Question

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

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

Answer 1

The correct option is C $4\sqrt{7}cm$

Given $△ABC$ with $D$ a point on $BC$ such that $3BD=BC$

$\therefore$ $BD=\frac{BC}{3}=4cm$

$AD\perp BC$

Let's take a point $E$ on $BC$ which makes a right angle triangle $ADE$ and $AEB$ at $E$ such that $BE=\frac{1}{2}BC=6cm$

$\therefore DE=BE-BD=2cm$.

$\therefore A{E}^2=A{B}^2-B{E}^2=144-36=108$

$\because AED={90}^o$

$\therefore A{D}^2=A{E}^2+D{E}^2=108+4=112$

$\therefore AD=4\sqrt{7}cm$.