Question

A wire when bent in the form of an equilateral triangle encloses an area of $121\sqrt{3}c{m}^2$. The same wire is bent to form a circle. Find the area enclosed by the circle.

Answer 1

Area of an equilateral $\Delta =\frac{\sqrt{3}}{4}\times a^2$ where a = side

Now, given that -

$\frac{\sqrt{3}}{4}\times a^2=121\sqrt{3}$

$\Rightarrow a^2=121\times 4$

$\Rightarrow a=11\times 2=22$

Therefore, length of wire = a + a + a = 66

Now, this wire is bent into a circle, so the circumference of the circle $=2\pi r=66$

$\Rightarrow 2\times \left(\frac{22}{7}\right)r=66$

$\Rightarrow r=\frac{21}{2}$

Now, the area of this circle $=\pi r^2$

$=\left(\frac{22}{7}\right)\times (\frac{21}{2}{)}^2=\frac{693}{2}=346.5cm²$