Question

The area of a circle is $A_1$ and the area of a regular pentagon inscribed in the circle is $A_2$. Then $A_1:A_2$ is

• A. $\frac{\pi }{5}\cos \frac{\pi }{10}$
• B. $\frac{2\pi }{10}\sec \frac{\pi }{10}$
• C. $\frac{2\pi }{5}cosec\frac{2\pi }{5}$
• D. none of these

Answer 1

Answer: (C)

$\frac{2\pi }{5}cosec\frac{2\pi }{5}$

As shown in figure, the regular pentagon is made up of 5 such $\Delta sO{A}_1{A}_2$
Area of $\Delta O{A}_1{A}_2=\frac{1}{2}ha$ where $OB=h$ and side of pentagon $=a$
Now, $\Delta O{A}_{1}B\cong \Delta O{A}_{2}B$ (RHS criterion)
$\Rightarrow A_{1}B=A_{2}B=\frac{a}{2}$ and $\angle A_{1}OB=\frac{\pi }{5}$
$\therefore \sin \frac{\pi }{5}=\frac{a}{2r}\Rightarrow a=2rsin\frac{\pi }{5}$
$\cos \frac{\pi }{5}=\frac{h}{r}\Rightarrow h=r\cos \frac{\pi }{5}$

Area of pentagon $A_2=5\times \Delta O{A}_1{A}_2=5\times \frac{1}{2}2rsin\frac{\pi }{5}\times rcos\frac{\pi }{5}$

$=\frac{5}{2}{r}^2\sin \frac{2\pi }{5}$

Area of circle, $A_1=\pi r^2$

$\therefore A_1:A_2=\frac{\pi r^2}{\frac{5}{2}{r}^2\sin \frac{2\pi }{5}}$

$=\frac{2\pi }{5}cosec\frac{2\pi }{5}$

Hence, D is correct