Question

In $\triangle ABC$, $D$ is a point on $BC$ such that $3BD=BC$. If each side of the triangle is $12\,$cm, then $AD$ equals 

• A. $4\sqrt{5}$
• B. $4\sqrt{6}$
• C. $4\sqrt{7}$
• D. $4\sqrt{11}$

Hint:

For equilateral side $s$, use coordinates $B(-\tfrac{s}{2},0)$, $C(\tfrac{s}{2},0)$, $A\!\left(0,\tfrac{\sqrt3}{2}s\right)$ to compute distances fast.

Answer 1

The Correct Option is C

Step 1: Since all sides are \(12\), \(\triangle ABC\) is equilateral; \(D\) on \(BC\) with \(3BD=BC=12\) gives \(BD=4\), \(DC=8\).

Step 2: By Stewart's theorem, \(b^2m+c^2n=a(AD^2+mn)\): \[ 144(4)+144(8)=12(AD^2+32) \]

Step 3: Solve: \[ 1728=12AD^2+384\;\Rightarrow\;AD^2=112\;\Rightarrow\;AD=\boxed{4\sqrt7} \]