Understanding Quadrilaterals : Exercise 3.2 (Mathematics NCERT Class 8th)

Q.1 Find x in the following figures.Sol. (a) We know that the sum of all exterior angles of any polygon is 360áµ’
Hence, 125áµ’ + 125áµ’ + x = 360áµ’
250áµ’ + x = 360áµ’
x = 110áµ’

(b) We know that the sum of all exterior angles of any polygon is 360áµ’
Hence, 60áµ’ + 90áµ’ + 70áµ’ + x + 90áµ’ = 360áµ’
310áµ’ + x = 360áµ’
x = 50áµ’

Q.2 Find the measure of each exterior angle of a regular polygon of
(i) 9 sides (ii) 15 sides
Sol. (i) We know that the sum of all exterior angles of any polygon = 360áµ’
For any regular polygon, the measure of each exterior angle is same.
Hence, measure of each exterior angle of a regular polygon having 9 sides = ${{360^\circ } \over 9} = 40^\circ$

(ii) We know that the sum of all exterior angles of any polygon = 360áµ’
For any regular polygon, the measure of each exterior angle is same.
Hence, measure of each exterior angle of a regular polygon having 15 sides = ${{360^\circ } \over {15}} = 24^\circ$

Q.3 How many sides does a regular polygon have if the measure of an exterior angle is 24°?
Sol. We know that the sum of all exterior angles of any polygon = 360áµ’
Given, measure of an exterior angle = 24áµ’
Therefore, number of sides of the regular polygon = ${{360^\circ } \over {24^\circ }} = 15$

Q.4 How many sides does a regular polygon have if each of its interior angles is How many sides does a regular polygon have if each of its interior angles is 165°?
Sol. Let the number of sides be x.
Here, the exterior angle will be (180° -165°) = 15°
We know that the sum of all exterior angles of any polygon = 360áµ’
Therefore, number of sides of the regular polygon = ${{360^\circ } \over {15^\circ }} = 24$

Q.5 (a) Is it possible to have a regular polygon with measure of each exterior angle as 22°?
(b) Can it be an interior angle of a regular polygon? Why?
Sol. (a) No.
It is not possible to have a regular polygon with measure of each exterior angle as 22°.
Reason: The sum of all exterior angles of any polygon is 360áµ’. Hence, each exterior angle has to be multiple of 360áµ’.
Here, 22° is not the multiple of 360ᵒ and hence such polygon is not possible.

(b) No.
It is not possible to have a regular polygon with measure of interior angle as 22°.
Reason: The sum of all exterior angles of any polygon is 360áµ’. Hence, each exterior angle has to be multiple of 360áµ’.
Here, interior angle = 22°
Therefore, exterior angle = 180° - 22° = 158°
Here, 158° is not the multiple of 360ᵒ and hence such polygon is not possible.

Q.6 (a) What is the minimum interior angle possible for a regular polygon? Why?
(b) What is the maximum exterior angle possible for a regular polygon?
Sol. (a) A regular polygon with lowest possible number of equal sides is 3 (i.e. equilateral triangle) which is 60áµ’
Since, sum of all the angles of a triangle = 180áµ’
Hence, x + x + x = 180áµ’
3x = 180áµ’
x = 60áµ’
Hence, minimum interior angle possible for a regular polygon is 60áµ’.

(b) A regular polygon with lowest possible number of sides is 3.
We know that the sum of all exterior angles of any polygon = 360áµ’
So, exterior angle of triangle = ${{360^\circ } \over 3} = 120^\circ$
Hence, maximum exterior angle possible for a regular polygon is 120áµ’.