A wire when bent in the form of an equilateral triangle encloses an area of

Solution:

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