Question

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

Answer 1

Given Area of circle$=154c{m}^2$

$\Rightarrow \pi r^2=154c{m}^2\Rightarrow r^2=\frac{154\times 7}{22}=49cm$

$\Rightarrow r=\sqrt{49}=7cm$

ABC is an equilateral $△$, h is the altitude of $△ABC$

$0$ is the incenter of $△ABC$, and this is the point of intersection of the angular bisectors. Hence, these bisectors are also the altitude and medians whose point of intersection divides the medians in the ratio $2:1$

$\therefore \angle ADB=90°$ & $OD=\frac{1}{3}AD$ & $OD$ is radius of circle. Then,

$r=\frac{h}{3}\Rightarrow h=3r=3\times 7=21cm$

Let each side of an equilateral triangle be 'a', then altitude of an equilateral triangle is $\left(\frac{\sqrt{3}}{2}\right)$ times its side. So that

$h=\frac{\sqrt{3}}{2}a\Rightarrow a=\frac{2h}{\sqrt{3}}=\frac{2\times 21}{\sqrt{3}}=\frac{2\times 21\sqrt{3}}{3}=14\sqrt{3}cm$

Perimeter of triangle $ABC=3a=3\times 14\sqrt{3}cm=42\sqrt{3}cm=42\times 1.73$

$=72.66c{m}^2$