If {c^2} = {a^2} + {b^2},then 4s(s - a)(s - b | vedclass

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(D) Given the condition ${c^2} = {a^2} + {b^2}$,the triangle is a right-angled triangle with the hypotenuse $c$ and $\angle C = 90^{\circ}$.By Heron's

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${a^2}{b^2}$

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(D) Given the condition ${c^2} = {a^2} + {b^2}$,the triangle is a right-angled triangle with the hypotenuse $c$ and $\angle C = 90^{\circ}$.By Heron's formula,the area of the triangle $\Delta$ is given by $\Delta = \sqrt{s(s - a)(s - b)(s - c)}$.Squaring both sides,we get $\Delta^2 = s(s - a)(s - b)(s - c)$,which implies $4\Delta^2 = 4s(s - a)(s - b)(s - c)$.For a right-angled triangle with legs $a$ and $b$,the area is $\Delta = \frac{1}{2}ab$.Substituting this into the expression,we get $4\Delta^2 = 4(\frac{1}{2}ab)^2 = 4(\frac{1}{4}a^2b^2) = {a^2}{b^2}$.

If ${c^2} = {a^2} + {b^2}$,then $4s(s - a)(s - b)(s - c) = $

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