Question

A wire in the shape of an equilateral triangle encloses an area 'S' $c{m}^2$ If the same wire is bent to form a circle then the area of the circle will be

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

Answer 1

The correct option is D $\frac{3S\sqrt{3}}{\pi }$

Let the side of equilateral triangle an area S sq cm

Then S=$\frac{\sqrt{3}}{4}{a}^2=S\Rightarrow a^2=\frac{4}{\sqrt{3}}S\Rightarrow a=\sqrt{\frac{4S}{\sqrt{3}}}$

Then perimeter of equilateral triangle =3$\sqrt{\frac{4S}{\sqrt{3}}}$

The same wire is band as circle

Then perimeter of equilateral triangle = perimeter of circle

$\therefore 2\pi r=3\sqrt{\frac{4S}{\sqrt{3}}}\Rightarrow r=\frac{3}{\pi }\sqrt{\frac{S}{\sqrt{3}}}$ cm