In an isosceles triangle with lateral side $a$ and base $b$ , what is the value of the incircle?

\frac{a b + (b^{2}) / 2}{2 \sqrt{a^{2} - (b^{2}) / 4}}

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\frac{b}{2} \sqrt{\frac{2 a + b}{2 a - b}}

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\frac{a b - (b^{2}) / 2}{2 \sqrt{a^{2} - (b^{2}) / 4}}

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\frac{b}{2} \sqrt{\frac{2 a - b}{2 a + b}}

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Solution

The correct options are
A $\frac{a b - (b^{2}) / 2}{2 \sqrt{a^{2} - (b^{2}) / 4}}$
B $\frac{b}{2} \sqrt{\frac{2 a - b}{2 a + b}}$
$r = (s - a) tan \frac{A}{2} = (s - a) \sqrt{\frac{1 - cos A}{1 + cos A}}$
$= \frac{b}{2} \sqrt{\frac{1 - \frac{b}{2 a}}{1 + \frac{b}{2 a}}}$
$= \frac{b}{2} \sqrt{\frac{2 a - b}{2 a + b}} = \frac{b}{2} \sqrt{\frac{(2 a - b)^{2}}{(2 a)^{2} - (b)^{2}}}$

$= \frac{a b - (b) / 2}{2 \sqrt{a^{2} - (b^{2}) / 4}}$

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