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

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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.

Updated On: Jul 16, 2026

$4\sqrt{5}$
$4\sqrt{6}$
$4\sqrt{7}$
$4\sqrt{11}$

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The Correct Option is C

Approach Solution - 1

“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 \[ AD=\sqrt{(0-(-2))^2+\big(6\sqrt3-0\big)^2} =\sqrt{4+108}=\sqrt{112}=4\sqrt7. \]

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Approach Solution -2

An alternate method uses Stewart's theorem for a cevian in a triangle, instead of coordinate geometry, and then checks each option.

Option A ($4\sqrt5$): By Stewart's theorem, for cevian $AD$ in $\triangle ABC$ with $BD=m$, $DC=n$, $BC=a=m+n$, $AB=c$, $AC=b$: \[ b^2m+c^2n=a\left(AD^2+mn\right). \] Here $a=BC=12$, $m=BD=4$, $n=DC=8$ (since $3BD=BC=12\Rightarrow BD=4$), and $b=c=12$ (equilateral triangle). Substituting: \[ 144(4)+144(8)=12\left(AD^2+32\right)\;\Rightarrow\;576+1152=12AD^2+384\;\Rightarrow\;12AD^2=1344\;\Rightarrow\;AD^2=112. \] So $AD=\sqrt{112}=4\sqrt7$, which does not equal $4\sqrt5$, ruling out this option.
Option B ($4\sqrt6$): This also does not match $AD^2=112$, so it is incorrect.
Option C ($4\sqrt7$): Since $AD=\sqrt{112}=\sqrt{16\times7}=4\sqrt7$, this matches exactly.
Option D ($4\sqrt{11}$): This does not match $AD^2=112$, so it is incorrect.

Stewart's theorem confirms $AD=4\sqrt7$.

Hence, the correct answer is option C: $4\sqrt7$.