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: From the incircle's area, \(\pi r^2=4\pi\), so \(r=2\ \text{cm}\).

Step 2: For an equilateral triangle of side \(a\), area \(=\dfrac{\sqrt{3}}{4}a^2\) and semi-perimeter \(s=\dfrac{3a}{2}\); since \(r=\dfrac{\text{Area}}{s}\), \(2=\dfrac{\frac{\sqrt{3}}{4}a^2}{\frac{3a}{2}}=\dfrac{\sqrt{3}a}{6}\), giving \(a=\dfrac{12}{\sqrt{3}}=4\sqrt{3}\ \text{cm}\).

Step 3: In an equilateral triangle the angle bisector from any vertex is also the altitude, \(AD=\dfrac{\sqrt{3}}{2}a=\dfrac{\sqrt{3}}{2}\times4\sqrt{3}=6\ \text{cm}\). \[ \boxed{6\ \text{cm}} \]