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: $AD=\dfrac{\sqrt{3}}{2}\,a = \dfrac{\sqrt{3}}{2}\cdot 4\sqrt{3}=\dfrac{4\cdot 3}{2}=6\ \text{cm}$. \[ \boxed{6\ \text{cm}} \]

Answer 2

For an equilateral triangle, there is a neat shortcut: the angle bisector (which is also the altitude and median) equals the sum of the circumradius and inradius, since the circumcentre, incentre and centroid all coincide.

1. 2 cm: If \(AD=2\), then the side would be \(a=\dfrac{2AD}{\sqrt{3}}=\dfrac{4}{\sqrt{3}}\), giving an inradius \(r=\dfrac{AD}{3}=\dfrac{2}{3}\), and an incircle area of \(\pi\left(\dfrac{2}{3}\right)^2=\dfrac{4\pi}{9}\), not the given \(4\pi\). This fails.
2. 4 cm: Here \(r=\dfrac{4}{3}\), giving an incircle area of \(\pi\left(\dfrac{4}{3}\right)^2=\dfrac{16\pi}{9}\), which is still not \(4\pi\).
3. 6 cm: Here \(r=\dfrac{6}{3}=2\), giving an incircle area of \(\pi(2)^2=4\pi\), which exactly matches the given area.
4. 10 cm: Here \(r=\dfrac{10}{3}\), giving an incircle area of \(\pi\left(\dfrac{10}{3}\right)^2=\dfrac{100\pi}{9}\), much larger than \(4\pi\).

Since the incircle area is \(4\pi\), the inradius is \(r=2\). For an equilateral triangle, the circumradius is always twice the inradius, so \(R=4\), and the angle bisector (which equals the altitude) is \(AD=R+r=4+2=6\).

So the correct answer is 6 cm.