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 $\sqrt{3}=1.73$ ]
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 $72.7 \mathrm{~cm}$.
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