Question

The area of a circle inscribed in an equilateral triangle is 154 cm². Find the perimeter of the triangle.

Answer 1

Let the radius of the inscribed circle be r cm.
Given:
Area of the circle = 154 ${\mathrm{cm}}^2$
We know:
Area of the circle =$\pi r^2$
$$\begin{gathered} \Rightarrow 154=\frac{22}{7}{r}^2 \\ \Rightarrow \frac{154\times 7}{22}=r^2 \\ \Rightarrow r^2=49 \\ \Rightarrow r=7 \end{gathered}$$
In a triangle, the centre of the inscribed circle is the point of intersection of the medians and altitudes of the triangle. The centroid divides the median of a triangle in the ratio 2:1.
Here,
AO:OD = 2:1

Now,
Let the altitude be h cm.
We have:
$$\begin{gathered} \angle ADB={90}^{\circ } \\ OD=\frac{1}{3}AD \\ OD=\frac{h}{3} \end{gathered}$$
$$\begin{gathered} \Rightarrow h=3r \\ \Rightarrow h=21 \end{gathered}$$

Let each side of the triangle be a cm.
$$\begin{gathered} \mathrm{In}\mathrm{the}\mathrm{right}-\mathrm{angled}∆\mathrm{ADB},\mathrm{we}\mathrm{have}: \\ A{B}^2=A{D}^2+D{B}^2 \\ {a}^2=h^2+{\left(\frac{a}{2}\right)}^2 \\ 4a^2=4h^2+a^2 \\ 3a^2=4h^2 \\ {a}^2=\frac{4h^2}{3} \\ a=\frac{2h}{\sqrt{3}} \\ a=\frac{42}{\sqrt{3}} \end{gathered}$$
∴ Perimeter of the triangle = 3a
​ $$\begin{gathered} =3\times \frac{42}{\sqrt{3}} \\ =\sqrt{3}\times 42 \\ =72.66\mathrm{cm} \end{gathered}$$