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.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and Its Properties – Exercise 6.2

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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 = 50^o + 70^o

= x = 120^o

(ii)

Solution:-

We know that,

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

= x = 65^o + 45^o

= x = 110^o

(iii)

Solution:-

We know that,

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

= x = 30^o + 40^o

= x = 70^o

(iv)

Solution:-

We know that,

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

= x = 60^o + 60^o

= x = 120^o

(v)

Solution:-

We know that,

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

= x = 50^o + 50^o

= x = 100^o

(vi)

Solution:-

We know that,

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

= x = 30^o + 60^o

= x = 90^o

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 + 50^o = 115^o

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

= x = 115^o – 50^o

= x = 65^o

(ii)

Solution:-

We know that,

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

= 70^o + x = 100^o

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

= x = 100^o – 70^o

= x = 30^o

(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 90^o.

= x + 90^o = 125^o

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

= x = 125^o – 90^o

= x = 35^o

(iv)

Solution:-

We know that,

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

= x + 60^o = 120^o

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

= x = 120^o – 60^o

= x = 60^o

(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 90^o.

= x + 30^o = 80^o

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

= x = 80^o – 30^o

= x = 50^o

(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 90^o.

= x + 35^o = 75^o

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

= x = 75^o – 35^o

= x = 40^o

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NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7 

NCERT Solutions