Sharma Class 9 Solution Chapter 16 Circles

Question 7.
Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm
Class 9 RD Sharma Solutions Chapter 16 Circles
(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.
(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.
(iv) Join AC.
∆ABC is the required triangle.

Question 8.
Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 6.4 cm.
(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.
(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.
(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
Class 9 Maths Chapter 16 Circles RD Sharma Solutions

Question 9.
Using ruler and compasses only, construct a ∆ABC, from the following data:
AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment PQ = 12 cm.
(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.
(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.
(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.
(v) Join AB and AC.
∆ABC is the required triangle.
RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles

Question 10.
Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11
Solution:
Steps of construction :
(i) Draw a line segment PQ =11 cm.
(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.
(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.
(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.
(v) Join XY and XZ.
Then ∆XYZ is the required triangle
RD Sharma Class 9 Book Chapter 16 Circles

Solutions of Chapter 16 Circles Ex 16.1 and 16.2 are given below:
Chapter 16 Circles Ex 16.3
Chapter 16 Circles Ex 16.4
Chapter 16 Circles Ex 16.5

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.1 1

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2

RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 1
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 2
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 3
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 4
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 5
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 6
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 7
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 8
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 9
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 10
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 11
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 12
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 13
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 14
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 16
RD Sharma Class 9 solutions Chapter 16 Circles Ex 16.2 17

All Chapter RD Sharma Solutions For Class 9 Maths

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

Class 9th solutions Chapter 16 CirclesIf these solutions have helped you, you can also share alarity.in to your friends.

If these solutions have helped you, you can also share alarity.in to your friends.

Leave a Reply

Your email address will not be published. Required fields are marked *