Qstnid -B-12- Ab, Bc And Cd Are Three Consecutive Sides Of A Regular Polygon. If Angle Bac = 20° ; Find : (I) Its Each Interior Angle, (Ii) Its Each Exterior Angle (Iii) The Number Of Sides In The Polygon. (I) Its Each Interior Angle, (Ii) Its Each Exterior Angle (Iii) The Number Of Sides In The Polygon. (I) Its Each Interior Angle, (Ii) Its Each Exterior Angle (Iii) The Number Of Sides In The Polygon. (Ii) Its Each Exterior Angle (Iii) The Number Of Sides In The Polygon. (Ii) Its Each Exterior Angle (Iii) The Number Of Sides In The Polygon. (I) Its Each Interior Angle, (Ii) Its Each Exterior Angle (Iii) The Number Of Sides In The Polygon. (Ii) Its Each Exterior Angle (Iii) The Number Of Sides In The Polygon. (Ii) Its Each Exterior Angle (Iii) The Number Of Sides In The Polygon. - 16: Understanding Shapes (Including Polygons) Class 8 Maths

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 5

Qstn# B-12 Prvs-Qstn Next-Qstn

#12
AB, BC and CD are three consecutive sides of a regular polygon. If angle BAC = 20° ; find : (i) its each interior angle, (ii) its each exterior angle (iii) the number of sides in the polygon. (i) its each interior angle, (ii) its each exterior angle (iii) the number of sides in the polygon. (i) its each interior angle, (ii) its each exterior angle (iii) the number of sides in the polygon. (ii) its each exterior angle (iii) the number of sides in the polygon. (ii) its each exterior angle (iii) the number of sides in the polygon. (i) its each interior angle, (ii) its each exterior angle (iii) the number of sides in the polygon. (ii) its each exterior angle (iii) the number of sides in the polygon. (ii) its each exterior angle (iii) the number of sides in the polygon.
Ans :
∵ Polygon is regular (Given)
∴ AB = BC
⇒ ∠BAC = ∠BAC [∠s opp. to equal sides]
But ∠BAC = 20˚
∴ ∠BCA = 20˚
i.e., In ΔABC,
∠B + ∠BAC + ∠BCA = 180˚
⇒ ∠B + 20˚ + 20˚ = 180˚
⇒ ∠B = 180˚ - 40˚
⇒ ∠B = 140˚ (i) each interior angle = 140˚ (ii) each exterior angle = 180˚ - 140˚ = 40˚ (iii) Let no. of sides = n
∴ 360˚/n = 40˚
⇒ n = 360˚/40˚ = 9
⇒ n = 9
∴ (i) 140˚ (ii) 9 (i) each interior angle = 140˚ (ii) each exterior angle = 180˚ - 140˚ = 40˚ (iii) Let no. of sides = n
∴ 360˚/n = 40˚
⇒ n = 360˚/40˚ = 9
⇒ n = 9
∴ (i) 140˚ (ii) 9 (i) each interior angle = 140˚ (ii) each exterior angle = 180˚ - 140˚ = 40˚ (iii) Let no. of sides = n
∴ 360˚/n = 40˚
⇒ n = 360˚/40˚ = 9
⇒ n = 9
∴ (i) 140˚ (ii) 9 (ii) each exterior angle = 180˚ - 140˚ = 40˚ (iii) Let no. of sides = n
∴ 360˚/n = 40˚
⇒ n = 360˚/40˚ = 9
⇒ n = 9
∴ (i) 140˚ (ii) 9 (ii) each exterior angle = 180˚ - 140˚ = 40˚ (iii) Let no. of sides = n
∴ 360˚/n = 40˚
⇒ n = 360˚/40˚ = 9
⇒ n = 9
∴ (i) 140˚ (ii) 9 (i) each interior angle = 140˚ (ii) each exterior angle = 180˚ - 140˚ = 40˚ (iii) Let no. of sides = n
∴ 360˚/n = 40˚
⇒ n = 360˚/40˚ = 9
⇒ n = 9
∴ (i) 140˚ (ii) 9 (ii) each exterior angle = 180˚ - 140˚ = 40˚ (iii) Let no. of sides = n
∴ 360˚/n = 40˚
⇒ n = 360˚/40˚ = 9
⇒ n = 9
∴ (i) 140˚ (ii) 9 (ii) each exterior angle = 180˚ - 140˚ = 40˚ (iii) Let no. of sides = n
∴ 360˚/n = 40˚
⇒ n = 360˚/40˚ = 9
⇒ n = 9
∴ (i) 140˚ (ii) 9

#12-i
its each interior angle, (ii) its each exterior angle (iii) the number of sides in the polygon. (ii) its each exterior angle (iii) the number of sides in the polygon.
Ans : each interior angle = 140˚ (ii) each exterior angle = 180˚ - 140˚ = 40˚ (iii) Let no. of sides = n
∴ 360˚/n = 40˚
⇒ n = 360˚/40˚ = 9
⇒ n = 9
∴ (i) 140˚ (ii) 9 (ii) each exterior angle = 180˚ - 140˚ = 40˚ (iii) Let no. of sides = n
∴ 360˚/n = 40˚
⇒ n = 360˚/40˚ = 9
⇒ n = 9
∴ (i) 140˚ (ii) 9

#12-ii
its each exterior angle
Ans : each exterior angle = 180˚ - 140˚ = 40˚

#12-iii
the number of sides in the polygon.
Ans : Let no. of sides = n
∴ 360˚/n = 40˚
⇒ n = 360˚/40˚ = 9
⇒ n = 9
∴ (i) 140˚ (ii) 9