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

s^{4}

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b^{2} c^{2}

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c^{2} a^{2}

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a^{2} b^{2}

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Solution

The correct option is D $a^{2} b^{2}$
Given, $c^{2} = a^{2} + b^{2} \Rightarrow ∠ C = \frac{π}{2}$
$∴ Δ = \frac{1}{2} a b sin C = \frac{1}{2} a b$
$\Rightarrow \sqrt{s (s - a) (s - b) (s - c)} = \frac{1}{2} a b \Rightarrow 4 s (s - a) (s - b) (s - c) = a^{2} b^{2}$

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Similar questions

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

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

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

Δ A B C

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

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

△ A B C

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

2 s = a + b = c

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

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

4 s (s - a) (s - b) (s - c) = a^{2} b^{2}

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