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

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 C $a^{2} b^{2}$
Given that $c^{2} = a^{2} + b^{2}$

Therefore, from the above equation is $c$ is the hypotenuse.

Hence, $a, b$ will be the base and height of the triangle.

we know that area of the triangle is $\frac{1}{2} \times b a s e \times h e i g h t = \frac{1}{2} a b$ -------(1)

The heron's formula for area of the triangle is $\sqrt{s (s - a) (s - b) (s - c)}$ -------(2)

equating (1) and (2) we get

$\sqrt{s (s - a) (s - b) (s - c)} = \frac{1}{2} a b$

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

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

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Similar questions

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

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

Δ A B C

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

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

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)

△ 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}, 2 s = a + b + c,

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

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