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.
Area of an equilateral triangle $=\frac{\sqrt{3}}{4} \times(\text { Side })^{2}$
$\Rightarrow 121 \sqrt{3}=\frac{\sqrt{3}}{4} \times(\text { Side })^{2}$
$\Rightarrow 121 \times 4=(\text { Side })^{2}$
$\Rightarrow$ Side $=22 \mathrm{~cm}$
Perimeter of an equilateral triangle $=3 \times$ Side
$=3 \times 22$
$=66 \mathrm{~cm}$
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
$2 \pi r=66$
$\Rightarrow 2 \times \frac{22}{7} \times r=66$
$\Rightarrow r=\frac{66 \times 7}{2 \times 22}$
$\Rightarrow r=\frac{21}{2}$
$\Rightarrow r=10.5$
Thus, we have:
Area of the circle $=\pi r^{2}$
$=\frac{22}{7} \times 10.5 \times 10.5$
$=346.5$ sq. $\mathrm{cm}$
Area enclosed by the circle = 346.5 cm2
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