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

“All sides $12$ cm" $\Rightarrow$ $\triangle ABC$ is equilateral, so place $B(-6,0)$, $C(6,0)$, $A\big(0,6\sqrt3\big)$. Since $3BD=BC=12$, we have $BD=4$. Moving $4$ units from $B$ to $C$ along $BC$ gives \[ D=(-6,0)+\frac{4}{12}(12,0)=(-2,0). \] Therefore \[ AD=\sqrt{(0-(-2))^2+\big(6\sqrt3-0\big)^2} =\sqrt{4+108}=\sqrt{112}=4\sqrt7. \]