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

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Solution

L.H.S $= 4 s (s - a) (s - b) (s - c) = 4 Δ^{2}$
we know that $s (s - a) (s - b) (s - c) = Δ$
Given $c^{2} = a^{2} + b^{2}$ that means it is a right angled triangle
$\Rightarrow Δ = \frac{1}{2} a b$
$∴$ L.H.S $= 4. \frac{1}{4} a^{2} b^{2} = a^{2} b^{2} =$ R.H.S

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c^{2} = a^{2} + b^{2}

4 s (s - a) (s - b) (s - c) = ?

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4 s (s - a) (s - b) (s - c)

c^{2} = a^{2} + b^{2}, 2 s = a + b + c,

4 s (s - a) (s - b) (s - c)

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