Question

Let $ABC$ be a triangle with $\angle B=90^{\circ }$. Let $AD$ be the bisector of $\angle A$ with $D$ on $BC$. Suppose $AC=6cm$ and the area of the $△ADC$ 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)

Given,
$ABC$ is right angled triangle with $B$ is $90^{\circ }$.

$AD$ is angle bisector of $\angle A$.
$\therefore \frac{AB}{AC}=\frac{BD}{DC}$
$\Rightarrow AB\cdot CD=BD\cdot AC$
Area of $△ADC=10$
$\Rightarrow \frac{1}{2}\times AB\cdot CD=10$
$\Rightarrow \frac{1}{2}\times BD\cdot AC=10$
$\to BD=\frac{20}{AC}$
$\Rightarrow BD=\frac{20}{6}[\because AC=6]$
$\Rightarrow BD=\frac{10}{3}$