Question

In an equilateral triangle $ABC$, if the area of its in-circle is $4\pi\ \text{cm}^2$, then find the length of the angle bisector $AD$?

• A. $2\ \text{cm}$
• B. $4\ \text{cm}$
• C. $6\ \text{cm}$
• D. $10\ \text{cm}$

Hint:

In an equilateral triangle, median = altitude = angle bisector = perpendicular bisector. Use $r=\frac{\sqrt{3}}{6}a$ and $h=\frac{\sqrt{3}}{2}a$ to move between inradius, side, and altitude quickly.

Answer 1

The Correct Option is C

Step 1: Find the inradius.
Area of in-circle $= \pi r^2 = 4\pi \Rightarrow r=2\ \text{cm}$. 

Step 2: Relate $r$ and side $a$ of an equilateral triangle.
For an equilateral triangle, $r=\dfrac{\sqrt{3}}{6}\,a \Rightarrow a=\dfrac{6r}{\sqrt{3}}=\dfrac{12}{\sqrt{3}}=4\sqrt{3}\ \text{cm}$. 

Step 3: Angle bisector equals altitude.
In an equilateral triangle, the angle bisector $AD$ is also the altitude: $AD=\dfrac{\sqrt{3}}{2}\,a = \dfrac{\sqrt{3}}{2}\cdot 4\sqrt{3}=\dfrac{4\cdot 3}{2}=6\ \text{cm}$. \[ \boxed{6\ \text{cm}} \]