### RS Aggarwal Solutions Class 9 Chapter 5 Congruence of Triangles and Inequalities in a Triangle

Here you can get solutions of RS Aggarwal Solutions Class 9 Chapter 5 Congruence of Triangles and Inequalities in a Triangle. These Solutions are part of RS Aggarwal Solutions Class 9. we have given RS Aggarwal Solutions Class 9 Chapter 5 Congruence of Triangles and Inequalities in a Triangle download pdf.

## RS Aggarwal Solutions Class 9 Chapter 5 Congruence of Triangles and Inequalities in a Triangle

**Exercise 5A**

**Question 1:**

**Question 2:**

Consider the isosceles triangle ∆ABC.

Since the vertical angle of ABC is 100° , we have, ∠A = 100°.

By angle sum property of a triangle, we have,

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**Question 3:**

**Question 4:**

**Question 5:**

In a right angled isosceles triangle, the vertex angle is ∠A = 90° and the other two base angles are equal.

Let x° be the base angle and we have, ∠B = ∠C = 90°.

By angle sum property of a triangle, we have

**Question 6:**

Given: ABC is an isosceles triangle in which AB=AC and BC

Is produced both ways,

**Question 7:**

Let be an equilateral triangle.

Since it is an equilateral triangle, all the angles are equiangular and the measure of each angle is 60°

The exterior angle of ∠A is ∠BAF

The exterior angle of ∠B is ∠ABD

The exterior angle of ∠C is ∠ACE

We can observe that the angles ∠A and ∠BAF, ∠B and ∠ABD, ∠C and ∠ACE and form linear pairs.

Therefore, we have

Similarly, we have

Also, we have

Thus, we have, ∠BAF = 120°, ∠ABD = 120°, ∠ACE = 120°

So, the measure of each exterior angle of an equilateral triangle is 120°.

**Question 8:**

**Question 9:**

**Question 10:**

**Question 11:**

**Question 12:**

**Question 13:**

**Question 14:**

**Question 15:**

**Question 16:**

**Question 17:**

**Question 18:**

**Question 19:**

**Question 20:**

**Question 21:**

**Question 22:**

**Question 23:**

**Question 24:**

**Question 25:**

**Question 26:**

**Question 27:**

**Question 28:**

**Question 29:**

**Question 30:**

**Question 31:**

**Question 32:**

**Question 33:**

Let AB be the breadth of a river. Now take a point M on that bank of the river where point B is situated. Through M draw a perpendicular and take point N on it such that point, A, O and N lie on a straight line where point O is the mid point of BM.

**Question 34**

**Question 35:**

**Question 36:**

**Question 37:**

**Question 38:**

**Question 39:**

**Question 40:**

**Question 41:**

**Question 42:**

**Question 43:**

**Question 44:**

Given : ABC is a triangle and O is appoint inside it.

To Prove : (i) AB+AC > OB +OC

(ii) AB+BC+CA > OA+OB+OC

(iii) OA+OB+OC > [latex]frac { 1 }{ 2 } [/latex] (AB+BC+CA)

Proof:

(i) In ∆ABC,

AB+AC > BC ….(i)

And in , ∆OBC,

OB+OC > BC ….(ii)

Subtracting (i) from (i) we get

(AB+AC) – (OB+OC) > (BC-BC)

i.e. AB+AC>OB+OC

(ii) AB+AC > OB+OC [proved in (i)]

Similarly, AB+BC > OA+OC

And AC+BC > OA +OB

Adding both sides of these three inequalities, we get

(AB+AC) + (AC+BC) + (AB+BC) > OB+OC+OA+OB+OA+OC

i.e. 2(AB+BC+AC) > 2(OA+OB+OC)

Therefore, we have

AB+BC+AC > OA+OB+OC

(iii) In ∆OAB

OA+OB > AB ….(i)

In ∆OBC,

OB+OC > BC ….(ii)

And, in ∆OCA,

OC+OA > CA

Adding (i), (ii) and (iii) we get

(OA+OB) + (OB+OC) + (OC+OA) > AB+BC+CA

i.e 2(OA+OB+OC) > AB+BC+CA

⇒ OA+OB+OC > [latex]frac { 1 }{ 2 } [/latex] (AB+BC+CA)

**Question 45:**

Since AB=3cm and BC=3.5 cm

∴ AB+BC=(3+3.5) cm =6.5 m

And CA=6.5 cm

So AB+BC=CA

A triangle can be drawn only when the sum of two sides is greater than the third side.

So, with the given lengths a triangle cannot be drawn.

**Complete RS Aggarwal Solutions Class 9**

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