Question

A regular pentagon and a regular decagon have the same perimeter, the ratio of their areas is

• A. $3: \sqrt{5}$
• B. $1: \sqrt{5}$
• C. $2: \sqrt{5}$ .
• D. $2: \sqrt{5}$

Answer 1

Answer: (D)

Area of pentagon $=5 \times \frac{1}{2} \times r \times r \cdot \sin \frac{4 \pi}{10}=\frac{5}{8} \frac{a^2 \sin \frac{4 \pi}{10}}{\sin ^2 \frac{2 \pi}{10}}$
$A_1=\frac{5}{4} a^2 \cot \frac{2 \pi}{10}$
$\cos \frac{4 \pi}{10}=\frac{r^2+r^2-a^2}{2 r^2} $
$ \Rightarrow \cos \frac{4 \pi}{10}=1-\frac{a^2}{2 r^2} $
$\Rightarrow \frac{a^2}{2 r^2}=2 \sin ^2 \frac{2 \pi}{10} $
$\Rightarrow r^2=\frac{a^2}{4 \sin ^2 \frac{2 \pi}{10}}$
For decagon , $\cos \frac{2 \pi}{10}=\frac{r_1^2+r_1^2-\left(\frac{a}{2}\right)^2}{2 r_1^2}$
$ \Rightarrow \frac{a^2}{8 r_1^2}=2 \sin ^2 \frac{\pi}{10}$
$ \Rightarrow r_1^2=\frac{a^2}{16 \sin ^2 \frac{\pi}{10}}$
Area of decagon, $A_2=10 \times \frac{1}{2} r_1^2 \sin \frac{2 \pi}{10}$
$ =5 \cdot \frac{a^2}{16 \sin ^2 \frac{\pi}{10}} \cdot \sin \frac{2 \pi}{10}$
$ \Rightarrow A_2=\frac{5}{8} a^2 \cot ^{\frac{\pi}{10}}$
$A_1: A_2=2 \cot \frac{2 \pi}{10}: \cot \frac{\pi}{10}=2 \cot \frac{\pi}{5}: \cot \frac{\pi}{10}$
$=\frac{2 \cos \frac{\pi}{5}}{\sin \frac{\pi}{5}} \cdot \frac{\sin \frac{\pi}{10}}{\cos \frac{\pi}{10}}=\frac{\sin \frac{2 \pi}{5} \sin \frac{\pi}{10}}{\sin ^2 \frac{\pi}{5} \cos \frac{\pi}{10}}=\frac{\cos \frac{\pi}{10} \sin \frac{\pi}{10}}{\sin ^2 \frac{\pi}{5} \cos \frac{\pi}{10}}=\frac{\frac{\sqrt{5}-1}{4}}{1-\left(\frac{\sqrt{5}+1}{4}\right)^2}=2: \sqrt{5}$