Question

A wire in the shape of an equilateral triangle encloses an area $Sc{m}^2$. If the same wire is bent to form a circle, the area of the circle will be

• A. $\frac{\pi S^2}{9}$
• B. $\frac{3S^2}{\pi }$
• C. $\frac{3S}{\pi }$
• D. $\frac{3\sqrt{3S}}{\pi }$

Answer 1

Answer: (D)

$\frac{3\sqrt{3S}}{\pi }$

If the side of the triangle be $a$ cm, then
$S=\frac{a^2\sqrt{3}}{4}$
or $a^2=\frac{4S}{\sqrt{3}}$
and the perimeter of the triangle $=3a$
$\therefore$ Circumference of the circle $=3a$
$\therefore$ If r be the radius of the circle, $2\pi r=3a$
$\therefore r=\frac{3a}{2\pi }$
$\therefore$ Area of the circle $=\pi r^2$
$=\pi \times \frac{9a^2}{4{\pi }^2}=\frac{9}{4\pi }\times \frac{4S}{\sqrt{3}}=\frac{3\sqrt{3S}}{\pi }$