Question
the angle in one regular polygon is to that in another as 3:2 and the number of the sides in first is twice that in second determine the the number of sides of 2 polygons
Answers
let the number of sides of the second be n
then the number of sides of the first is 2n
size of angle of the second = 180(n-2)/n
size of angle of the first = 180(2n-2)/(2n)
but that ratio is 3:2
[180(n-2)/n] / [180(2n-2)/(2n)] = 2/3
after some canceling and cross-multiplying ...
n = 4
one polygon is a square , the other is an octogon
check:
(4 sides and 8 sides)
each angle in a square is 90°, each angle in an octogon is 135°
135 : 90 = 3 : 2
then the number of sides of the first is 2n
size of angle of the second = 180(n-2)/n
size of angle of the first = 180(2n-2)/(2n)
but that ratio is 3:2
[180(n-2)/n] / [180(2n-2)/(2n)] = 2/3
after some canceling and cross-multiplying ...
n = 4
one polygon is a square , the other is an octogon
check:
(4 sides and 8 sides)
each angle in a square is 90°, each angle in an octogon is 135°
135 : 90 = 3 : 2
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