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

Solution:

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, $B D=\left(\frac{B C}{2}\right)=\left(\frac{9}{2}\right)=4.5 \mathrm{~cm}$

In right angled ΔADB, we have:

$A B^{2}=A D^{2}+B D^{2}$

$\Rightarrow A D^{2}=A B^{2}-B D^{2}$

$\Rightarrow A D=\sqrt{A B^{2}-B D^{2}}$

$=\sqrt{(9)^{2}-\left(\frac{9}{2}\right)^{2}} \mathrm{~cm}$

$=\frac{9 \sqrt{3}}{2} \mathrm{~cm}$

In the equilateral triangle, the centroid and circumcentre coincide and AG : GD = 2 : 1.

Now, radius $=A G=\frac{2}{3} A D$

$\Rightarrow A G=\left(\frac{2}{3} \times \frac{9 \sqrt{3}}{2}\right)=3 \sqrt{3} \mathrm{~cm}$

$\therefore$ The radius of the circle is $3 \sqrt{3} \mathrm{~cm}$.