If $a + b + c = 2 s$ , then the value of $(s - a)^{2} + (s - b)^{2} + (s - c)^{2}$ will be

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

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a^{2} + b^{2} + c^{2} - s^{2}

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s^{2} - a^{2} - b^{2} - c^{2}

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4 s^{2} - a^{2} - b^{2} - c^{2}

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Solution

The correct option is C $a^{2} + b^{2} + c^{2} - s^{2}$
$a + b + c = 2 s . . . . . (1)$
$(s - a)^{2} + (s - b)^{2} + (s - c)^{2}$
$= s^{2} + a^{2} - 2 a s + s^{2} + b^{2} - 2 s b + s^{2} + c^{2} - 2 s c$
$= 3 s^{2} + a^{2} + b^{2} + c^{2} - 2 s (a + b + c)$
$= 3 s^{2} + a^{2} + b^{2} + c^{2} - 2 s (2 s)$ [From $(1)$ ]
$= 3 s^{2} + a^{2} + b^{2} + c^{2} - 4 s^{2} = a^{2} + b^{2} + c^{2} - s^{2}$ .

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Similar questions

a + b + c = 2 s

\frac{(s - a)^{2} + (s - b)^{2} + (s - c)^{2} + s^{2}}{a^{2} + b^{2} + c^{2}}

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 = 0

\frac{1}{b^{2} + c^{2} - a^{2}} + \frac{1}{c^{2} + a^{2} - b^{2}} + \frac{1}{a^{2} + b^{2} - c^{2}}

|\begin{matrix} - a (b^{2} + c^{2} - a^{2}) & 2 b^{3} & 2 c^{3} \\ 2 a^{3} & - b (c^{2} + a^{2} - b^{2}) & 2 c^{3} \\ 2 a^{3} & 2 b^{3} & - c (a^{2} + b^{2} - c^{2}) \end{matrix}| = a b c {(a^{2} + b^{2} + c^{2})}^{3}

\frac{a^{2} - {(b - c)}^{2}}{{(a + c)}^{2} - b^{2}} + \frac{b^{2} - {(a - c)}^{2}}{{(a + b)}^{2} - c^{2}} + \frac{c^{2} - {(a - b)}^{2}}{{(b + c)}^{2} - a^{2}}

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