Regular Polygon Interior Angle problem 1 by Dhaval Furia

Geometry Level pending

Let $A$ and $B$ be two regular polygons having $a$ and $b$ sides, respectively. If $b = 2a$ and each interior angle of $B$ is $3/2$ times each interior angle of $A$ , then each interior angle, in degrees, of a regular polygon with $a + b$ sides is $_____$

The answer is 150.

1 solution

Dhaval Furia
May 14, 2020

Fact: Each interior angle (in degrees) of a regular polygon with n sides is (n-2)*180/n

Here, each interior angle of regular polygon with a and b sides, respectively is (a-2) 180/a and (b-2) 180/b

Accordingly, (b-2) 180/b = (3/2) (a-2)*180/a

Using the fact that b = 2a,

b-2 = 3(a-2)

2a - 2 = 3a - 6

a = 4 & b = 8

a + b = 12

So each interior angle of a regular polygon with a + b = 12 sides is (12-2)*180/12 = 150

Regular Polygon Interior Angle problem 1 by Dhaval Furia

The answer is 150.

1 solution

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