Question

An equilateral triangle $ABC,$ whose side is $6cm,$ is inscribed in a circle. Find the radius of the circle.

• A. $2\sqrt{3}cm$
• B. $2\sqrt{2}cm$
• C. $3\sqrt{3}cm$
• D. $3cm$

Answer 1

The correct option is A

$2\sqrt{3}cm$

Given, that $ABC$ is an equilateral triangle of side $6cm$
Let 'r' be the radius of the circle.
Construct an altitude from point 'A' to point 'D' in $\Delta ABC$

In an equilateral triangle the altitute will also be the perpendicular bisector. Perpendicular bisectors intersect at circumcentre. So, in the below fig. 'O' is the centre of the circle.

In $\Delta ABD$
$A{B}^2=B{D}^2+A{D}^2$ (by Pythagoras theorem)
$\Rightarrow 6^2=3^2+{(r+x)}^2$
$\Rightarrow 36=9+{(r+x)}^2$
$\Rightarrow r+x=3\sqrt{3}$ ... (i)

OD = $x$ = $(r+x)-r$
= $3\sqrt{3}-r$

In $△BOD,$
${BO}^2={OD}^2+{BD}^2$
$\Rightarrow r^2={(3\sqrt{3}-r)}^2+3^2$
$\Rightarrow r^2=27-6\sqrt{3}r+r^2+9$
$\Rightarrow 6\sqrt{3}r=36$
$\Rightarrow r=\frac{36}{6\sqrt{3}}$
$\Rightarrow r=2\sqrt{3}\text{cm}$
which is the radius of the circle.

Alternate:
In $△BOD,\angle OBD={30}^{\circ }$
$cos{30}^{\circ }=\frac{BD}{OB}$
$\because cos{30}^{\circ }=\frac{\sqrt{3}}{2}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{3}{r}$
$\Rightarrow r=\frac{6}{\sqrt{3}}$
$\Rightarrow r=2\sqrt{3}$ cm