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

Solution:

Let a cm be the side of the equilateral triangle.
Now,

Area of the equilateral triangle $=\frac{\sqrt{3}}{4} a^{2}$

We have:

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

$\Rightarrow \frac{a^{2}}{4}=121$

$\Rightarrow a^{2}=484$

 

$\Rightarrow a=22$

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 $=2 \pi r$

$\Rightarrow 2 \times \frac{22}{7} \times r=66$

$\Rightarrow r=\left(66 \times \frac{7}{44}\right) \mathrm{cm}$

 

$\Rightarrow r=\frac{21}{2} \mathrm{~cm}$

Also,

Area of the circle $=\pi r^{2}$

$=\left(\frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}\right) \mathrm{cm}^{2}$

$=\frac{693}{2} \mathrm{~cm}^{2}$

 

$=346.5 \mathrm{~cm}^{2}$