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