Question

An equilateral $\Delta$ABC of side 6 cm is inscribed in a circle. Find the radius of the circle.

Answer 1

Given: $\Delta$ABC is an equilateral triangle of side 6 cm.
Construct an altitude from vertex A.

We know that altitude of an equilateral triangle bisects the opposite side i.e., BD = DC = 3 cm.
Altitude bisects the chord BC, so it will pass through the centre of the circle.

In $\Delta ABD$
${\left(AB\right)}^2={\left(BD\right)}^2+{\left(AD\right)}^2$
$⟹{\left(6\right)}^2={\left(3\right)}^2+{(r+x)}^2$
$⟹36=9+{(r+x)}^2$
$⟹(r+x)=3\sqrt{3})$ ... (1)
In $\Delta BOD$
$⟹{\left(BO\right)}^2={\left(OD\right)}^2+{\left(BD\right)}^2$
$⟹{\left(r\right)}^2={\left(x\right)}^2+{\left(3\right)}^2$
$⟹{\left(r\right)}^2-{\left(x\right)}^2=9$
$⟹(r+x)(r-x)=9$

We know that $(r+x)=3\sqrt{3}$
$\therefore (r-x)=\frac{9}{r+x}=\frac{9}{3\sqrt{3}}$
$⟹r-x=\sqrt{3}$ ... (2)

Adding (1) and (2), we get
$r=2\sqrt{3}\text{cm}$