Question

If a regular pentagon and a regular decagon have the same perimeters, prove that their areas are as $2:\sqrt{5}$.

Answer 1

Given Perimeter of regular pentagon and regular decagon is same

Let the perimeter be $10a$

number of sides in regular pentagon is $n_1=5$

number of sides in regular decagon is $n_2=10$

As we know that

Perimeter of regular polygon with $n$ sides of length $a$ is $na$

Hence

Length of a side of regular pentagon is $\frac{10a}{5}=2a$

Length of a side of regular decagon is $\frac{10a}{10}=a$

As we know that

Area of regular polygon with $n$ sides of length $a$ is $\frac{n{a}^2}{4}\cot \frac{{180}^{\circ }}{n}$

Hence

Area of regular pentagon is $5\times \frac{(2a{)}^2}{4}\times \cot \frac{{180}^{\circ }}{5}=5a^2\cot {36}^{\circ }=5a^2\times \frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}$

Area of regular decagon is $10\times \frac{(a{)}^2}{4}\times \cot \frac{{180}^{\circ }}{10}=\frac{5a^2}{2}\cot {18}^{\circ }=\frac{5a^2}{2}\times \frac{\sqrt{10+2\sqrt{5}}}{\sqrt{5}-1}$

Ratio of areas of regular pentagon and decagon $=5a^2\times \frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}}:\frac{5a^2}{2}\times \frac{\sqrt{10-2\sqrt{5}}}{\sqrt{5}-1}$

$=\frac{2\times (\sqrt{5}+1)(\sqrt{5}-1)}{\left(\sqrt{10-2\sqrt{5}}\right)(\sqrt{10+2\sqrt{5}}}$

$=\frac{2\times \left(\right(\sqrt{5}{)}^2-\left(1{)}^2\right)}{\sqrt{{10}^2-(2\sqrt{5}{)}^2}}$

$=\frac{2\times (5-1)}{\sqrt{100-20}}$ $(\because a^2-b^2=(a+b\left)\right(a-b\left)\right)$

$=\frac{8}{\sqrt{80}}=\frac{2}{\sqrt{5}}$

$\therefore$ Ratio of areas of regular pentagon and decagon is $2:\sqrt{5}$

Hence Proved