Question

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

Answer 1

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{a}\mathrm{n}\mathrm{equilateral}\mathrm{t}\mathrm{riangle}=\frac{\sqrt{3}}{4}\times {\left(\mathrm{Side}\right)}^2 \\ \Rightarrow 121\sqrt{3}=\frac{\sqrt{3}}{4}\times {\left(\mathrm{Side}\right)}^2 \\ \Rightarrow 121\times 4={\left(\mathrm{Side}\right)}^2 \\ \Rightarrow \mathrm{Side}=22\mathrm{cm} \end{gathered}$$

$\mathrm{Perimeter}\mathrm{of}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}=3\times \mathrm{Side}$
$$\begin{gathered} =3\times 22 \\ =66\mathrm{cm} \end{gathered}$$

Length of the wire$=66\mathrm{cm}$
Now, let the radius of the circle be r cm.
We know:
Circumference of the circle = Length of the wire
$$\begin{gathered} 2\mathrm{\pi }r\mathit{}=66 \\ \Rightarrow 2\times \frac{22}{7}\times r=66 \\ \Rightarrow r=\frac{66\times 7}{2\times 22} \\ \Rightarrow r=\frac{21}{2} \\ \Rightarrow r=10.5 \end{gathered}$$
Thus, we have:
Area of the circle =$\mathrm{\pi }{r}^{\mathit{2}}$
$$\begin{gathered} =\frac{22}{7}\times 10.5\times 10.5 \\ =346.5\mathrm{sq}.\mathrm{cm} \end{gathered}$$
Area enclosed by the circle = 346.5 cm²