 # NCERT Solutions for Class 7 Maths Exercise 6.2 Chapter 6 The Triangle and its Properties

NCERT Solutions for Class 7 Maths Exercise 6.2 Chapter 6 The Triangle and its Properties in simple PDF are provided here. An exterior angle of a triangle and its property is the only topic covered in this exercise of NCERT Solutions for Maths Class 7 Chapter 6. As this is the second exercise, this will help the students to understand the main concepts of the triangle and its properties as well. By practising the NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties, students will be able to score good marks in Maths.

## Download the PDF of NCERT Solutions For Class 7 Maths Chapter 6 The Triangle and its Properties – Exercise 6.2     ### Access Other Exercises of NCERT Solutions For Class 7 Maths Chapter 6 – The Triangle and its Properties

Exercise 6.1 Solutions

Exercise 6.3 Solutions

Exercise 6.4 Solutions

Exercise 6.5 Solutions

### Access Answers to NCERT Class 7 Maths Chapter 6 – The Triangle and its Properties Exercise 6.2

1. Find the value of the unknown exterior angle x in the following diagram:

(i) Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 50o + 70o

= x = 120o

(ii) Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 65o + 45o

= x = 110o

(iii) Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 30o + 40o

= x = 70o

(iv) Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 60o + 60o

= x = 120o

(v) Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 50o + 50o

= x = 100o

(vi) Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x = 30o + 60o

= x = 90o

2. Find the value of the unknown interior angle x in the following figures:

(i) Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x + 50o = 115o

By transposing 50o from LHS to RHS it becomes – 50o

= x = 115o – 50o

= x = 65o

(ii) Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= 70o + x = 100o

By transposing 70o from LHS to RHS it becomes – 70o

= x = 100o – 70o

= x = 30o

(iii) Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

The given triangle is a right angled triangle. So the angle opposite to the x is 90o.

= x + 90o = 125o

By transposing 90o from LHS to RHS it becomes – 90o

= x = 125o – 90o

= x = 35o

(iv) Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

= x + 60o = 120o

By transposing 60o from LHS to RHS it becomes – 60o

= x = 120o – 60o

= x = 60o

(v) Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

The given triangle is a right angled triangle. So the angle opposite to the x is 90o.

= x + 30o = 80o

By transposing 30o from LHS to RHS it becomes – 30o

= x = 80o – 30o

= x = 50o

(vi) Solution:-

We Know That,

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

The given triangle is a right angled triangle. So the angle opposite to the x is 90o.

= x + 35o = 75o

By transposing 35o from LHS to RHS it becomes – 35o

= x = 75o – 35o

= x = 40o