Two regular polygons are such that the ratio between their number of sides is 1:2 and the ratio of measures - Maths - Quadrilaterals

Question

Two regular polygons are such that the ratio between their
number of sides is 1:2 and the ratio of measures of their interior
angles is 3:4 The number of sides of each polygon is

Harleen Birdi · Accepted Answer

Dear Student!! Let n1 be the number of sides of 1st polygon And n2 be the number of sides of 2nd polygon $∴ \frac{n_{1}}{n_{2}} = \frac{1}{2}$ (1) $Also, \frac{{{Measure of interior angle of 1}^{st} polygon}^{}}{{Measure of interior angle of 2}^{nd} polygon} = \frac{3}{4} (2)$ We know, for a regular polygon: $Measure of each interior angle= \frac{Sum of interior angles}{Number of sides} = \frac{(n - 2)}{n} \times 180 °, where n = number of sides$ Thus, $(2) \Rightarrow \frac{{(n}_{1} - 2) \times 180 °}{n_{1}} \div \frac{(n_{2} - 2) \times 180 °}{n_{2}} = \frac{3}{4} \Rightarrow \frac{{(n}_{1} - 2) n_{2}}{(n_{2} - 2) n_{1}} = \frac{3}{4} \Rightarrow (\frac{n_{1} - 2}{n_{2} - 2}) \times (\frac{2}{1}) = \frac{3}{4} [(1) \Rightarrow \frac{n_{1}}{n_{2}} = \frac{1}{2} \Rightarrow \frac{n_{2}}{n_{1}} = \frac{2}{1}] \Rightarrow \frac{n_{1} - 2}{n_{2} - 2} = \frac{3}{8} \Rightarrow 8 (n_{1} - 2) = 3 (n_{2} - 2) \Rightarrow 8 n_{1} + 16 = 3 (2 n_{1}) - 6 ((1) \Rightarrow n_{2} = 2 n_{1}) \Rightarrow 8 n_{1} - 6 n_{1} = - 6 + 16 \Rightarrow n_{1} = 5 Thus , n_{2} = n_{1} = 2 \times 5 = 10$ Hence, number of sides of each polygon is 5 and 10 respectively Cheers !!

Two regular polygons are such that the ratio between their number of sides is 1:2 and the ratio of measures of their interior angles is 3:4. The number of sides of each polygon is