# NCERT Solution class 7 Mathematics EXERCISE- 6.3

# CHAPTER – 6

# EXERCISE – 6.3

**Q1. Find the value of the unknown x in the following diagrams:**

**SOLUTOIN:-**

**(i) **We know that,

The sum of all the interior angles of a triangle is 180^{0}

Then,

= x + 50^{0} + 60^{0} = 180^{0}

= x + 110^{0 } = 180^{0}

= x = 180^{0} – 110^{0}

= x = 70^{0}

**(ii) **We know that,

The sum of al the interior angles of a triangle is 180^{0}.

The given triangle is a right – angled triangle , So the Angle QPR is 90^{0}

= 90^{0} + 30^{0} + x = 180^{0}

= 120^{0} + x = 180^{0}

= x

^{0}– 120

^{0}

= x = 60^{0}

**(iii) **We know that ,

The sum of interior angles of a triangle is 180^{0}

= 110^{0} + 30^{0} + x = 180^{0}

= 140^{0} + x = 180^{0}

= x = 180^{0} – 140^{0}

= x = 40^{0}

**(iv) **We know that,

The sum of all the interior angle of a triangle is 180^{0}.

Then,

= 50^{0} + x + x = 180^{0}

= 50^{0} + 2x = 180^{0}

= 2x = 180^{0} – 50^{0}

= 2x = 130^{0}

= x = 130^{0}/2

= x = 65^{0}

**(v) **We know that,

The sum of all interior angles of a triangle is 180^{0}

Then,

= x + x + x = 180^{0}

= 3x = 180^{0}

= x = 180^{0}/3

= x = 60^{0}

**(vi) **We know that,

The sum of all the interior angles of a triangle is 180^{0}.

Then,

= 90^{0} + x + 2x = 180^{0}

= 90^{0} + 3x = 180^{0}

= 3x = 180^{0} – 90^{0}

= 3x = 90^{0}

= x = 90^{0}/3

= x = 30^{0}

Then,

= 2x = 2 x 30^{0} = 60^{0}

**Q2. Find the values of the unknown x and y in the following diagrams:**

** **

**SOLUTION:-**

**(i) **We know that,

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

Then,

= 50^{0} + x = 120^{0}

By transposing 50^{0} from LHS to RHS it becomes 50^{0}

= x = 120^{0} – 50^{0}

= x = 70^{0}

We also know that,

The sum of all interior angles of a triangle is 180^{0}.

Then,

= 50^{0} + x + y = 180^{0}

= 50^{0} + 70^{0} + y = 180^{0}

= 120^{0} + y = 180^{0}

= y = 180^{0} – 120^{0}

= y = 60^{0}

**(ii) **From the rule of vertically opposite angles,

= y = 80^{0}

Then,

We know that,

The sum of all the interior angles of a triangle is 180^{0}

Then,

We know that,

The sum of all the interior angles of a triangle is 180^{0}

Then,

= 50^{0} + 80^{0} + x = 180^{0}

= 130^{0} + x = 180^{0}

By transposing 130^{0} from LHS to RHS it becomes 130^{0}

= x = 180^{0} – 130^{0}

= x = 50^{0}

**(iii) **We know that,

The sum of all the interior angles of a triangle is 180^{0}

Then,

= 50^{0} + 60^{0} + y = 180^{0}

= 110^{0} + y = 180^{0}

By transposing 110^{0} from LHS to RHS it becomes 110^{0}

= y = 180^{0} – 110^{0}

= y = 70^{0}

Now,

From the rule of linear pair,

= x + y = 180^{0}

= x + 70^{0} = 180^{0}

= x = 180^{0} – 70^{0}

= x = 110^{0}

**(iv) **From the rule of vertically opposite angles,

= x = 60^{0}

Then,

We know that,

The sum of all the interior angles of a triangle is 180^{0}

Then,

= 30^{0} + x + y = 180^{0}

= 30^{0} + 60^{0} + y = 180^{0}

= 90^{0} + y = 180^{0}

= y = 180^{0} – 90^{0}

= y = 90^{0}

**(v) **From the rule of vertically opposite angles,

= y = 90^{0}

Then,

We know that,

The sum of all the interior angles of a triangle is 180^{0}.

Then,

= x + x + y = 180^{0}

= 2x + 90^{0} = 180^{0}

= 2x = 180^{0} – 90^{0}

= 2x = 90^{0}

= x = 90^{0}/2

= x = 45^{0}

**(vi) **From the rule of vertically opposite angles,

= x = y

Then,

We know that ,

The sum of all the interior angles of a triangle is 180^{0}

Then,

= x + x + x = 180^{0}

= 3x = 180^{0}

= x = 180^{0}/3

= x = 60^{0}