Question

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

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

Answer 1

The correct option is A

$2\sqrt{3}\text{cm}$

Construct an altitude from point A to point D in $\Delta$ABC.

Given: ABC is an equilateral triangle of side 6 cm,
We know that, perpendicular drawn from the cneter of the circle bisects the chord.
Here, BC is a chord of the circle, as OD is perpendiular from the centre to BC. Hence it bisects BC
i.e., BD = DC = 3cm

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 $(r+x)=3\sqrt{3}$
$\therefore (r-x)=\frac{9}{r+x}=\frac{9}{3\sqrt{3}}$
$⟹r-x=\sqrt{3}--------\left(2\right)$

Adding (1) and (2), we get
$r=2\sqrt{3}\text{cm}$
which is the radius of the circle.