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ICSE-VIII-Mathematics

16: Understanding Shapes (Including Polygons) Class 8 Maths

with Solutions - page 3
Qstn# B-5-i Prvs-QstnNext-Qstn
  • #5-i
    80° (ii) 40% of a right angle. (ii) 40% of a right angle. (ii) 40% of a right angle.
    Ans : Let no. of sides = n each exterior angle = 80°
    360˚/n = 80˚
    ⇒ n = 360˚/80˚
    ⇒ n = 9/2
    Which is not a whole number.
    Hence it is not possible to have a regular polygon whose each exterior angle is of 80°. (ii) Let number of sides = n
    Each exterior angle = 40% of a right angle
    = 40/100 × 90
    = 36˚
    n = 360˚/36˚
    ⇒ n = 10
    Which is a whole number.
    Hence it is possible to have a regular polygon whose each exterior angle is 40% of a right angle. (ii) Let number of sides = n
    Each exterior angle = 40% of a right angle
    = 40/100 × 90
    = 36˚
    n = 360˚/36˚
    ⇒ n = 10
    Which is a whole number.
    Hence it is possible to have a regular polygon whose each exterior angle is 40% of a right angle. (ii) Let number of sides = n
    Each exterior angle = 40% of a right angle
    = 40/100 × 90
    = 36˚
    n = 360˚/36˚
    ⇒ n = 10
    Which is a whole number.
    Hence it is possible to have a regular polygon whose each exterior angle is 40% of a right angle.
  • #5-ii
    40% of a right angle.
    Ans : Let number of sides = n
    Each exterior angle = 40% of a right angle
    = 40/100 × 90
    = 36˚
    n = 360˚/36˚
    ⇒ n = 10
    Which is a whole number.
    Hence it is possible to have a regular polygon whose each exterior angle is 40% of a right angle.