Question

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

Answer 1

Given area of inscribed circle = 154 $c{m}^2$
Let the radius of the incircle be r.
⇒ Area of this circle $=\pi r^2=154$
$\left(\frac{22}{7}\right)\times r^2=154$
⇒$r^2=154\times \left(\frac{7}{22}\right)=49$
∴ r = 7 cm
Recall that incentre of a circle is the point of intersection of the angular bisectors.
Given ABC is an equilateral triangle and AD = h be the altitude.
Hence these bisectors are also the altitudes and medians whose point of intersection divides the medians in the ratio 2: 1
∠ADB = 90° and OD = $\left(\frac{1}{3}\right)$ AD
That is $r=\left(\frac{h}{3}\right)$
$h=3r=3\times 7=21cm$
Let each side of the triangle be a, then
Altitude of an equilateral triangle is $\left(\frac{\sqrt{3}}{2}\right)$ times its side
ie, $h=\left(\frac{\sqrt{3}a}{2}\right)$ ​​​​​​​

$a=\frac{2h}{\sqrt{3}}=\frac{2\times 21}{\sqrt{3}}=\frac{2\times 21\sqrt{3}}{3}=14\sqrt{3}$
​​​​​​​
∴ $a=14\sqrt{3}cm$
We know that perimeter of an equilateral triangle = 3a
$=3\times 14\sqrt{3}=42\sqrt{3}$
$=42\times 1.73=72.66$ cm