The ratio between the number of sides of two regular polygons is 3 : 4 and the ratio between the sum of their interior angles is 2 : 3. Find the number of sides in each polygon.

It is given that the ratio between the number of sides of two regular polygons is 3 : 4.

Let the number of sides of the first polygon be 3n and the number of sides of the second polygon be 4n.

The sum of the interior angles of a polygon is (2n - 4) x 90°.

Sum of interior angles of the first polygon = (2 x 3n - 4) x 90°

= (6n - 4) x 90°

Sum of interior angles of the second polygon = (2 x 4n - 4) x 90°

= (8n - 4) x 90°

It is also given that the ratio between the sum of their interior angles is 2 : 3.

$⇒ \dfrac{(6n - 4) \times 90°}{(8n - 4) \times 90°} = \dfrac{2}{3}\\[1em] ⇒ \dfrac{(6n - 4) \times \cancel{90°}}{(8n - 4) \times \cancel{90°}} = \dfrac{2}{3}\\[1em] ⇒ \dfrac{6n - 4}{8n - 4} = \dfrac{2}{3}\\[1em]$

By cross multiplying, we get

⇒ 3(6n - 4) = 2(8n - 4)

⇒ 18n - 12 = 16n - 8

⇒ 18n - 16n = 12 - 8

⇒ 2n = 4

⇒ n = $\dfrac{4}{2}$

⇒ n = 2

The number of sides of the first polygon is = 3n = 3 x 2 = 6

The number of sides of the second polygon is = 4n = 4 x 2 = 8

Hence, the number of sides are 6 and 8.