Question

An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle.

Answer 1

Let ΔABC be an equilateral triangle of side 9 cm.
Let AD be one of its median.

Then, AD ⊥ BC (ΔABC is an equilateral triangle)
Also, $BD\mathit{=}\left({\displaystyle \frac{BC}{2}}\right)=\left(\frac{9}{2}\right)=4.5\mathrm{cm}$
In right angled ΔADB, we have:
AB² = AD² + BD²
⇒ AD² = AB² - BD²
$\Rightarrow AD=\sqrt{A{B}^2-B{D}^2}$
$=\sqrt{{\left(9\right)}^2-{\left(\frac{9}{2}\right)}^2}\mathrm{cm}$
$=\frac{9\sqrt{3}}{2}\mathrm{cm}$
In the equilateral triangle, the centroid and circumcentre coincide and AG : GD = 2 : 1.
Now, radius = $AG\mathit{=}\frac{\mathit{2}}{\mathit{3}}AD$
$\Rightarrow AG=\left(\frac{2}{3}\times \frac{9\sqrt{3}}{2}\right)=3\sqrt{3}\mathrm{cm}$
∴ The radius of the circle is $3\sqrt{3}\mathrm{cm}$.