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 $=5\times \frac{1}{2}\times r\times r\cdot \sin \frac{4\pi }{10}=\frac{5}{8}\frac{a^2\sin \frac{4\pi }{10}}{{\sin }^2\frac{2\pi }{10}}$
$A_1=\frac{5}{4}{a}^2\cot \frac{2\pi }{10}$
$\cos \frac{4\pi }{10}=\frac{r^2+r^2-a^2}{2r^2}$
$\Rightarrow \cos \frac{4\pi }{10}=1-\frac{a^2}{2r^2}$
$\Rightarrow \frac{a^2}{2r^2}=2{\sin }^2\frac{2\pi }{10}$
$\Rightarrow r^2=\frac{a^2}{4{\sin }^2\frac{2\pi }{10}}$
For decagon , $\cos \frac{2\pi }{10}=\frac{r_1^2+r_1^2-{\left(\frac{a}{2}\right)}^2}{2r_1^2}$
$\Rightarrow \frac{a^2}{8r_1^2}=2{\sin }^2\frac{\pi }{10}$
$\Rightarrow r_1^2=\frac{a^2}{16{\sin }^2\frac{\pi }{10}}$
Area of decagon, $A_2=10\times \frac{1}{2}{r}_1^2\sin \frac{2\pi }{10}$
$=5\cdot \frac{a^2}{16{\sin }^2\frac{\pi }{10}}\cdot \sin \frac{2\pi }{10}$
$\Rightarrow A_2=\frac{5}{8}{a}^2{\cot }^{\frac{\pi }{10}}$
$A_1:A_2=2\cot \frac{2\pi }{10}:\cot \frac{\pi }{10}=2\cot \frac{\pi }{5}:\cot \frac{\pi }{10}$
$=\frac{2\cos \frac{\pi }{5}}{\sin \frac{\pi }{5}}\cdot \frac{\sin \frac{\pi }{10}}{\cos \frac{\pi }{10}}=\frac{\sin \frac{2\pi }{5}\sin \frac{\pi }{10}}{{\sin }^2\frac{\pi }{5}\cos \frac{\pi }{10}}=\frac{\cos \frac{\pi }{10}\sin \frac{\pi }{10}}{{\sin }^2\frac{\pi }{5}\cos \frac{\pi }{10}}=\frac{\frac{\sqrt{5}-1}{4}}{1-{\left(\frac{\sqrt{5}+1}{4}\right)}^2}=2:\sqrt{5}$