Question

The numbers of sides of two regular sides of two regular polygons are in the ratio 2: 1 and their interior angles are in the ratio 4: 3, find the sum of the number of sides of the polygons.

• A. 10
• B. 12
• C. 15
• D. 18

Answer 1

The correct option is C

15

Let the number of sided in polygon A and B be x and y respectively.

As per question, ratio of their side is 1: 2

So, x : y = 1: 2

x = 2y ------- (i)

The ratio of their interior angle is 4: 3.

Interior angle of polygon A = $2\left(\frac{x-2}{x}\right)\times {90}^{\circ }$

Interior angle of polygon B = $2\left(\frac{y-2}{y}\right)\times {90}^{\circ }$

$\frac{\left(2\right(\frac{x-2}{x})\times {90}^{\circ })}{\left(2\right(\frac{y-2}{y})\times {90}^{\circ })}$ = $\frac{4}{3}$

$\Rightarrow$ $\frac{(x-2)y}{(y-2)x}$ = $\frac{4}{3}$

$\Rightarrow$ 3 yx – 6y = 4xy – 8x

Substitute the value of x from(i)

$\Rightarrow$ 6y${}^2$​ - 6y = 8y${}^2$​ -16y

⇒ 2y${}^2$​ = 10y

⇒ y = 5

x = 2y = 10

Hence, sum of the sides of polygons = 10 + 5 = 15.