NCERT Class 9 Maths Chapter 7 Exercise 7.2

NCERT Class 9  Maths Chapter 7 Exercise 7.2

Here you will find complete solutions to all questions of maths class 9-chapter 7 exercise 7.2 along with the video. You can watch videos related to maths exercise 7.2 class 9 which will help you to understand the topic and questions of 9th class maths chapter 7 exercise 7.2.

Exercise 7.2 Solutions
Question- 1
Question-2
Question -3
Question- 4
Question-5
Question-6
Question-7
Questions-8
Video-solution

Class 9 Maths Chapter 7 Exercise 7.2 Question 1

Q1.In an isosceles triangle ABC, with AB=AC, the bisectors of `\right angle B` and `\right angle C` intersects each other at O. Join A to O. show that –

(i) OB=OC    (ii) AO bisects `\right angle A`

Exercise 7.2 Class 9 Question 1 Solution

 In ABC, we have: AB=AC

∠B=∠C (∵ Angles opposite to equal sides are equal)

or  `\frac{1}{2}∠B=frac{1}{2}∠C` 

∠OBC=∠OCB ………….(i)

 (∵OB and OC bisects ∠s  B and C respectively)

∴∠OBC=`\frac{1}{2}∠B and  ∠OCB=frac{1}{2}∠C` 

⇒OB=OC   ………….(ii)

(∵sides opposite to equal∠s are equal)

(ii)Now, in ∆s ABO and ACO, we have:

    AB=AC (given)

⟹∠ABC= ∠ACB (Angles opposite equal sides)

∠OBC= ∠OCB (from (i))

∠ABO= ∠ACO (∠ABO= ∠ABC-∠OBC)

(∠ACO= ∠ACB-∠OCB)

OB=OC  [(∵∠OBC=∠OCB)]

⇒OB=OC(sides opp equal angles) 

∴ By SAS criteriono of congruence ,we have:

∆ ABO=∆ACO 

∠BAO=∠CAO (CPCT)

Thus AO bisects `\angle A`

Class 9 Maths Chapter 7 Exercise 7.2 Question 2

In ∆ABC,AD is the perpendicular bisector of BC. Show that ∆ABC is an isosceles triangle in which AB=AC

Exercise 7.2 Class 9 Question 2 Solution 

In ∆s ABD  and ACD,we have:

DB=DC(given)

∠ADB=∠ADC (∵AD⊥BC.so each=90°)

AD=AD (common)

∴ By SAS criterion of congruence, we have:

∆ABD=∆ACD 

So, AB=AC

(∵Corresponding parts of congruent triangles are equal)

Hence, ∆ABC is an isosceles

Class 9 Maths Chapter 7 Exercise 7.2 Question 3

ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB respectively. Show that these altitudes are equal.

Exercise 7.2 Class 9 Question 3 Solution

Let BE⊥AC and CF⊥AB

In ∆s ABE and ACF, we have:

∠AEB=∠AFC     (∵each=90°)

∠A=∠A               (common)

And, AB=AC     (given)

By AAS criterion of congruence, we have:

∆ABE=∆ACF 

BE=CF 

(∵Corresponding parts of congruent triangles are equal.)

 

Class 9 Maths Chapter 7 Exercise 7.2 Question 4

ABC is a triangle in which altitude BE and CF to sides AC and AB are equal. Show that 

(i)∆ABE≅∆ACF

(ii).AB=AC.ie.ABC is an isosceles triangles.

Exercise 7.2 Class 9 Question 4 Solution

In ∆s ABE and ACF, we have:

∠AEB=∠AFC       (∵each=90°)

∠BAE=∠CAF      (common)

BE=CF                 (given )

∴ By AAS criterion of congruence, we have:

∆ABE=∆ACF 

So,  AB=AC

(∵corresponding parts of congruent triangles are equal)

Hence, ∆ABC is isosceles.

Class 9 Maths Chapter 7 Exercise 7.2 Question 5

ABC and DBC  are two isosceles triangles on the same base BC. Show that ∠ABD=∠ACD.

Exercise 7.2 Class 9 Question 5 Solution

In ∆ABC, we have∶AB=AC

∠ABC=∠ACB ………………..(i)

(∵Angles opposite to equal sides are equal)

In ∆DBC ,we have:BD=CD

∠DBC=∠DCB ………………..(ii)

(∵Angles opposite to equal sides are equal)

Adding (i) and (ii), we have

∠ABC+∠DBC=∠ACB+ ∠DCB 

⇒∠ABD=∠ACD 

Class 9 Maths Chapter 7 Exercise 7.2 Question 6

∆ABC is an isosceles triangle in which AB=AC, side BA is produced to D such that AD=AB. Show that∠BCD  is a right angle.

Exercise 7.2 Class 9 Question 6 Solution

In ∆ABC, we have: AB=AC

∠ACB=∠ABC ………………….(i)

(∵Angles opposite to equal sides are equal) 

Now,AB=AD        (given)

∴ AD=AC            (∵AB=AC)

Thus in ∆ADC, we have: AD=AC

∠ACD=∠ADC ………………….(ii)

 (∵Angles opposite to equal sides are equal)

Adding (i) and (ii), we get

∠ACB+∠ACD=∠ABC+ ∠ADC 

Or ∠BCD=∠ABC+ ∠BDC (∵∠ADC=∠BDC)

So, ∠BCD+∠BCD=∠ABC+∠BDC+∠BCD

(Adding ∠BCD on both sides)

∴2 ∠BCD=180° 

 ∠BCD=90°

Hence, ∠BCD is a right angle. 

Class 9 Maths Chapter 7 Exercise 7.2 Question 7

ABC is a right-angle triangle in which ∠A=90° and AB=AC. Find ∠B and ∠C.

Exercise 7.2 Class 9 Question 7 Solution

Given

AB=AC and ∠A=90° 

∴   ∠C=∠B (Angles opposite  to equal sides of a triangle) 

in  ∆ABC,   ∠A+∠B+∠C=180°    (Angle sum property)

⟹   90°+∠B+∠C=180° 

⟹   2∠B=180°-90° 

⟹   2∠B=90° 

⟹   ∠B=`\frac{90°}{2}=45°` 

 ∴ ∠B=∠C=45° 

Class 9 Maths Chapter 7 Exercise 7.2 Question 8

Show that the angles of an equilateral triangle are 60° each.

Exercise 7.2 Class 9 Question 8 Solution

Let us take ∆ABC is an equilateral triangle

Then AB=BC=AC

Now, AB=AC⟹∠C=∠B (Angle opposite to equal sides of a triangle)

Again AC=BC ⟹∠B=∠A   (Angle opposite to equal sides of a triangle)

So we have ∠A=∠B=∠C

Now in ∆ABC  ∠A+∠B+∠C=180° (Angle sum property)

⟹      ∠A+∠A+∠A=180° 

⟹      3∠A=180° 

⟹      3∠A=`\frac{180°}{3}=60°` 

    ∴∠A+∠B+∠C=60° 

Hence, all interior angles in an equilateral triangle are of 60°

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Exercise 7.2 Class 9 Maths Video

Related  Topics 

1. Exercise 7.1

2.Exercise 7.3