Question

A wire in the shape of an equilateral triangle encloses an area of $s$ $c{m}^2$. 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

Answer: (D)

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

Let the side of the equilateral triangle $=a$ cm. Hence, its area s $=\frac{\sqrt{3}}{4}{a}^2{cm}^2=>a^2=\frac{4}{\sqrt{3}}s$
Let the radius of the circle formed $=r$ cm.
Since, the same wire was used to make both the shapes, their perimeters are the same.
Perimeter of the triangle $=$ Perimeter of the circle
$\therefore 3a=2\pi r$
$\therefore r=\frac{3a}{2\pi }$
Area of the circle $=\pi r^2=\pi {\left(\frac{3a}{2\pi }\right)}^2=\frac{3\times 3\times a^2}{4\pi }$
Substituting the value of $a^2$ , we get
$\therefore$ Area of the circle $=\frac{3\times 3\times \frac{4}{\sqrt{3}}s}{4\pi }=\frac{3\sqrt{3}s}{\pi }$