Question

In a triangle $ABC$, the angle bisector BD of $\angle B$ intersects AC in D. Suppose $BC=2,CD=1$ and $BD=\frac{3}{\sqrt{2}}$. The perimeter of the triangle $ABC$ is

• A. $\frac{17}{2}$
• B. $\frac{15}{2}$
• C. $\frac{17}{4}$
• D. $\frac{15}{4}$

Answer 1

Answer: (B)

$\frac{2ac}{a+c}\cos \frac{B}{2}=\frac{3}{\sqrt{2}}$
$\Rightarrow \frac{4c}{2+c}\left[\frac{4+\frac{9}{2}-1}{2\times 2\times \frac{3}{\sqrt{2}}}\right]=\frac{3}{\sqrt{2}}$
$\Rightarrow c=3$
Now, $\frac{c}{a}=\frac{AD}{DC}\Rightarrow AD=\frac{3}{2}$
$\Rightarrow b=\frac{5}{2}$
$\Rightarrow \text{ Perimeter }=\frac{15}{2}$