Question

A wire in the shape of an equilateral triangle encloses an area $s$ sq. cm. If the same wire is bent to form circle, the area of the circle will be

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

Answer 1

Answer: (D)

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

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