## 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 > (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 > (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.

Praveen Chaubey says

The solutions of Rs aggarwal class 9 are from old book.please give from New book.

Santy says

Please don’t change keep it up

Because we have old books

Anand says

These are the solutions from fifth edition.

Please give from sixth edition.

Thanks

Divya says

Plzz upload your solution of rsagarwal class-9 from new adition book

Request

Kumar Nikhil says

Give the solution of new edition’s book

Aman says

Thanks for solution

Anshum says

Please give the solutions of the new book it do not contain all questions please please please

Rajnish Gupta says

Very nice.

Rajnish Gupta says

Very nice. .????????????

Anmol says

Pls upload the exercises of rs aggarwal latest addition

Arya says

Ye bhai new addition add kar ya bolo ki nahi ata hai

Tanisha says

Plz try to give solutions of the questions from the new edition of RSA . I will be a great thanks to u

Aditya keshri says

Very nice ????????????????

rs g says

new book solutions plz…no spaming

Bittu Kumar says

Thank you really very much for this wonderful solution

Tamanna Shree says

IT is nice to have solutions… but please have old ones also, and New ones too..????????????

Aakriti says

Very useful for us .if we have any doubt we can see this solution that is why we don’t need any teacher.

Aakriti says

Very useful for us if we have any doubt in question we can easily see it that is why we don’t need a teacher

Ry says

Thkew

Raj says

Give solution of sixth edition because we have sixth edition

Anish bhaskar says

Please give result of formative assessment

Anand says

Thanks