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If \( a ^ { 2 } = a ^ { 2 } + b ^ { 2 } , \) write the value \( \operatorname { oftg } ( s ) = a \{ ( s - b ) ( s - c ) \text { in terms of a and b } \)

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Similar Questions

Q1

If

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

then find the value of

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

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Q2

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)

(c) c² + a² − b² + 2ca = 4s (s − b)

(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)

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Q3

If

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

, then

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

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Q4

If

2 s = a + b + c,

prove that

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} & {(s - a)}^{2} & {(s - a)}^{2} {(s - b)}^{2} & b^{2} & {(s - b)}^{2} {(s - c)}^{2} & {(s - c)}^{2} & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

= 2 s^{3} (s - a) (s - b) (s - c) .

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Q5

Q. If a+b+c = 25, then prove that

|\begin{matrix} a^{2} & {(s - a)}^{2} & {(s - a)}^{2} \\ {(s - b)}^{2} & b^{2} & {(s - b)}^{2} \\ {(s - c)}^{2} & {(s - c)}^{2} & c^{2} \end{matrix}| = 25^{3} (s - a) (s - b) (s - c)

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