Question

If the radius of the circumcircle of an equilateral triangle is 6 cm, then find the radius of its incircle.

Answer 1

Let a be the side of the equilateral triangle ABC.

$\therefore$ AB$=$BC$=$AC$=$a

and since $\Delta$ABC is an equilateral triangle

$\angle A=\angle B=\angle C={60}^o$

Given, radius of the circumcircle, $r=6cm$

Since the triangle ABC is equilateral, its perpendicular bisector i.e. the median meet at the same point O which is the centre of the in circle

$\therefore$ AD is the perpendicular bisector of BC.

$\Rightarrow$ BD$=$DC$=\frac{1}{2}$BC

$=\frac{a}{2}$

and OB is the angle bisector of $\angle B$.

$\therefore \angle$ABO $=\angle$ OBD $=\frac{{60}^o}{2}={30}^o$

In rt $\Delta$ ODB

$\sin {30}^o=\frac{OD}{OB}$

$\Rightarrow \frac{1}{2}=\frac{OD}{r}$

$\Rightarrow OD=\frac{r}{2}$

$\Rightarrow OD=\frac{6}{2}$

$\Rightarrow OD=3$cm

$\therefore$ Radius of the incircle $=3$cm.