Question

If in $∆ABC$, $\sin \frac{A}{2}\sin \frac{C}{2}=\sin \frac{B}{2}$ and $2s$ is the perimeter of the triangle, then $s$ is equal to:

• A. $2b$
• B. $b$
• C. $3b$
• D. $4b$

Answer 1

The correct option is A

$2b$

Explanation for the correct option:

Find the value of $s$:

We know that in a $∆ABC$,

$\sin \frac{A}{2}=\sqrt{\frac{\left(s-b\right)\left(s-c\right)}{bc}},\sin \frac{B}{2}=\sqrt{\frac{\left(s-a\right)\left(s-c\right)}{ac}},\mathrm{and}\sin \frac{C}{2}=\sqrt{\frac{\left(s-a\right)\left(s-b\right)}{ab}}$

It is given that,

$\begin{array}{rcl}\sin \frac{A}{2}\sin \frac{C}{2}& =& \sin \frac{B}{2}\\ & \Rightarrow & \sqrt{\frac{\left(s-b\right)\left(s-c\right)}{bc}}\times \sqrt{\frac{\left(s-a\right)\left(s-b\right)}{ab}}=\sqrt{\frac{\left(s-a\right)\left(s-c\right)}{ac}}\end{array}$

We can write it as,

$\begin{array}{rcl}\frac{\sqrt{\left(s-b\right)}\times \sqrt{\left(s-c\right)}}{\sqrt{bc}}\times \frac{\sqrt{\left(s-a\right)}\times \sqrt{\left(s-b\right)}}{\sqrt{ab}}& =& \frac{\sqrt{\left(s-a\right)}\times \sqrt{\left(s-c\right)}}{\sqrt{ac}}\\ \Rightarrow \frac{\sqrt{\left(s-b\right)}\times \sqrt{\left(s-b\right)}}{\sqrt{b}\times \sqrt{b}}& =& 1\\ \Rightarrow s-b& =& b\\ \Rightarrow s& =& 2b\end{array}$

Hence, option A is correct.