Question

An equilateral triangle of side 9cm inscribed in a circle, the radius of the circle is
$$\begin{gathered} (a)\text{ 3cm} \\ (b)\sqrt{3}\text{cm} \\ (c)\text{ 3}\sqrt{3}\text{cm} \\ (d){\displaystyle \frac{3\sqrt{3}}{2}}\text{cm} \end{gathered}$$

Answer 1

Hint – In this question join the center O with the vertices of the triangle and draw a perpendicular from vertex A onto the side BC. This perpendicular will bisect the side BC. Also OB, OC and OA will be the angle bisector of angle A, B and C. Use this concept to get the value of sides OB, OC or OA as they depict the radius of the circle.

Complete step-by-step solution -

Let ABC be the equilateral triangle inscribed in a circle with center O.
The side of the equilateral triangle is 9 cm (given).
Now join point OA, OB and OC which is the radius of the circle.
Let the radius of the circle be r cm.
Therefore OA = OB = OC = r cm.
Now draw the perpendicular from point A on side BC which cuts the line BC at point D as shown in figure.
Therefore, $BD={\displaystyle \frac{BC}{2}}={\displaystyle \frac{9}{2}}=4.5$ cm
Now as we know that all the angles in the equilateral triangle are equal = 60 degree.
$\Rightarrow \mathrm{\angle }A=\mathrm{\angle }B=\mathrm{\angle }C={60}^0$
Now OB is the bisector of angle B.
$\Rightarrow \mathrm{\angle }OBD={\displaystyle \frac{\mathrm{\angle }B}{2}}={30}^0$
Now in the triangle OBD cos is the ratio of base to hypotenuse.
$\Rightarrow \cos {30}^0={\displaystyle \frac{4.5}{r}}={\displaystyle \frac{\sqrt{3}}{2}}$, $\left[\because \cos {30}^0={\displaystyle \frac{\sqrt{3}}{2}}\right]$
Now simplify the above equation we have,
$\Rightarrow r={\displaystyle \frac{9}{\sqrt{3}}}$
Now multiply and divide by $\sqrt{3}$ we have,
$\Rightarrow r={\displaystyle \frac{9}{\sqrt{3}}}\times {\displaystyle \frac{\sqrt{3}}{\sqrt{3}}}={\displaystyle \frac{9\sqrt{3}}{3}}=3\sqrt{3}$ cm.
So the radius of the circle is $3\sqrt{3}$ cm.
Hence option (C) is correct.

Note – An equilateral triangle is one in which all the sides are equal, the interior angles of an equilateral triangle are all equal and in magnitude is of ${60}^0$, this happens as the angles opposite to equal sides are equal and the angle sum property of a triangle is ${180}^0$. Diagrammatic representation of the information provided is helpful as it helps figuring out the geometry and the triangles involved.