In a triangle ABC, the angle bisector BD of ∠B intersects AC in D. Suppose BC=2, CD=1 and BD=3√2. The perimeter of the triangle ABC is

Correct answer: (B) 15/2. 2ac/(a+c)cosB/2=(3√2 ⇒4)/(c2+c4+92-12×2×3√2=3√2 ⇒c=3 Now, ca)=AD/DC⇒AD=3/2 ⇒b=5/2 ⇒Perimeter=15/2

Q55 / 91 · KVPY 2020 (SB/SX) · Mathematics

In a triangle $A B C,$ the angle bisector $B D$ of $∠ B$ intersects $A C$ in $D .$ Suppose $B C = 2, C D = 1$ and $B D = \frac{3}{\sqrt{2}} .$ The perimeter of the triangle
$A B C$ is

$\frac{2 a c}{a + c} \cos \frac{B}{2} = \frac{3}{\sqrt{2}}$

$\Rightarrow \frac{4 c}{2 + c} [\frac{4 + \frac{9}{2} - 1}{2 \times 2 \times \frac{3}{\sqrt{2}}}] = \frac{3}{\sqrt{2}}$

$\Rightarrow c = 3$

Now, $\frac{c}{a} = \frac{A D}{D C} \Rightarrow A D = \frac{3}{2}$

$\Rightarrow b = \frac{5}{2}$

$\Rightarrow$ Perimeter $= \frac{15}{2}$

Hence, the correct answer is option: 'B'.