Qstnid -B-2- Find The Number Of Sides In A Regular Polygon, If Its Each Interior Angle Is : (I) 160° (Ii) 135° (Iii) 1 1/5 Of A Right-angle - 16: Understanding Shapes (Including Polygons) Class 8 Maths - Icse-viii-mathematics

Home Learn Fast select Course

Show Unit/Chapter Change Unit Change Subject Last Menu

ICSE-VIII-Mathematics

16: Understanding Shapes (Including Polygons) Class 8 Maths

with Solutions - page 4

Qstn# B-2 Prvs-Qstn Next-Qstn

#2
Find the number of sides in a regular polygon, if its each interior angle is : (i) 160° (ii) 135° (iii) 1 1/5 of a right-angle
Ans : (i) Let no. of sides of regular polygon be n.
Each interior angle = 160˚
∴ (n - 2)/n × 180˚ = 360˚
⇒ 180˚n - 360˚ = 160n
⇒ 20n = 360˚
⇒ n = 18 (ii) No. of sides = n
Each interior angle = 135˚
(n - 2)/n × 180˚ = 135˚
⇒ 180n - 360˚ = 135n
⇒ 180n - 135n = 360˚
⇒ 45n = 360˚
⇒ n = 8 (iii) No. of sides = n
Each interior angle = 1 1/5 right angles
= 6/5 × 90˚
= 108˚
∴ (n - 2)/n × 180˚ = 108˚
⇒ 180n - 360 ˚ = 108n
⇒ 180n -108n = 360˚
⇒ 72n = 360˚
⇒ n = 5

#2-i
160°
Ans : Let no. of sides of regular polygon be n.
Each interior angle = 160˚
∴ (n - 2)/n × 180˚ = 360˚
⇒ 180˚n - 360˚ = 160n
⇒ 20n = 360˚
⇒ n = 18

#2-ii
135°
Ans : No. of sides = n
Each interior angle = 135˚
(n - 2)/n × 180˚ = 135˚
⇒ 180n - 360˚ = 135n
⇒ 180n - 135n = 360˚
⇒ 45n = 360˚
⇒ n = 8

#2-iii
1 1/5 of a right-angle
Ans : No. of sides = n
Each interior angle = 1 1/5 right angles
= 6/5 × 90˚
= 108˚
∴ (n - 2)/n × 180˚ = 108˚
⇒ 180n - 360 ˚ = 108n
⇒ 180n -108n = 360˚
⇒ 72n = 360˚
⇒ n = 5