Question

An equilateral triangle of side 9cm inscribed in a circle, the radius of the circle is
\[
  (a){\text{ 3cm}} \\
  (b){\text{ }}\sqrt 3 {\text{cm}} \\
  (c){\text{ 3}}\sqrt 3 {\text{cm}} \\
  (d){\text{ }}\dfrac{{3\sqrt 3 }}{2}{\text{cm}} \\
 \]

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, $BD = \dfrac{{BC}}{2} = \dfrac{9}{2} = 4.5$ cm
Now as we know that all the angles in the equilateral triangle are equal = 60 degree.
$ \Rightarrow \angle A = \angle B = \angle C = {60^0}$
Now OB is the bisector of angle B.
$ \Rightarrow \angle OBD = \dfrac{{\angle B}}{2} = {30^0}$
Now in the triangle OBD cos is the ratio of base to hypotenuse.
$ \Rightarrow \cos {30^0} = \dfrac{{4.5}}{r} = \dfrac{{\sqrt 3 }}{2}$, $\left[ {\because \cos {{30}^0} = \dfrac{{\sqrt 3 }}{2}} \right]$
Now simplify the above equation we have,
$ \Rightarrow r = \dfrac{9}{{\sqrt 3 }}$
Now multiply and divide by $\sqrt 3 $ we have,
$ \Rightarrow r = \dfrac{9}{{\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }} = \dfrac{{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.