Question

In $△ABC$ if ${\sin }^{2}A+{\sin }^{2}B={\sin }^{2}C$ and $l\left(AB\right)=10$, then the maximum value of the area of $△ABC$ is

• A. $50$
• B. $10\sqrt{2}$
• C. $25$
• D. $25\sqrt{2}$

Answer 1

The correct option is C $25$

${\sin }^{2}A+{\sin }^{2}B={\sin }^{2}C$
$\Rightarrow a^2+b^2=c^2$ (Sine Rule)
$A\left(△ABC\right)=\frac{1}{2}ab.....\left(1\right)$
From sine rule $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$
$\Rightarrow \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{10}{1}$
$\Rightarrow a=10\sin A,b=10\sin B$
Using equation $\left(1\right)$

$A\left(△ABC\right)=\frac{1}{2}\left(10\sin A\right)\left(10\sin B\right)$
$=50\sin A\sin B$
But maximum value of $\sin A\sin B=\frac{1}{2}$
$\therefore$ Maximum value of $A\left(△ABC\right)=50\times \frac{1}{2}=25$
OR
$\angle C={90}^{\circ }\Rightarrow ABC$ is right angled triangle
$\therefore$ Area of $△$ is maximum when it is ${45}^{\circ }-{45}^{\circ }-{90}^{\circ }△$.
$\therefore A\left(△ABC\right)=\frac{1}{2}\times 5\sqrt{2}\times 5\sqrt{2}=25$