# NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Exercise 6.3

NCERT Solutions for Class 7 Maths Exercise 6.3 Chapter 6 The Triangle and its Properties are available here in simple PDF. Angle sum property of a triangle is the only topic covered in this exercise of NCERT Solutions for Class 7 Maths Chapter 6. It is an essential material as it offers a wide range of questions that test the studentsâ€™ understanding of concepts. Students who aspire to score good marks in Maths refer to the NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties.

## NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties â€“ Exercise 6.3

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### Access answers to Maths NCERT Solutions for Class 7 Chapter 6 â€“ The Triangle and its Properties Exercise 6.3

1. Find the value of the unknown x in the following diagrams:

(i)

Solution:-

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= âˆ BAC + âˆ ABC + âˆ BCA = 180o

= x + 50o + 60o = 180o

= x + 110o = 180o

By transposing 110o from LHS to RHS it becomes â€“ 110o

= x = 180oÂ â€“ 110o

= x = 70o

(ii)

Solution:-

We know that,

The sum of all the interior angles of a triangle is 180o.

The given triangle is a right angled triangle. So the âˆ QPR is 90o.

Then,

= âˆ QPR + âˆ PQR + âˆ PRQ = 180o

= 90o + 30o + x = 180o

= 120o + x = 180o

By transposing 110o from LHS to RHS it becomes â€“ 110o

= x = 180oÂ â€“ 120o

= x = 60o

(iii)

Solution:-

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= âˆ XYZ + âˆ YXZ + âˆ XZY = 180o

= 110o + 30o + x = 180o

= 140o + x = 180o

By transposing 140o from LHS to RHS it becomes â€“ 140o

= x = 180oÂ â€“ 140o

= x = 40o

(iv)

Solution:-

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= 50o + x + x = 180o

= 50o + 2x = 180o

By transposing 50o from LHS to RHS it becomes â€“ 50o

= 2x = 180oÂ â€“ 50o

= 2x = 130o

= x = 130o/2

= x = 65o

(v)

Solution:-

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= x + x + x = 180o

= 3x = 180o

= x = 180o/3

= x = 60o

âˆ´The given triangle is an equiangular triangle.

(vi)

Solution:-

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= 90o + 2x + x = 180o

= 90o + 3x = 180o

By transposing 90o from LHS to RHS it becomes â€“ 90o

= 3x = 180oÂ â€“ 90o

= 3x = 90o

= x = 90o/3

= x = 30o

Then,

= 2x = 2 Ã— 30o = 60o

2. Find the values of the unknowns x and y in the following diagrams:

(i)

Solution:-

We Know That,

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

Then,

= 50o + x = 120o

By transposing 50o from LHS to RHS it becomes â€“ 50o

= x = 120o â€“ 50o

= x = 70oÂ

We also know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= 50o + x + y = 180o

= 50o + 70oÂ + y = 180o

= 120o + y = 180o

By transposing 120o from LHS to RHS it becomes â€“ 120o

= y = 180oÂ â€“ 120oÂ

= y = 60o

(ii)

Solution:-

From the rule of vertically opposite angles,

= y = 80o

Then,

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= 50o + 80o + x = 180o

= 130oÂ + x = 180o

By transposing 130o from LHS to RHS it becomes â€“ 130o

= x = 180oÂ â€“ 130oÂ

= x = 50o

(iii)

Solution:-

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= 50o + 60o + y = 180o

= 110oÂ + y = 180o

By transposing 110o from LHS to RHS it becomes â€“ 110o

= y = 180oÂ â€“ 110oÂ

= y = 70o

Now,

From the rule of linear pair,

= x + y = 180o

= x + 70o = 180o

By transposing 70o from LHS to RHS it becomes â€“ 70o

= x = 180o â€“ 70

= x = 110o

(iv)

Solution:-

From the rule of vertically opposite angles,

= x = 60o

Then,

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= 30o + x + y = 180o

= 30o + 60o + y = 180o

= 90oÂ + y = 180o

By transposing 90o from LHS to RHS it becomes â€“ 90o

= y = 180oÂ â€“ 90oÂ

= y = 90o

(v)

Solution:-

From the rule of vertically opposite angles,

= y = 90o

Then,

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= x + x + y = 180o

= 2x + 90o = 180o

By transposing 90o from LHS to RHS it becomes â€“ 90o

= 2x = 180oÂ â€“ 90oÂ

= 2x = 90o

= x = 90o/2

= x = 45o

(vi)

Solution:-

From the rule of vertically opposite angles,

= x = y

Then,

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= x + x + x = 180o

= 3x = 180o

= x = 180o/3

= x = 60o

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