Question

An equilateral triangle of side $9\mathrm{cm}$ is inscribed in a circle then the radius of the circle is?

Answer 1

Step 1. Find the length of the median of the equilateral triangle.

Let $△\mathrm{ABC}$ be an equilateral triangle with side $9\mathrm{cm}$.

Let $\mathrm{AD}$ be one of its medians. Then,

$\mathrm{AD}\perp \mathrm{BC}$ and $\mathrm{BD}=\frac{1}{2}\times \mathrm{BC}$

Since, $\mathrm{BC}=9\mathrm{cm}$.

Therefore,

$\mathrm{BD}=\frac{1}{2}\times 9=4.5\mathrm{cm}$

Apply Pythagoras Theorem in $△\mathrm{ABD}$:

${\mathrm{AB}}^2={\mathrm{AD}}^2+{\mathrm{BD}}^2$

Substitute the values of $\mathrm{AB}$ and $\mathrm{BD}$ in the above equation:

$\begin{array}{rcl}{\left(9\right)}^2& =& {\mathrm{AD}}^2+{\left(\frac{9}{2}\right)}^2\\ & \Rightarrow & 81={\mathrm{AD}}^2+\frac{81}{4}\\ & \Rightarrow & {\mathrm{AD}}^2=81-\frac{81}{4}\\ & \Rightarrow & {\mathrm{AD}}^2=\frac{\left(81\times 4\right)-81}{4}\\ & \Rightarrow & {\mathrm{AD}}^2=\frac{81\left(4-1\right)}{4}\\ & \Rightarrow & \mathrm{AD}=\sqrt{\frac{81\times 3}{4}}\\ & \Rightarrow & \mathrm{AD}=\frac{9\sqrt{3}}{2}\mathrm{cm}\end{array}$

Step 2. Find the radius of the circumcircle of the triangle.

The centroid and circumcenter coincide in an equilateral triangle. Hence, $\mathrm{AG}$ is the radius of the circumcircle.

Also, the centroid divides the median in the ratio $2:1$ i.e.,

$\mathrm{AG}:\mathrm{GD}=2:1$

Hence, the radius can be expressed as $\mathrm{AG}=\frac{2}{3}\mathrm{AD}$.

Substitute the value of $\mathrm{AD}$:

$\begin{array}{rcl}\mathrm{AG}& =& \frac{2}{3}\times \frac{9\sqrt{3}}{2}\\ & \Rightarrow & \mathrm{AG}=3\sqrt{3}\mathrm{cm}\end{array}$

Hence, the radius of the circle is $3\sqrt{3}\mathrm{cm}$.