Question

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 1

Let n and m be the number of sides in two regular polygon respectively.

We know that each angle of n - sided regular polygon is $\frac{(2n-4)}{n}$ right angles.

Now,

According to the question,

$\frac{\left(\frac{2n-4}{n}\right)\times {90}^{\circ }}{\left(\frac{2m-4}{m}\right)\times {90}^{\circ }}=\frac{3}{2}$
$\Rightarrow \frac{(2n-4)m}{(2m-4)n}=\frac{3}{2}$ . . .(i)

Also,

n = 2m . . .(ii) [given]

Put (ii) in (i), we get

$\frac{(4m-4)m}{(2m-4)2m}=\frac{3}{2}$

$\Rightarrow$ 4m - 4 = 6m - 12

$\Rightarrow$ 2m = 8

$\therefore$ m = 4

From (ii)

n = 2m

$=2\times 4=8$

$\therefore$ n = 8, m = 4