Question

In $△ABC,$ if $a=10$ and $b\cot B+c\cot C=2(r+R),$ then the maximum area(in $\text{sq.units}$) of $△ABC$ is
(Here usual notations are used for $△ABC$)

Answer 1

Given :
$b\cot B+c\cot C=2(r+R)\Rightarrow 2R\sin B\cdot \frac{\cos B}{\sin B}+2R\sin C\cdot \frac{\cos C}{\sin C}=2(r+R)\Rightarrow \cos B+\cos C=1+\frac{r}{R}\Rightarrow \cos B+\cos C=1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\Rightarrow \cos B+\cos C=\cos A+\cos B+\cos C\therefore \cos A=0\Rightarrow A=\frac{\pi }{2}$
So, $a^2=b^2+c^2$
$\Rightarrow b^2+c^2=100\cdots \left(i\right);a=10$
Using $A.M\ge G.M.$
$\Rightarrow \frac{b^2+c^2}{2}\ge \sqrt{b^2{c}^2}\Rightarrow bc\le 50\Rightarrow \frac{bc}{2}\le 25$
So, maximum area of triangle is $25\text{sq.units}$