With usual notations, if in $\triangle \mathrm{ABC}, 8$ is semi perimeter and…
With usual notations, if in $\triangle \mathrm{ABC}, 8$ is semi perimeter and $(\mathrm{s}-a)(\mathrm{s}-\mathrm{b})=\mathrm{s}(\mathrm{s}-\mathrm{c})$,
then $\triangle A B C$ is
an equilateral triangle
an obtuse angle triangle
a right angled triangle
an acute angle triangle
Solution
We have
$\begin{array}{l}
\sin \frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{a b}} \Rightarrow \sin ^{2} \frac{C}{2}=\frac{(s-a)(s-b)}{a b} \text { and } \\
\cos \frac{C}{2}=\sqrt{\frac{s(s-c)}{a b}} \Rightarrow \cos ^{2} \frac{C}{2}=\frac{s(s-c)}{a b} \\
\text { Given }(s-a)(s-b)=s(s-c) \\
\therefore \quad a b \sin ^{2} \frac{C}{2}=a b \cos ^{2} \frac{C}{2}
\end{array}$
$\therefore \tan ^{2} \frac{C}{2}=1 \Rightarrow \tan \frac{C}{2}=1 \Rightarrow \frac{C}{2}=45^{\circ} \Rightarrow C=90^{\circ}$
$\therefore \triangle \mathrm{ABC}$ is a right angled triangle