Question

If the radius of the circumcircle of an equilateral triangle is 6 cm, then find the radius of its incircle.

Answer 1

Hint: In this question, first of all draw the diagram it will gives us a clear if idea of what we have to find and draw a perpendicular bisector to the base of the equilateral triangle and also construct a circle which is inside the triangle and find the radius of that incircle.

Complete step-by-step answer:

Let \[a\] be the side of the equilateral triangle \[ABC\]
Therefore, \[AB = BC = AC = a\].
Since \[\Delta ABC\] is an equilateral triangle each angle is equal to \[{60^0}\].
Therefore, \[\angle A = \angle B = \angle C = {60^0}\].
Given radius of circumcircle \[r = 6{\text{ cm}}\]
Since the \[\Delta ABC\] is equilateral, its perpendicular bisector i.e., the median meet at the same point \[O\] which is the centre of the incircle as shown in the below figure:

Hence, \[AD\] is the perpendicular bisector of \[BC\].
\[ \Rightarrow BD = DC = \dfrac{1}{2}BC = \dfrac{a}{2}{\text{ }}\left[ {\because BC = a} \right]\]
And \[OB\] is the angular bisector of \[\angle B\].
Therefore, \[\angle ABO = \angle OBD = \dfrac{{{{60}^0}}}{2} = {30^0}\]
From the figure clearly \[OD\] is the
In the right-angled triangle \[OBD\], consider
\[
   \Rightarrow \sin {30^0} = \dfrac{{OD}}{{OB}} \\
   \Rightarrow \dfrac{1}{2} = \dfrac{{OD}}{r}{\text{ }}\left[ {\because OB = r} \right] \\
   \Rightarrow OD = \dfrac{r}{2} \\
   \Rightarrow OD = \dfrac{6}{2}{\text{ }}\left[ {r = 6{\text{ cm}}} \right] \\
  \therefore OD = 3{\text{ cm}} \\
\]
Thus, the radius of the incircle is 3 cm.

Note: For a given equilateral triangle all sides are equal and all angles are equal to \[{60^0}\]. Here the equilateral triangle is divided into two right-angled triangles with the drawn perpendicular in which one of the angles is equal to \[{90^0}\].