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With the usual meaning for a, b, c and s if $Δ$ be the area of a triangle then the error in $Δ$ resulting from a small error in the measurement of c, is

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$\frac{Δ}{4} (\frac{1}{s} + \frac{1}{s - a} + \frac{1}{s - b} - \frac{1}{s - c}) δ c$

$(\frac{1}{s} + \frac{1}{s - a} + \frac{1}{s - b} + \frac{1}{s - c}) δ c$

$\frac{Δ}{4} (\frac{1}{s} + \frac{1}{s - a} + \frac{1}{s - b} + \frac{1}{s - c})$

$\frac{1}{4} (\frac{1}{s} + \frac{1}{s - a} + \frac{1}{s - b} + \frac{1}{s - c}) δ c$

answer is A.

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Detailed Solution

$Δ = \sqrt{s (s - a) (s - b) (s - c)}$

$\log Δ = \frac{1}{2} [(\log s + \log (s - a) + \log (s - b) + \log (s - c)]$

$S = \frac{a + b + c}{2} \Rightarrow δ S = \frac{1}{2} δ C Now$

$\frac{1}{Δ} δ Δ = \frac{1}{2} [\frac{1}{S} δ S + \frac{1}{S - a} δ S + \frac{1}{S - b} δ S + \frac{1}{S - C} (δ S - δ C)]$

$\Rightarrow \frac{Δ}{4} (\frac{1}{s} + \frac{1}{s - a} + \frac{1}{s - b} - \frac{1}{s - c}) δ c$

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