Qstnid -A-3- Find The Number Of Sides In A Polygon If The Sum Of Its Interior Angles Is : (I) 900° (Ii) 1620° (Iii) 16 Right-angles (Iv) 32 Right-angles. (I) 900° (Ii) 1620° (Iii) 16 Right-angles (Iv) 32 Right-angles. - 16: Understanding Shapes (Including Polygons) Class 8 Maths - Icse-viii-mathematics

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ICSE-VIII-Mathematics

16: Understanding Shapes (Including Polygons) Class 8 Maths

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#3
Find the number of sides in a polygon if the sum of its interior angles is : (i) 900° (ii) 1620° (iii) 16 right-angles (iv) 32 right-angles. (i) 900° (ii) 1620° (iii) 16 right-angles (iv) 32 right-angles.
Ans : (i) Let no. of sides = n
Sum of angles of polygon = 900˚(n - 2) × 180˚ = 900˚
⇒ n - 2 = 900/180
⇒ n - 2 = 5
⇒ n = 5 + 2
⇒ n = 7 (ii) Let no. of sides = n
Sum of angles of polygon = 1620˚
(n - 2) × 180˚ = 1620˚
⇒ n - 2 = 1620/180
⇒ n - 2 = 9
⇒ n = 9 + 2
⇒ n = 11 (iii) Let no. of sides = n
Sum of angles of polygon = 16 right = 16 × 90 = 1440˚
(n - 2) × 180˚ = 1440˚
⇒ n - 2 = 1440/180˚
⇒ n - 2 = 8
⇒ n = 8 + 2
⇒ n = 10 (iv) Let no. of sides = n
Sum of angles of polygon = 32 right angles = 32 × 90 = 2880˚
(n × 2) × 180˚ = 2880
n - 2 = 2880/180
n - 2 = 16
n = 16 + 2
n =18 (i) Let no. of sides = n
Sum of angles of polygon = 900˚(n - 2) × 180˚ = 900˚
⇒ n - 2 = 900/180
⇒ n - 2 = 5
⇒ n = 5 + 2
⇒ n = 7 (ii) Let no. of sides = n
Sum of angles of polygon = 1620˚
(n - 2) × 180˚ = 1620˚
⇒ n - 2 = 1620/180
⇒ n - 2 = 9
⇒ n = 9 + 2
⇒ n = 11 (iii) Let no. of sides = n
Sum of angles of polygon = 16 right = 16 × 90 = 1440˚
(n - 2) × 180˚ = 1440˚
⇒ n - 2 = 1440/180˚
⇒ n - 2 = 8
⇒ n = 8 + 2
⇒ n = 10 (iv) Let no. of sides = n
Sum of angles of polygon = 32 right angles = 32 × 90 = 2880˚
(n × 2) × 180˚ = 2880
n - 2 = 2880/180
n - 2 = 16
n = 16 + 2
n =18

#3-i
900°
Ans : Let no. of sides = n
Sum of angles of polygon = 900˚(n - 2) × 180˚ = 900˚
⇒ n - 2 = 900/180
⇒ n - 2 = 5
⇒ n = 5 + 2
⇒ n = 7

#3-ii
1620°
Ans : Let no. of sides = n
Sum of angles of polygon = 1620˚
(n - 2) × 180˚ = 1620˚
⇒ n - 2 = 1620/180
⇒ n - 2 = 9
⇒ n = 9 + 2
⇒ n = 11

#3-iii
16 right-angles
Ans : Let no. of sides = n
Sum of angles of polygon = 16 right = 16 × 90 = 1440˚
(n - 2) × 180˚ = 1440˚
⇒ n - 2 = 1440/180˚
⇒ n - 2 = 8
⇒ n = 8 + 2
⇒ n = 10

#3-iv
32 right-angles.
Ans : Let no. of sides = n
Sum of angles of polygon = 32 right angles = 32 × 90 = 2880˚
(n × 2) × 180˚ = 2880
n - 2 = 2880/180
n - 2 = 16
n = 16 + 2
n =18