 # The area of a circle inscribed in an equilateral Question:

The area of a circle inscribed in an equilateral triangle is $154 \mathrm{~cm}^{2}$. Find the perimeter of the triangle. [Use $\pi=22 / 7$ and $\sqrt{3}=1.73$ ]

Solution:

It is given that the area A of circle inscribed in an equilateral triangle is 154 cm2.

We know that the area A of circle inscribed in an equilateral triangle is

$A=\pi r^{2}$

Now, we will find the value of r.

Substituting the value of area,

$154=3.14 \times r^{2}$

$r^{2}=\frac{154}{3.14}$

$r^{2}=49$

$r=\sqrt{49}$

$r=7 \mathrm{~cm}$

Let the height of triangle be h. Then

$r=\frac{h}{3}$

$h=3 r$

$=3 \times 7$

$=21 \mathrm{~cm}$

If a is the side of triangle, then

$h=\frac{\sqrt{3}}{2} a$

$a=\frac{2 h}{\sqrt{3}}$

Substituting the value of h,

$a=\frac{2 \times 21}{\sqrt{3}}$

$=14 \sqrt{3} \mathrm{~cm}$

perimeter of triangle $=3 a$

$=3 \times 14 \sqrt{3}$

$=42 \times 1.732$

$=72.7 \mathrm{~cm}$

Hence perimeter of triangle is $72.7 \mathrm{~cm}$.