Question

In $\Delta$ ABC, D is a point on BC such that $3$BD=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. $4\sqrt{11}$

Answer 1

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

$\because 3x=12\Rightarrow x=4$
$AE=\frac{\sqrt{3}}{2}\left(3x\right)$
$DE=AE-AD=\frac{3x}{2}-x=\frac{x}{2}$
$AD=\sqrt{A{E}^2+D{E}^2}=\sqrt{\frac{27}{4}{x}^2+\frac{x^2}{4}}$
$=\sqrt{\frac{28x^2}{4}}=\sqrt{7x^2}=\sqrt{7}x=4\sqrt{7}$