# If c^2=a^2+b^2 ; a+b+c=2s, then 4s(s-a)(s-b)(s-c) equals

Solution:     $c^{2}=a^{2}+b^{2}\Rightarrow \angle C=90^{\circ}$

We know that

$\Delta =\sqrt{s(s-a)(s-b)(s-c)}$       $.......(1)$

and       $\Delta=\frac{1}{2}ab\sin C$                          $........(2)$

From   $(1)$   $4\Delta ^{2}=4s(s-a)(s-b)(s-c)$

From     $(2)$   $2\Delta =ab \sin 90^{\circ}\Rightarrow 4\Delta ^{2}=a^{2}b^{2}$

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