Question

Let $A B C$ be a triangle with $\angle B=90^{\circ}$. Let $A D$ be the bisector of $\angle A$ with $D$ on $B C$. Suppose $A C=6 \,cm$ and the area of the $\triangle A D C$ is $10\, cm ^{2}$. Then, the length of $B D$ 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,
$A B C$ is right angled triangle with $B$ is $90^{\circ}$.

$A D$ is angle bisector of $\angle A$.
$\therefore \frac{A B}{A C}=\frac{B D}{D C}$
$\Rightarrow A B \cdot C D=B D \cdot A C$
Area of $\triangle A D C=10$
$\Rightarrow \frac{1}{2} \times A B \cdot C D=10$
$\Rightarrow \frac{1}{2} \times B D \cdot A C=10$
$\rightarrow B D=\frac{20}{A C}$
$\Rightarrow B D=\frac{20}{6} [\because A C=6]$
$\Rightarrow B D=\frac{10}{3}$