If $s$ is the semi-perimeter of $\triangle A B C$ and if $\frac{s-a}{4}=\frac{s-b}{5}=\frac{s-c}{6}$, then…

If $s$ is the semi-perimeter of $\triangle A B C$ and if $\frac{s-a}{4}=\frac{s-b}{5}=\frac{s-c}{6}$, then $\sum \sin ^2\left(\frac{A}{2}\right)=$
  1. $\frac{74}{25}$
  2. $\frac{25}{74}$
  3. $\frac{74}{33}$
  4. $\frac{25}{33}$

Solution

Let $\frac{s-a}{4}=\frac{s-b}{5}=\frac{s-c}{6}=k$ $\Rightarrow b+c-a=8 k, a+c-b=10 k$ and $a+b-c=12 k$ So, $a+b+c=30 k \Rightarrow s=15 k, a=11 k, b=10 k$ and $c=9 k$ $ \text { Now, } \begin{aligned} \Sigma \sin ^2\left(\frac{A}{2}\right)=\frac{(s-b)(s-c)}{b c}+\frac{(s-c)(s-a)}{c a} & \\ + & \frac{(s-a)(s-b)}{a b} \end{aligned} $ $\begin{aligned} & =\frac{30 k^2}{90 k^2}+\frac{24 k^2}{99 k^2}+\frac{20 k^2}{110 k^2} \\ & =\frac{1}{3}+\frac{8}{33}+\frac{2}{11}=\frac{11+8+6}{33}=\frac{25}{33} .\end{aligned}$

Asked in: AP EAMCET 2018 (23 Apr Shift 2)

Practice more Properties of Triangles questions on Aicharya