Question

A wire when bent in the form of an equilateral triangle encloses an area of $121\sqrt{3}{\mathrm{cm}}^2$. If the same wire is bent into the form of a circle, what will be the area of the circle?

Answer 1

Let a cm be the side of the equilateral triangle.
Now,
Area of the equilateral triangle$=\frac{\sqrt{3}}{4}{a}^2$
We have:
$$\begin{gathered} \frac{\sqrt{3}}{4}{a}^2=121\sqrt{3} \\ \Rightarrow \frac{a^2}{4}=121 \\ \Rightarrow a^2=484 \\ \Rightarrow a=22 \end{gathered}$$

Perimeter of the triangle = Circumference of the circle
Perimeter of the triangle = (22 + 22 + 22) cm
= 66 cm
Now, let r cm be the radius of the circle.
We know:
Circumference of the circle$=2\mathrm{\pi }r$
$$\begin{gathered} \Rightarrow 2\times \frac{22}{7}\times r=66 \\ \Rightarrow r=\left(66\times \frac{7}{44}\right)\mathrm{cm} \\ \Rightarrow r=\frac{21}{2}\mathrm{cm} \end{gathered}$$
Also,
Area of the circle$=\mathrm{\pi }{r}^2$
$$\begin{gathered} =\left(\frac{22}{7}\times \frac{21}{2}\times \frac{21}{2}\right){\mathrm{cm}}^2 \\ =\frac{693}{2}{\mathrm{cm}}^2 \\ =346.5{\mathrm{cm}}^2 \end{gathered}$$