GSEB Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2

Gujarat BoardĀ GSEB Textbook Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2

Question 1.
Find x in the following figures.
GSEB Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2
Solution:
(a) Sum of all the exterior angles of a polygon = 360Ā°
āˆ“ 125Ā° + 125Ā° + x = 360Ā°
or 250Ā° + x = 360Ā°
or x = 360Ā° – 250Ā° = 110Ā°

(b) āˆµ x + 90Ā° + 60Ā° + 90Ā° + 70Ā° = 360Ā°
or x + 310Ā° = 360Ā°
or x = 360Ā° – 310Ā° = 50Ā°

GSEB Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2

Question 2.
Find the measure of each exterior angle of a regular polygon of
(i) 9 sides
(ii) 15 sides
Solution:
(i) Number of sides (n) = 9
Number of exterior angles = 9
Since, sum of all the exterior angles = 360Ā°
āˆ“ The given polygon is a regular polygon.
āˆµ All the exterior angles are equal.
āˆ“ Measure of an exterior angle = \(\frac { 360Ā° }{ 9 }\) = 40Ā°

(ii) Number of sides of regular polygon =15
āˆ“ Number of equal exterior angles =15
The sum of all the exterior angles = 360Ā°
āˆ“ The measure of each exterior angle
= \(\frac { 360Ā° }{ 15 }\)

GSEB Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2

Question 3.
How many sides does a regular polygon have if the measure of an exterior angle is 24Ā°?
Solution:
For a regular polygon, measure of each angle is equal.
āˆ“ Sum of all the exterior angles = 360Ā°
Measure of an exterior angle = 24Ā°
āˆ“ Number of angles = \(\frac { 360Ā° }{ 24Ā° }\) =15
Thus, there are 15 sides of the polygon.

Question 4.
How many sides does a regular polygon have if each of its interior angles is 165Ā°?
Solution:
The given polygon is regular polygon.
Each interior angle = 165Ā°
Each exterior angle = 180Ā° – 165Ā° =15Ā°
Number of sides = \(\frac { 360Ā° }{ 15Ā° }\) = 24
Thus, there are 24 sides of the polygon.

GSEB Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2

Question 5.
(a) Is it possible to have a regular polygon with measure of each exterior angle 22Ā°?
(b) Can it be an interior angle of a regular polygon? Why?
Solution:
(a) Each exterior angle = 22Ā°
Number of sides = \(\frac { 360Ā° }{ 22Ā° }\) = \(\frac { 180 }{ 11 }\)
If it is a regular polygon, then its number of sides must be a whole number.
Here, \(\frac { 180 }{ 11 }\) is not a whole number.
āˆ“ 22Ā° cannot be an exterior angle of a regular polygon.

(b) If 22Ā° is an interior angle, then 180Ā° -22Ā°, i.e., 158Ā° is an exterior angle.
āˆ“ Number of sides = \(\frac { 360Ā° }{ 158Ā° }\) = \(\frac { 180Ā° }{ 79 }\)
which is not a whole number.
Thus, 22Ā° cannot be an interior angle of a regular polygon.

GSEB Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2

Question 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?
Solution:
(a) The minimum number of sides of a polygon 3
The regular polygon of 3-sides is an equilateral triangle.
āˆ“ Each interior angle of an equilateral triangle = 60Ā°
Hence, the minimum possible interior angle of a polygon = 60Ā°

(b) āˆµ The sum of an exterior angle and its corresponding interior angle is 180Ā°
And minimum interior angle of a regular polygon = 60Ā°
The maximum exterior angle of a regular polygon – 180Ā° – 60Ā° = 120Ā°

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