The angle in one regular polygon is to that in another as 3:2 and the

1.	The angle in one regular polygon is to that in another as 3:2 and the number of sides in first is twice that in the second. Determine the number of sides of two polygons.
Answer» Suppose the number of sides in the first polygon be 2x and The number of sides in the second polygon be x. As we know that, angle of an n-sided regular polygon = [(n - 2)/n]π radian The angle of the first polygon = [(2x - 2)/2x]π = [(x - 1)/x]π radian The angle of the second polygon = [(x - 2)/x]π radian Hence, [(x - 1)/x]π/[(x - 2)/x]π = 3/2 (x - 1)/(x - 2) = 3/2 Now, upon cross-multiplication we get, 2x – 2 = 3x – 6 3x - 2x = 6 - 2 x = 4 ∴ Number of sides in the first polygon = 2x = 2(4) = 8 Thus, the number of sides in the second polygon = x = 4

The angle in one regular polygon is to that in another as 3:2 and the number of sides in first is twice that in the second. Determine the number of sides of two polygons.

Answer»

Suppose the number of sides in the first polygon be 2x and

The number of sides in the second polygon be x.

As we know that, angle of an n-sided regular polygon = [(n - 2)/n]π radian

The angle of the first polygon = [(2x - 2)/2x]π = [(x - 1)/x]π radian

The angle of the second polygon = [(x - 2)/x]π radian

Hence,

[(x - 1)/x]π/[(x - 2)/x]π = 3/2
(x - 1)/(x - 2) = 3/2

Now, upon cross-multiplication we get,

2x – 2 = 3x – 6

3x - 2x = 6 - 2

x = 4

∴ Number of sides in the first polygon = 2x = 2(4) = 8

Thus, the number of sides in the second polygon = x = 4