Question

A regular pentagon and regular decagon have the same perimeter then ratio of their areas is?

• A. $1:\sqrt{5}$
• B. $2:\sqrt{5}$
• C. $3:\sqrt{5}$
• D. $4:\sqrt{5}$

Answer 1

Answer: (B)

$2:\sqrt{5}$

Area of polygon $=\frac{n{a}^2\cot \left(\frac{\pi }{n}\right)}{4}$

$n=$ number of sides of polygon

Let each side of regular pentagon $=a$ unit

Let each side of regular decagon $=b$ unit

So perimeter of pentagon $=5a$ perimeter of decagon $=106$

Both have equal perimete $5a=106$ $a=2b-\cdots \left(1\right)$

Hence side of a pentagon is twice the side of decagon

Area of pentagon $=\frac{5a^2}{4}$ cot $\left(\frac{\pi }{5}\right)$

Area of $\mathrm{decagon}=\frac{{106}^2}{4}$ cot $\left(\frac{\pi }{10}\right)$

Put $a=2b$ from eq, 1 ,

Ratio $=\frac{5\times {46}^2\cot {36}^{\circ }}{{106}^2\cot {18}^{\circ }}=\frac{2\cot {36}^{\circ }}{\cot {18}^{\circ }}$

Ratio $=2\cos {36}^{\circ }\times \sin {18}^{\circ }(\sin 2\theta =2\sin \theta \cos \theta )$

$\sin {36}^{\circ }\times \cos {18}^{\circ }$

$\frac{2\cos {36}^{\circ }\times \sin {18}^{\circ }}{2\sin {18}^{\circ }\times \cos {18}^{\circ }\times \cos {18}^{\circ }}=2\cos {36}^{\circ }$

Use $2{\cos }^2\theta =1+\cos 2\theta$

Ratio $=2\cos {36}^{\circ }$ $1+\cos {36}^{\circ }$ $\cos {36}^{\circ }=\sqrt{5}+1$

Ratio $=\frac{2\left(\frac{\sqrt{5}+1}{4}\right)}{1+\frac{\sqrt{5}+1}{4}}=2\left(\frac{\sqrt{5}+1}{4}\right)\times \frac{4}{5+\sqrt{5}}$

Ratio $=\frac{2}{\sqrt{5}}$

$2:\sqrt{5}$

$B$ is correct answer.