Question

Let $ABC$ be a triangle with $\angle B=90^{\circ }$. Let $AD$ be the external bisector of $\angle A$ with $D$ on $BC$. If $AC=6cm$ and the area of triangle $ADC$ is $10c{m}^2$ then the length of $BD$ in $cm$ is equal to

• A. $\frac{3}{5}$
• B. $\frac{3}{10}$
• C. $\frac{5}{3}$
• D. $\frac{10}{3}$

Answer 1

Answer: (D)

From angle bisector theorem,
$\frac{p}{q}=\frac{r}{6}$
$\Rightarrow qr=6p$...(1)

$\Rightarrow \frac{1}{2}(qr)=10$
$\Rightarrow qr=20$
$\therefore$ From (1) and (2),
we get $p=\frac{10}{3}=BD$ ]