Question

The area of a circle inscribed in an equilateral triangle is $154c{m}^2$. Find the perimeter of the triangle. Take $\sqrt{3}=1.73$

Answer 1

Area of circle =$\pi r^2$ (where r be the radius of circle)

=$\frac{22}{7}\times r^2$

Given: area of a circle =$154c{m}^2$

$\frac{22}{7}\times r^2=154$

$r^2=154\times \frac{7}{22}$

or $r=7$

Radius of the circle is $7cm$.

Let us consider each side of the equilateral triangle is $“a”$ and height is $h$, then

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

or $h=3r=3\times 7=21$

Height of the triangle is $21cm$

From right triangle APC:

$h^2=a^2-(\frac{a}{2}{)}^2=\frac{3a^2}{4h}$

$h=\frac{\sqrt{3}}{2}\left(a\right)$

$21=\frac{\sqrt{3}}{2}\times a$ (putting value of $h$ )

$a=14\sqrt{3}cm$

Perimeter of the triangle = Sum of all the sides$=3a=3\times 14\sqrt{3}=42\times 1.73=72.66cm$