RD Sharma Class 9 Solutions Chapter 11 Coordinate Geometry

In this chapter, we provide RD Sharma Solutions for Class 9 Solutions Chapter 11 Coordinate Geometrys 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 Solutions Chapter 11 Coordinate Geometry pdf, free RD Sharma Solutions for Class 9 Solutions Chapter 11 Coordinate Geometry book pdf download. Now you will get step by step solution to each question.

RD Sharma Class 9 Solutions Chapter 11 Coordinate Geometry

RD Sharma Class 9 Solutions Chapter 11 Coordinate Geometry Ex 11.1

Question 1.
In a ∆ABC, if ∠A = 55°, ∠B = 40°, find ∠C.
Solution:
∵ Sum of three angles of a triangle is 180°
∴ In ∆ABC, ∠A = 55°, ∠B = 40°
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
RD Sharma Class 9 Chapter 11 Coordinate Geometry
⇒ 55° + 40° + ∠C = 180°
⇒ 95° + ∠C = 180°
∴ ∠C= 180° -95° = 85°

Question 2.
If the angles of a triangle are in the ratio 1:2:3, determine three angles.
Solution:
Ratio in three angles of a triangle =1:2:3
Let first angle = x
Then second angle = 2x
and third angle = 3x
∴ x + 2x + 3x = 180° (Sum of angles of a triangle)
⇒6x = 180°
⇒x = (frac { { 180 }^{ circ } }{ 6 })  = 30°
∴ First angle = x = 30°
Second angle = 2x = 2 x 30° = 60°
and third angle = 3x = 3 x 30° = 90°
∴ Angles are 30°, 60°, 90°

Question 3.
The angles of a triangle are (x – 40)°, (x – 20)° and ((frac { 1 }{ 2 }) x – 10)°. Find the value of x.
Solution:
∵ Sum of three angles of a triangle = 180°
∴ (x – 40)° + (x – 20)° + ((frac { 1 }{ 2 })x-10)0 = 180°
⇒ x – 40° + x – 20° + (frac { 1 }{ 2 })x – 10° = 180°
⇒ x + x+ (frac { 1 }{ 2 })x – 70° = 180°
⇒ (frac { 5 }{ 2 })x = 180° + 70° = 250°
⇒ x = (frac { { 250 }^{ circ }x 2 }{ 5 })  = 100°
∴ x = 100°

Question 4.
Two angles of a triangle are equal and the third angle is greater than each of those angles by 30°. Determine all the angles of the triangle.
Solution:
Let each of the two equal angles = x
Then third angle = x + 30°
But sum of the three angles of a triangle is 180°
∴ x + x + x + 30° = 180°
⇒ 3x + 30° = 180°
⇒3x = 150° ⇒x = (frac { { 150 }^{ circ } }{ 3 }) = 50°
∴ Each equal angle = 50°
and third angle = 50° + 30° = 80°
∴ Angles are 50°, 50° and 80°

Question 5.
If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.
Solution:
In the triangle ABC,
RD Sharma Class 9 Solutions Chapter 11 Coordinate Geometry
∠B = ∠A + ∠C
But ∠A + ∠B + ∠C = 180°
⇒∠B + ∠A + ∠C = 180°
⇒∠B + ∠B = 180°
⇒2∠B = 180°
∴ ∠B = (frac { { 180 }^{ circ } }{ 2 }) = 90°
∵ One angle of the triangle is 90°
∴ ∆ABC is a right triangle.

Question 6.
Can a triangle have:
(i) Two right angles?
(ii) Two obtuse angles?
(iii) Two acute angles?
(iv) All angles more than 60°?
(v) All angles less than 60°?
(vi) All angles equal to 60°?
Justify your answer in each case.
Solution:
(i) In a triangle, two right-angles cannot be possible. We know that sum of three angles is 180° and if there are two right-angles, then the third angle will be zero which is not possible.
(ii) In a triangle, two obtuse angle cannot be possible. We know that the sum of the three angles of a triangle is 180° and if there are
two obtuse angle, then the third angle will be negative which is not possible.
(iii) In a triangle, two acute angles are possible as sum of three angles of a trianlge is 180°.
(iv) All angles more than 60°, they are also not possible as the sum will be more than 180°.
(v) All angles less than 60°. They are also not possible as the sum will be less than 180°.
(vi) All angles equal to 60°. This is possible as the sum will be 60° x 3 = 180°.

Question 7.
The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angle is 10°, find the three angles.
Solution:
Let three angles of a triangle be x°, (x + 10)°, (x + 20)°
But sum of three angles of a triangle is 180°
∴ x + (x+ 10)° + (x + 20) = 180°
⇒ x + x+10°+ x + 20 = 180°
⇒ 3x + 30° = 180°
⇒ 3x = 180° – 30° = 150°
∴ x = (frac { { 180 }^{ circ } }{ 2 }) = 50°
∴ Angle are 50°, 50 + 10, 50 + 20
i.e. 50°, 60°, 70°

Question 8.
ABC is a triangle is which ∠A = 72°, the internal bisectors of angles B and C meet in O. Find the magnitude of ∠BOC.
Solution:
In ∆ABC, ∠A = 12° and bisectors of ∠B and ∠C meet at O
RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry
Now ∠B + ∠C = 180° – 12° = 108°
∵ OB and OC are the bisectors of ∠B and ∠C respectively
∴ ∠OBC + ∠OCB = (frac { 1 }{ 2 }) (B + C)
= (frac { 1 }{ 2 }) x 108° = 54°
But in ∆OBC,
∴ ∠OBC + ∠OCB + ∠BOC = 180°
⇒ 54° + ∠BOC = 180°
∠BOC = 180°-54°= 126°
OR
According to corollary,
∠BOC = 90°+ (frac { 1 }{ 2 }) ∠A
= 90+ (frac { 1 }{ 2 }) x 72° = 90° + 36° = 126°

Question 9.
The bisectors of base angles of a triangle cannot enclose a right angle in any case.
Solution:
In right ∆ABC, ∠A is the vertex angle and OB and OC are the bisectors of ∠B and ∠C respectively
To prove : ∠BOC cannot be a right angle
Proof: ∵ OB and OC are the bisectors of ∠B and ∠C respectively
RD Sharma Class 9 PDF Chapter 11 Coordinate Geometry
∴ ∠BOC = 90° x (frac { 1 }{ 2 }) ∠A
Let ∠BOC = 90°, then
(frac { 1 }{ 2 }) ∠A = O
⇒∠A = O
Which is not possible because the points A, B and C will be on the same line Hence, ∠BOC cannot be a right angle.

Question 10.
If the bisectors of the base angles of a triangle enclose an angle of 135°. Prove that the triangle is a right triangle.
Solution:
Given : In ∆ABC, OB and OC are the bisectors of ∠B and ∠C and ∠BOC = 135°
Coordinate Geometry Class 9 RD Sharma Solutions
To prove : ∆ABC is a right angled triangle
Proof: ∵ Bisectors of base angles ∠B and ∠C of the ∆ABC meet at O
∴ ∠BOC = 90°+ (frac { 1 }{ 2 })∠A
But ∠BOC =135°
∴ 90°+ (frac { 1 }{ 2 }) ∠A = 135°
⇒ (frac { 1 }{ 2 })∠A= 135° -90° = 45°
∴ ∠A = 45° x 2 = 90°
∴ ∆ABC is a right angled triangle

Question 11.
In a ∆ABC, ∠ABC = ∠ACB and the bisectors of ∠ABC and ∠ACB intersect at O such that ∠BOC = 120°. Show that ∠A = ∠B = ∠C = 60°.
Solution:
Given : In ∠ABC, BO and CO are the bisectors of ∠B and ∠C respectively and ∠BOC = 120° and ∠ABC = ∠ACB
RD Sharma Class 9 Solution Chapter 11 Coordinate Geometry
To prove : ∠A = ∠B = ∠C = 60°
Proof : ∵ BO and CO are the bisectors of ∠B and ∠C
∴ ∠BOC = 90° + (frac { 1 }{ 2 })∠A
But ∠BOC = 120°
∴ 90°+ (frac { 1 }{ 2 }) ∠A = 120°
∴ (frac { 1 }{ 2 }) ∠A = 120° – 90° = 30°
∴ ∠A = 60°
∵ ∠A + ∠B + ∠C = 180° (Angles of a triangle)
∠B + ∠C = 180° – 60° = 120° and ∠B = ∠C
∵ ∠B = ∠C = (frac { { 120 }^{ circ } }{ 2 }) = 60°
Hence ∠A = ∠B = ∠C = 60°

Question 12.
If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.
Solution:
In a ∆ABC,
Let ∠A < ∠B + ∠C
Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry
⇒∠A + ∠A < ∠A + ∠B + ∠C
⇒ 2∠A < 180°
⇒ ∠A < 90° (∵ Sum of angles of a triangle is 180°)
Similarly, we can prove that
∠B < 90° and ∠C < 90°
∴ Each angle of the triangle are acute angle.

RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry Ex 11.2

Question 1.
The exterior angles obtained on producing the base of a triangle both ways are 104° and 136°. Find all the angles of the triangle.
Solution:
In ∆ABC, base BC is produced both ways to D and E respectivley forming ∠ABE = 104° and ∠ACD = 136°
Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry

Question 2.
In the figure, the sides BC, CA and AB of a ∆ABC have been produced to D, E and F respectively. If ∠ACD = 105° and ∠EAF = 45°, find all the angles of the ∆ABC.
Class 9 Maths Chapter 11 Coordinate Geometry RD Sharma Solutions
Solution:
In ∆ABC, sides BC, CA and BA are produced to D, E and F respectively.
∠ACD = 105° and ∠EAF = 45°
∠ACD + ∠ACB = 180° (Linear pair)
⇒ 105° + ∠ACB = 180°
⇒ ∠ACB = 180°- 105° = 75°
∠BAC = ∠EAF (Vertically opposite angles)
= 45°
But ∠BAC + ∠ABC + ∠ACB = 180°
⇒ 45° + ∠ABC + 75° = 180°
⇒ 120° +∠ABC = 180°
⇒ ∠ABC = 180°- 120°
∴ ∠ABC = 60°
Hence ∠ABC = 60°, ∠BCA = 75°
and ∠BAC = 45°

Question 3.
Compute the value of x in each of the following figures:
RD Sharma Book Class 9 PDF Free Download Chapter 11 Coordinate Geometry
RD Sharma Class 9 Book Chapter 11 Coordinate Geometry
Solution:
(i) In ∆ABC, sides BC and CA are produced to D and E respectively
Coordinate Geometry With Solutions PDF RD Sharma Class 9 Solutions
(ii) In ∆ABC, side BC is produced to either side to D and E respectively
∠ABE = 120° and ∠ACD =110°
∵ ∠ABE + ∠ABC = 180° (Linear pair)
RD Sharma Class 9 Maths Book Questions Chapter 11 Coordinate Geometry
(iii) In the figure, BA || DC
RD Sharma Mathematics Class 9 Solutions Chapter 11 Coordinate Geometry

Question 4.
In the figure, AC ⊥ CE and ∠A: ∠B : ∠C = 3:2:1, find the value of ∠ECD.
Solution Of Rd Sharma Class 9 Chapter 11 Coordinate Geometry
Solution:
In ∆ABC, ∠A : ∠B : ∠C = 3 : 2 : 1
BC is produced to D and CE ⊥ AC
∵ ∠A + ∠B + ∠C = 180° (Sum of angles of a triangles)
Let∠A = 3x, then ∠B = 2x and ∠C = x
∴ 3x + 2x + x = 180° ⇒ 6x = 180°
⇒ x = (frac { { 180 }^{ circ } }{ 6 })  = 30°
∴ ∠A = 3x = 3 x 30° = 90°
∠B = 2x = 2 x 30° = 60°
∠C = x = 30°
In ∆ABC,
Ext. ∠ACD = ∠A + ∠B
⇒ 90° + ∠ECD = 90° + 60° = 150°
∴ ∠ECD = 150°-90° = 60°

Question 5.
In the figure, AB || DE, find ∠ACD.
RD Sharma Math Solution Class 9 Chapter 11 Coordinate Geometry
Solution:
In the figure, AB || DE
AE and BD intersect each other at C ∠BAC = 30° and ∠CDE = 40°
∵ AB || DE
∴ ∠ABC = ∠CDE (Alternate angles)
RD Sharma Class 9 Questions Chapter 11 Coordinate Geometry
⇒ ∠ABC = 40°
In ∆ABC, BC is produced
Ext. ∠ACD = Int. ∠A + ∠B
= 30° + 40° = 70°

Question 6.
Which of the following statements are true (T) and which are false (F):
(i) Sum of the three angles of a triangle is 180°.
(ii) A triangle can have two right angles.
(iii) All the angles of a triangle can be less than 60°.
(iv) All the angles of a triangle can be greater than 60°.
(v) All the angles of a triangle can be equal to 60°.
(vi) A triangle can have two obtuse angles.
(vii) A triangle can have at most one obtuse angles.
(viii) If one angle of a triangle is obtuse, then it cannot be a right angled triangle.
(ix) An exterior angle of a triangle is less than either of its interior opposite angles.
(x) An exterior angle of a triangle is equal to the sum of the two interior opposite angles.
(xi) An exterior angle of a triangle is greater than the opposite interior angles.
Solution:
(i) True.
(ii) False. A right triangle has only one right angle.
(iii) False. In this, the sum of three angles will be less than 180° which is not true.
(iv) False. In this, the sum of three angles will be more than 180° which is not true.
(v) True. As sum of three angles will be 180° which is true.
(vi) False. A triangle has only one obtuse angle.
(vii) True.
(viii)True.
(ix) False. Exterior angle of a triangle is always greater than its each interior opposite angles.
(x) True.
(xi) True.

Question 7.
Fill in the blanks to make the following statements true:
(i) Sum of the angles of a triangle is ………
(ii) An exterior angle of a triangle is equal to the two …….. opposite angles.
(iii) An exterior angle of a triangle is always …….. than either of the interior opposite angles.
(iv) A triangle cannot have more than ………. right angles.
(v) A triangles cannot have more than ……… obtuse angles.
Solution:
(i) Sum of the angles of a triangle is 180°.
(ii) An exterior angle of a triangle is equal to the two interior opposite angles.
(iii) An exterior angle of a triangle is always greater than either of the interior opposite angles.
(iv) A triangle cannot have more than one right angles.
(v) A triangles cannot have more than one obtuse angles.

Question 8.
In a ∆ABC, the internal bisectors of ∠B and ∠C meet at P and the external bisectors of ∠B and ∠C meet at Q. Prove that ∠BPC + ∠BQC = 180°.
Solution:
Given : In ∆ABC, sides AB and AC are produced to D and E respectively. Bisectors of interior ∠B and ∠C meet at P and bisectors of exterior angles B and C meet at Q.
Maths RD Sharma Class 9 Chapter 11 Coordinate Geometry
To prove : ∠BPC + ∠BQC = 180°
Proof : ∵ PB and PC are the internal bisectors of ∠B and ∠C
∠BPC = 90°+ (frac { 1 }{ 2 }) ∠A …(i)
Similarly, QB and QC are the bisectors of exterior angles B and C
∴ ∠BQC = 90° + (frac { 1 }{ 2 }) ∠A …(ii)
Adding (i) and (ii),
∠BPC + ∠BQC = 90° + (frac { 1 }{ 2 }) ∠A + 90° – (frac { 1 }{ 2 }) ∠A
= 90° + 90° = 180°
Hence ∠BPC + ∠BQC = 180°

Question 9.
In the figure, compute the value of x.
RD Sharma Class 9 Chapter 11 Coordinate Geometry
Solution:
In the figure,
∠ABC = 45°, ∠BAD = 35° and ∠BCD = 50° Join BD and produce it E
RD Sharma Class 9 Solutions Chapter 11 Coordinate Geometry

Question 10.
In the figure, AB divides ∠D AC in the ratio 1 : 3 and AB = DB. Determine the value of x.
RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry
Solution:
In the figure AB = DB
RD Sharma Class 9 PDF Chapter 11 Coordinate Geometry
Coordinate Geometry Class 9 RD Sharma Solutions

Question 11.
ABC is a triangle. The bisector of the exterior angle at B and the bisector of ∠C intersect each other at D. Prove that ∠D = (frac { 1 }{ 2 }) ∠A.
Solution:
Given : In ∠ABC, CB is produced to E bisectors of ext. ∠ABE and into ∠ACB meet at D.
RD Sharma Class 9 Solution Chapter 11 Coordinate Geometry
Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry

Question 12.
In the figure, AM ⊥ BC and AN is the bisector of ∠A. If ∠B = 65° and ∠C = 33°, find ∠MAN.
Class 9 Maths Chapter 11 Coordinate Geometry RD Sharma Solutions
Solution:
RD Sharma Book Class 9 PDF Free Download Chapter 11 Coordinate Geometry
RD Sharma Class 9 Book Chapter 11 Coordinate Geometry

Question 13.
In a AABC, AD bisects ∠A and ∠C > ∠B. Prove that ∠ADB > ∠ADC.
Solution:
Given : In ∆ABC,
∠C > ∠B and AD is the bisector of ∠A
Coordinate Geometry With Solutions PDF RD Sharma Class 9 Solutions
To prove : ∠ADB > ∠ADC
Proof: In ∆ABC, AD is the bisector of ∠A
∴ ∠1 = ∠2
In ∆ADC,
Ext. ∠ADB = ∠l+ ∠C
⇒ ∠C = ∠ADB – ∠1 …(i)
Similarly, in ∆ABD,
Ext. ∠ADC = ∠2 + ∠B
⇒ ∠B = ∠ADC – ∠2 …(ii)
From (i) and (ii)
∵ ∠C > ∠B (Given)
∴ (∠ADB – ∠1) > (∠ADC – ∠2)
But ∠1 = ∠2
∴ ∠ADB > ∠ADC

Question 14.
In ∆ABC, BD ⊥ AC and CE ⊥ AB. If BD and CE intersect at O, prove that ∠BOC = 180°-∠A.
Solution:
Given : In ∆ABC, BD ⊥ AC and CE⊥ AB BD and CE intersect each other at O
RD Sharma Class 9 Maths Book Questions Chapter 11 Coordinate Geometry
To prove : ∠BOC = 180° – ∠A
Proof: In quadrilateral ADOE
∠A + ∠D + ∠DOE + ∠E = 360° (Sum of angles of quadrilateral)
⇒ ∠A + 90° + ∠DOE + 90° = 360°
∠A + ∠DOE = 360° – 90° – 90° = 180°
But ∠BOC = ∠DOE (Vertically opposite angles)
⇒ ∠A + ∠BOC = 180°
∴ ∠BOC = 180° – ∠A

Question 15.
In the figure, AE bisects ∠CAD and ∠B = ∠C. Prove that AE || BC.
RD Sharma Mathematics Class 9 Solutions Chapter 11 Coordinate Geometry
Solution:
Given : In AABC, BA is produced and AE is the bisector of ∠CAD
∠B = ∠C
Solution Of Rd Sharma Class 9 Chapter 11 Coordinate Geometry
To prove : AE || BC
Proof: In ∆ABC, BA is produced
∴ Ext. ∠CAD = ∠B + ∠C
⇒ 2∠EAC = ∠C + ∠C (∵ AE is the bisector of ∠CAE) (∵ ∠B = ∠C)
⇒ 2∠EAC = 2∠C
⇒ ∠EAC = ∠C
But there are alternate angles
∴ AE || BC

RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry VSAQS

Question 1.
Define a triangle.
Solution:
A figure bounded by three lines segments in a plane is called a triangle.

Question 2.
Write the sum of the angles of an obtuse triangle.
Solution:
The sum of angles of an obtuse triangle is 180°.

Question 3.
In ∆ABC, if ∠B = 60°, ∠C = 80° and the bisectors of angles ∠ABC and ∠ACB meet at a point O, then find the measure of ∠BOC.
Solution:
In ∆ABC, ∠B = 60°, ∠C = 80°
OB and OC are the bisectors of ∠B and ∠C
∵ ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ ∠A + 60° + 80° = 180°
⇒ ∠A + 140° = 180°
∴ ∠A = 180°- 140° = 40°
RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry
= 90° + – x 40° = 90° + 20° = 110°

Question 4.
If the angles of a triangle are in the ratio 2:1:3. Then find the measure of smallest angle.
Solution:
Sum of angles of a triangle = 180°
Ratio in the angles = 2 : 1 : 3
Let first angle = 2x
Second angle = x
and third angle = 3x
∴ 2x + x + 3x = 180° ⇒ 6x = 180°
∴ x = (frac { { 180 }^{ circ } }{ 6 })  = 30°
∴ First angle = 2x = 2 x 30° = 60°
Second angle = x = 30°
and third angle = 3x = 3 x 30° = 90°
Hence angles are 60°, 30°, 90°

Question 5.
State exterior angle theorem.
Solution:
Given : In ∆ABC, side BC is produced to D
RD Sharma Class 9 PDF Chapter 11 Coordinate Geometry
To prove : ∠ACD = ∠A + ∠B
Proof: In ∆ABC,
∠A + ∠B + ∠ACB = 180° …(i) (Sum of angles of a triangle)
and ∠ACD + ∠ACB = 180° …(ii) (Linear pair)
From (i) and (ii)
∠ACD + ∠ACB = ∠A + ∠B + ∠ACB
∠ACD = ∠A + ∠B
Hence proved.

Question 6.
The sum of two angles of a triangle is equal to its third angle. Determine the measure of the third angle.
Solution:
In ∆ABC,
∠A + ∠C = ∠B
Coordinate Geometry Class 9 RD Sharma Solutions
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
∴ ∠B + ∠A + ∠C = 180°
⇒ ∠B + ∠B = 180°
⇒ 2∠B = 180°
⇒ ∠B = (frac { { 180 }^{ circ } }{ 2 })  = 90°
∴ Third angle = 90°

Question 7.
In the figure, if AB || CD, EF || BC, ∠BAC = 65° and ∠DHF = 35°, find ∠AGH.
RD Sharma Class 9 Solution Chapter 11 Coordinate Geometry
Solution:
Given : In figure, AB || CD, EF || BC ∠BAC = 65°, ∠DHF = 35°
Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry
∵ EF || BC
∴ ∠A = ∠ACH (Alternate angle)
∴ ∠ACH = 65°
∵∠GHC = ∠DHF
(Vertically opposite angles)
∴ ∠GHC = 35°
Now in ∆GCH,
Ext. ∠AGH = ∠GCH + ∠GHC
= 65° + 35° = 100°

Question 8.
In the figure, if AB || DE and BD || FG such that ∠FGH = 125° and ∠B = 55°, find x and y.
Class 9 Maths Chapter 11 Coordinate Geometry RD Sharma Solutions
Solution:
In the figure, AB || DF, BD || FG
RD Sharma Book Class 9 PDF Free Download Chapter 11 Coordinate Geometry
∠FGH = 125° and ∠B = 55°
∠FGH + FGE = 180° (Linear pair)
⇒ 125° + y – 180°
⇒ y= 180°- 125° = 55°
∵ BA || FD and BD || FG
∠B = ∠F = 55°
Now in ∆EFG,
∠F + ∠FEG + ∠FGE = 180°
(Angles of a triangle)
⇒ 55° + x + 55° = 180°
⇒ x+ 110°= 180°
∴ x= 180°- 110° = 70°
Hence x = 70, y = 55°

Question 9.
If the angles A, B and C of ∆ABC satisfy the relation B – A = C – B, then find the measure of ∠B.
Solution:
In ∆ABC,
∠A + ∠B + ∠C= 180° …(i)
and B – A = C – B
RD Sharma Class 9 Book Chapter 11 Coordinate Geometry
⇒ B + B = A + C ⇒ 2B = A + C
From (i),
B + 2B = 180° ⇒ 3B = 180°
∠B = (frac { { 180 }^{ circ } }{ 3 }) = 60°
Hence ∠B = 60°

Question 10.
In ∆ABC, if bisectors of ∠ABC and ∠ACB intersect at O at angle of 120°, then find the measure of ∠A.
Solution:
In ∆ABC, bisectors of ∠B and ∠C intersect at O and ∠BOC = 120°
Coordinate Geometry With Solutions PDF RD Sharma Class 9 Solutions
But ∠BOC = 90°+ (frac { 1 }{ 2 })
90°+ (frac { 1 }{ 2 }) ∠A= 120°
⇒ (frac { 1 }{ 2 }) ∠A= 120°-90° = 30°
∴ ∠A = 2 x 30° = 60°

Question 11.
If the side BC of ∆ABC is produced on both sides, then write the difference between the sum of the exterior angles so formed and ∠A.
Solution:
In ∆ABC, side BC is produced on both sides forming exterior ∠ABE and ∠ACD
Ext. ∠ABE = ∠A + ∠ACB
and Ext. ∠ACD = ∠ABC + ∠A
RD Sharma Class 9 Maths Book Questions Chapter 11 Coordinate Geometry
Adding we get,
∠ABE + ∠ACD = ∠A + ∠ACB + ∠A + ∠ABC
⇒ ∠ABE + ∠ACD – ∠A = ∠A 4- ∠ACB + ∠A + ∠ABC – ∠A (Subtracting ∠A from both sides)
= ∠A + ∠ABC + ∠ACB = ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)

Question 12.
In a triangle ABC, if AB = AC and AB is produced to D such that BD = BC, find ∠ACD: ∠ADC.
Solution:
In ∆ABC, AB = AC
AB is produced to D such that BD = BC
DC are joined
In ∆ABC, AB = AC
∴ ∠ABC = ∠ACB
In ∆ BCD, BD = BC
∴ ∠BDC = ∠BCD
and Ext. ∠ABC = ∠BDC + ∠BCD = 2∠BDC (∵ ∠BDC = ∠BCD)
⇒ ∠ACB = 2∠BCD (∵ ∠ABC = ∠ACB)
Adding ∠BDC to both sides
⇒ ∠ACB + ∠BDC = 2∠BDC + ∠BDC
⇒ ∠ACB + ∠BCD = 3 ∠BDC (∵ ∠BDC = ∠BCD)
⇒ ∠ACB = 3∠BDC

RD Sharma Mathematics Class 9 Solutions Chapter 11 Coordinate Geometry

Question 13.
In the figure, side BC of AABC is produced to point D such that bisectors of ∠ABC and ∠ACD meet at a point E. If ∠BAC = 68°, find ∠BEC.
Solution Of Rd Sharma Class 9 Chapter 11 Coordinate Geometry
Solution:
In the figure,
RD Sharma Math Solution Class 9 Chapter 11 Coordinate Geometry
side BC of ∆ABC is produced to D such that bisectors of ∠ABC and ∠ACD meet at E
∠BAC = 68°
In ∆ABC,
Ext. ∠ACD = ∠A + ∠B
⇒ (frac { 1 }{ 2 }) ∠ACD = (frac { 1 }{ 2 }) ∠A + (frac { 1 }{ 2 }) ∠B
⇒ ∠2= (frac { 1 }{ 2 }) ∠A + ∠1 …(i)
But in ∆BCE,
Ext. ∠2 = ∠E + ∠l
⇒ ∠E + ∠l = ∠2 = (frac { 1 }{ 2 }) ∠A + ∠l [From (i)]
⇒ ∠E = (frac { 1 }{ 2 }) ∠A = (frac { { 68 }^{ circ } }{ 2 })  =34°

Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry MCQS

Mark the correct alternative in each of the following:
Question 1.
If all the three angles of a triangle are equal, then each one of them is equal to
(a) 90°
(b) 45°
(c) 60°
(d) 30°
Solution:
∵ Sum of three angles of a triangle = 180°
∴ Each angle = (frac { { 180 }^{ circ } }{ 3 })  = 60° (c)

Question 2.
If two acute angles of a right triangle are equal, then each acute is equal to
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
In a right triangle, one angle = 90°
∴ Sum of other two acute angles = 180° – 90° = 90°
∵ Both angles are equal
∴ Each angle will be = (frac { { 90 }^{ circ } }{ 2 })  = 45° (b)

Question 3.
An exterior angle of a triangle is equal to 100° and two interior opposite angles are equal. Each of these angles is equal to
(a) 75°
(b) 80°
(c) 40°
(d) 50°
Solution:
In a triangle, exterior angles is equal to the sum of its interior opposite angles
∴ Sum of interior opposite angles = 100°
∵ Both angles are equal
∴ Each angle will be = (frac { { 100 }^{ circ } }{ 2 })  = 50° (d)

Question 4.
If one angle of a triangle is equal to the sum of the other two angles, then the triangle is
(a) an isosceles triangle
(b) an obtuse triangle
(c) an equilateral triangle
(d) a right triangle
Solution:
Let ∠A, ∠B, ∠C be the angles of a ∆ABC and let ∠A = ∠B + ∠C
RD Sharma Class 9 PDF Chapter 11 Coordinate Geometry
But ∠A + ∠B + ∠C = 180°
( Sum of angles of a triangle)
∴ ∠A + ∠A = 180° ⇒ 2∠A = 180°
⇒ ∠A = (frac { { 180 }^{ circ } }{ 2 })  = 90°
∴ ∆ is a right triangle (d)

Question 5.
Side BC of a triangle ABC has been produced to a point D such that ∠ACD = 120°. If ∠B = (frac { 1 }{ 2 })∠A, then ∠A is equal to
(a) 80°
(b) 75°
(c) 60°
(d) 90°
Solution:
Side BC of ∆ABC is produced to D, then
Coordinate Geometry Class 9 RD Sharma Solutions
Ext. ∠ACB = ∠A + ∠B
(Exterior angle of a triangle is equal to the sum of its interior opposite angles)
RD Sharma Class 9 Solution Chapter 11 Coordinate Geometry

Question 6.
In ∆ABC, ∠B = ∠C and ray AX bisects the exterior angle ∠DAC. If ∠DAX = 70°, then ∠ACB =
(a) 35°
(b) 90°
(c) 70°
(d) 55°
Solution:
In ∆ABC, ∠B = ∠C
AX is the bisector of ext. ∠CAD
∠DAX = 70°
Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry
∴ ∠DAC = 70° x 2 = 140°
But Ext. ∠DAC = ∠B + ∠C
= ∠C + ∠C (∵ ∠B = ∠C)
= 2∠C
∴ 2∠C = 140° ⇒ ∠C = (frac { { 140 }^{ circ } }{ 2 }) = 70°
∴ ∠ACB = 70° (c)

Question 7.
In a triangle, an exterior angle at a vertex is 95° and its one of the interior opposite angle is 55°, then the measure of the other interior angle is
(a) 55°
(b) 85°
(c) 40°
(d) 9.0°
Solution:
In ∆ABC, BA is produced to D such that ∠CAD = 95°
and let ∠C = 55° and ∠B = x°
Class 9 Maths Chapter 11 Coordinate Geometry RD Sharma Solutions
∵ Exterior angle of a triangle is equal to the sum of its opposite interior angle
∴ ∠CAD = ∠B + ∠C ⇒ 95° = x + 55°
⇒ x = 95° – 55° = 40°
∴ Other interior angle = 40° (c)

Question 8.
If the sides of a triangle are produced in order, then the sum of the three exterior angles so formed is
(a) 90°
(b) 180°
(c) 270°
(d) 360°
Solution:
In ∆ABC, sides AB, BC and CA are produced in order, then exterior ∠FAB, ∠DBC and ∠ACE are formed
RD Sharma Book Class 9 PDF Free Download Chapter 11 Coordinate Geometry
We know an exterior angles of a triangle is equal to the sum of its interior opposite angles
∴ ∠FAB = ∠B + ∠C
∠DBC = ∠C + ∠A and
∠ACE = ∠A + ∠B Adding we get,
∠FAB + ∠DBC + ∠ACE = ∠B + ∠C + ∠C + ∠A + ∠A + ∠B
= 2(∠A + ∠B + ∠C)
= 2 x 180° (Sum of angles of a triangle)
= 360° (d)

Question 9.
In ∆ABC, if ∠A = 100°, AD bisects ∠A and AD⊥ BC. Then, ∠B =
(a) 50°
(b) 90°
(c) 40°
(d) 100°
Solution:
In ∆ABC, ∠A = 100°
AD is bisector of ∠A and AD ⊥ BC
RD Sharma Class 9 Book Chapter 11 Coordinate Geometry
Now, ∠BAD = (frac { { 100 }^{ circ } }{ 2 }) = 50°
In ∆ABD,
∠BAD + ∠B + ∠D= 180°
(Sum of angles of a triangle)
⇒ ∠50° + ∠B + 90° = 180°
∠B + 140° = 180°
⇒ ∠B = 180° – 140° ∠B = 40° (c)

Question 10.
An exterior angle of a triangle is 108° and its interior opposite angles are in the ratio 4:5. The angles of the triangle are
(a) 48°, 60°, 72°
(b) 50°, 60°, 70°
(c) 52°, 56°, 72°
(d) 42°, 60°, 76°
Solution:
In ∆ABC, BC is produced to D and ∠ACD = 108°
Coordinate Geometry With Solutions PDF RD Sharma Class 9 Solutions
Ratio in ∠A : ∠B = 4:5
∵ Exterior angle of a triangle is equal to the sum of its opposite interior angles
∴ ∠ACD = ∠A + ∠B = 108°
Ratio in ∠A : ∠B = 4:5
RD Sharma Class 9 Maths Book Questions Chapter 11 Coordinate Geometry

Question 11.
In a ∆ABC, if ∠A = 60°, ∠B = 80° and the bisectors of ∠B and ∠C meet at O, then ∠BOC =
(a) 60°
(b) 120°
(c) 150°
(d) 30°
Solution:
In ∆ABC, ∠A = 60°, ∠B = 80°
∴ ∠C = 180° – (∠A + ∠B)
= 180° – (60° + 80°)
= 180° – 140° = 40°
Bisectors of ∠B and ∠C meet at O
RD Sharma Mathematics Class 9 Solutions Chapter 11 Coordinate Geometry
Solution Of Rd Sharma Class 9 Chapter 11 Coordinate Geometry

Question 12.
Line segments AB and CD intersect at O such that AC || DB. If ∠CAB = 45° and ∠CDB = 55°, then ∠BOD =
(a) 100°
(b) 80°
(c) 90°
(d) 135°
Solution:
In the figure,
RD Sharma Math Solution Class 9 Chapter 11 Coordinate Geometry
AB and CD intersect at O
and AC || DB, ∠CAB = 45°
and ∠CDB = 55°
∵ AC || DB
∴ ∠CAB = ∠ABD (Alternate angles)
In ∆OBD,
∠BOD = 180° – (∠CDB + ∠ABD)
= 180° – (55° + 45°)
= 180° – 100° = 80° (b)

Question 13.
In the figure, if EC || AB, ∠ECD = 70° and ∠BDO = 20°, then ∠OBD is
RD Sharma Class 9 Questions Chapter 11 Coordinate Geometry
(a) 20°
(b) 50°
(c) 60°
(d) 70°
Solution:
In the figure, EC || AB
∠ECD = 70°, ∠BDO = 20°
Maths RD Sharma Class 9 Chapter 11 Coordinate Geometry
∵ EC || AB
∠AOD = ∠ECD (Corresponding angles)
⇒ ∠AOD = 70°
In ∆OBD,
Ext. ∠AOD = ∠OBD + ∠BDO
70° = ∠OBD + 20°
⇒ ∠OBD = 70° – 20° = 50° (b)

Question 14.
In the figure, x + y =
(a) 270
(b) 230
(c) 210
(d) 190°
RD Sharma Class 9 Chapter 11 Coordinate Geometry
Solution:
In the figure
RD Sharma Class 9 Solutions Chapter 11 Coordinate Geometry
Ext. ∠OAE = ∠AOC + ∠ACO
⇒ x = 40° + 80° = 120°
Similarly,
Ext. ∠DBF = ∠ODB + ∠DOB
y = 70° + ∠DOB
[(∵ ∠AOC = ∠DOB) (vertically opp. angles)]
= 70° + 40° = 110°
∴ x+y= 120°+ 110° = 230° (b)

Question 15.
If the measures of angles of a triangle are in the ratio of 3 : 4 : 5, what is the measure of the smallest angle of the triangle?
(a) 25°
(b) 30°
(c) 45°
(d) 60°
Solution:
Ratio in the measures of the triangle =3:4:5
Sum of angles of a triangle = 180°
Let angles be 3x, 4x, 5x
Sum of angles = 3x + 4x + 5x = 12x
∴ Smallest angle = (frac { 180 x 3x }{ 12x }) = 45° (c)

Question 16.
In the figure, if AB ⊥ BC, then x =
(a) 18
(b) 22
(c) 25
(d) 32
RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry
Solution:
In the figure, AB ⊥ BC
∠AGF = 32°
∴ ∠CGB = ∠AGF (Vertically opposite angles)
= 32°
RD Sharma Class 9 PDF Chapter 11 Coordinate Geometry
In ∆GCB, ∠B = 90°
∴ ∠CGB + ∠GCB = 90°
⇒ 32° + ∠GCB = 90°
⇒ ∠GCB = 90° – 32° = 58°
Now in ∆GDC,
Ext. ∠GCB = ∠CDG + ∠DGC
⇒ 58° = x + 14° + x
⇒ 2x + 14° = 58°
⇒ 2x = 58 – 14° = 44
⇒ x = (frac { 44 }{ 2 }) = 22°
∴ x = 22° (b)

Question 17.
In the figure, what is ∠ in terms of x and y?
(a) x + y + 180
(b) x + y – 180
(c) 180° -(x+y)
(d) x+y + 360°
Coordinate Geometry Class 9 RD Sharma Solutions
Solution:
In the figure, BC is produced both sides CA and BA are also produced
In ∆ABC,
∠B = 180° -y
and ∠C 180° – x
RD Sharma Class 9 Solution Chapter 11 Coordinate Geometry
∴ z = ∠A = 180° – (B + C)
= 180° – (180 – y + 180 -x)
= 180° – (360° – x – y)
= 180° – 360° + x + y = x + y – 180° (b)

Question 18.
In the figure, for which value of x is l1 || l2?
(a) 37
(b) 43
(c) 45
(d) 47
Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry
Solution:
In the figure, l1 || l2
∴ ∠EBA = ∠BAH (Alternate angles)
∴ ∠BAH = 78°
Class 9 Maths Chapter 11 Coordinate Geometry RD Sharma Solutions
⇒ ∠BAC + ∠CAH = 78°
⇒ ∠BAC + 35° = 78°
⇒ ∠BAC = 78° – 35° = 43°
In ∆ABC, ∠C = 90°
∴ ∠ABC + ∠BAC = 90°
⇒ x + 43° = 90° ⇒ x = 90° – 43°
∴ x = 47° (d)

Question 19.
In the figure, what is y in terms of x?
RD Sharma Book Class 9 PDF Free Download Chapter 11 Coordinate Geometry
Solution:
In ∆ABC,
∠ACB = 180° – (x + 2x)
= 180° – 3x …(i)
and in ∆BDG,
∠BED = 180° – (2x + y) …(ii)
∠EGC = ∠AGD (Vertically opposite angles)
= 3y
RD Sharma Class 9 Book Chapter 11 Coordinate Geometry
In quad. BCGE,
∠B + ∠ACB + ∠CGE + ∠BED = 360° (Sum of angles of a quadrilateral)
⇒ 2x+ 180° – 3x + 3y + 180°- 2x-y = 360°
⇒ -3x + 2y = 0
⇒ 3x = 2y ⇒ y = (frac { 3 }{ 2 })x (a)

Question 20.
In the figure, what is the value of x?
(a) 35
(b) 45
(c) 50
(d) 60
Coordinate Geometry With Solutions PDF RD Sharma Class 9 Solutions
Solution:
In the figure, side AB is produced to D
∴ ∠CBA + ∠CBD = 180° (Linear pair)
⇒ 7y + 5y = 180°
⇒ 12y = 180°
⇒ y = (frac { 180 }{ 12 }) = 15
and Ext. ∠CBD = ∠A + ∠C
⇒ 7y = 3y + x
⇒ 7y -3y = x
⇒ 4y = x
∴ x = 4 x 15 = 60 (d)

Question 21.
In the figure, the value of x is
(a) 65°
(b) 80°
(c) 95°
(d) 120°
RD Sharma Class 9 Maths Book Questions Chapter 11 Coordinate Geometry
Solution:
In the figure, ∠A = 55°, ∠D = 25° and ∠C = 40°
RD Sharma Mathematics Class 9 Solutions Chapter 11 Coordinate Geometry
Now in ∆ABD,
Ext. ∠DBC = ∠A + ∠D
= 55° + 25° = 80°
Similarly, in ∆BCE,
Ext. ∠DEC = ∠EBC + ∠ECB
= 80° + 40° = 120° (d)

Question 22.
In the figure, if BP || CQ and AC = BC, then the measure of x is
(a) 20°
(b) 25°
(c) 30°
(d) 35°
Solution Of Rd Sharma Class 9 Chapter 11 Coordinate Geometry
Solution:
In the figure, AC = BC, BP || CQ
RD Sharma Math Solution Class 9 Chapter 11 Coordinate Geometry
∵ BP || CQ
∴ ∠PBC – ∠QCD
⇒ 20° + ∠ABC = 70°
⇒ ∠ABC = 70° – 20° = 50°
∵ BC = AC
∴ ∠ACB = ∠ABC (Angles opposite to equal sides)
= 50°
Now in ∆ABC,
Ext. ∠ACD = ∠B + ∠A
⇒ x + 70° = 50° + 50°
⇒ x + 70° = 100°
∴ x = 100° – 70° = 30° (c)

Question 23.
In the figure, AB and CD are parallel lines and transversal EF intersects them at P and Q respectively. If ∠APR = 25°, ∠RQC = 30° and ∠CQF = 65°, then
(a) x = 55°, y = 40°
(b) x = 50°, y = 45°
(c) x = 60°, y = 35°
(d) x = 35°, y = 60°
RD Sharma Class 9 Questions Chapter 11 Coordinate Geometry
Solution:
In the figure,
∵ AB || CD, EF intersects them at P and Q respectively,
∠APR = 25°, ∠RQC = 30°, ∠CQF = 65°
∵ AB || CD
∴ ∠APQ = ∠CQF (Corresponding anlges)
⇒ y + 25° = 65°
⇒ y = 65° – 25° = 40°
and APQ + PQC = 180° (Co-interior angles)
Maths RD Sharma Class 9 Chapter 11 Coordinate Geometry
y + 25° + ∠1 +30°= 180°
40° + 25° + ∠1 + 30° = 180°
⇒ ∠1 + 95° = 180°
∴ ∠1 = 180° – 95° = 85°
Now, ∆PQR,
∠RPQ + ∠PQR + ∠PRQ = 180° (Sum of angles of a triangle)
⇒ 40° + x + 85° = 180°
⇒ 125° + x = 180°
⇒ x = 180° – 125° = 55°
∴ x = 55°, y = 40° (a)

Question 24.
The base BC of triangle ABC is produced both ways and the measure of exterior angles formed are 94° and 126°. Then, ∠BAC = ?
(a) 94°
(b) 54°
(c) 40°
(d) 44°
Solution:
In ∆ABC, base BC is produced both ways and ∠ACD = 94°, ∠ABE = 126°
RD Sharma Class 9 Chapter 11 Coordinate Geometry
Ext. ∠ACD = ∠BAC + ∠ABC
⇒ 94° = ∠BAC + ∠ABC
Similarly, ∠ABE = ∠BAC + ∠ACB
⇒ 126° = ∠BAC + ∠ACB
Adding,
94° + 126° = ∠BAC + ∠ABC + ∠ACB + ∠BAC
220° = 180° + ∠BAC
∴ ∠BAC = 220° -180° = 40° (c)

Question 25.
If the bisectors of the acute angles of a right triangle meet at O, then the angle at O between the two bisectors is
(a) 45°
(b) 95°
(c) 135°
(d) 90°
Solution:
In right ∆ABC, ∠A = 90°
RD Sharma Class 9 Solutions Chapter 11 Coordinate Geometry
Bisectors of ∠B and ∠C meet at O, then 1
∠BOC = 90° + (frac { 1 }{ 2 }) ∠A
= 90°+ (frac { 1 }{ 2 }) x 90° = 90° + 45°= 135° (c)

Question 26.
The bisects of exterior angles at B and C of ∆ABC, meet at O. If ∠A = .x°, then ∠BOC=
RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry
RD Sharma Class 9 PDF Chapter 11 Coordinate Geometry
Solution:
In ∆ABC, ∠A = x°
and bisectors of ∠B and ∠C meet at O.
Coordinate Geometry Class 9 RD Sharma Solutions

Question 27.
In a ∆ABC, ∠A = 50° and BC is produced to a point D. If the bisectors of ∠ABC and ∠ACD meet at E, then ∠E =
(a) 25°
(b) 50°
(c) 100°
(d) 75°
Solution:
In ∆ABC, ∠A = 50°
BC is produced
RD Sharma Class 9 Solution Chapter 11 Coordinate Geometry
Bisectors of ∠ABC and ∠ACD meet at ∠E
∴ ∠E = (frac { 1 }{ 2 }) ∠A = (frac { 1 }{ 2 }) x 50° = 25° (a)

Question 28.
The side BC of AABC is produced to a point D. The bisector of ∠A meets side BC in L. If ∠ABC = 30° and ∠ACD =115°,then ∠ALC =
(a) 85°
(b) 72(frac { 1 }{ 2 }) °
(c) 145°
(d) none of these
Solution:
In ∆ABC, BC is produced to D
∠B = 30°, ∠ACD = 115°
Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry

Question 29.
In the figure , if l1 || l2, the value of x is
(a) 22 (frac { 1 }{ 2 })
(b) 30
(c) 45
(d) 60
Class 9 Maths Chapter 11 Coordinate Geometry RD Sharma Solutions
Solution:
In the figure, l1 || l2
EC, EB are the bisectors of ∠DCB and ∠CBA respectively EF is the bisector of ∠GEB
RD Sharma Book Class 9 PDF Free Download Chapter 11 Coordinate Geometry
∵ EC and EB are the bisectors of ∠DCB and ∠CBA respectively
∴ ∠CEB = 90°
∴ a + b = 90° ,
and ∠GEB = 90° (∵ ∠CEB = 90°)
2x = 90° ⇒ x = (frac { 90 }{ 2 }) = 45 (c)

Question 30.
In ∆RST (in the figure), what is the value of x?
(a) 40°
(b) 90°
(c) 80°
(d) 100°

RD Sharma Class 9 Book Chapter 11 Coordinate Geometry
Solution:
Coordinate Geometry With Solutions PDF RD Sharma Class 9 Solutions
RD-Sharma-Class-9-Solutions-Chapter-11-Coordinate-Geometry-Ex-11.1-Q-1
RD-Sharma-Class-9-Solutions-Chapter-11-Coordinate-Geometry-Ex-11.1-Q-2
RD-Sharma-Class-9-Solutions-Chapter-11-Coordinate-Geometry-Ex-11.1-Q-3
RD-Sharma-Class-9-Solutions-Chapter-11-Coordinate-Geometry-Ex-11.1-Q-4

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.

Leave a Comment

Your email address will not be published. Required fields are marked *