A circle touches two of the smaller sides of a $△ A B C (a < b < c)$ and has its center on the greatest side. The radius of the circle is

\frac{a - b - c}{2}

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\frac{a b c}{2}

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\frac{2 Δ}{a + b}

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None of these

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Solution

The correct option is D $\frac{2 Δ}{a + b}$
Let $O$ be the centre and $r$ be the radius of the circle.
Then, area $(△ A O C) +$ area $(△ A O C) =$ area $(△ A B C)$
$\Rightarrow (a + b) \frac{r}{2} = \sqrt{s (s - a) (s - b) (s - c)}$
$\Rightarrow r = \frac{2 \sqrt{s (s - a) (s - b) (s - c)}}{a + b} = \frac{2 △}{a + b}$
where, $s = \frac{a + b + c}{2}$

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A circle touches two of the smaller sides of a △ABC(a<b<c) and has its center on the greatest side. The radius of the circle is

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A circle touches two of the smaller sides of a $△ A B C (a < b < c)$ and has its center on the greatest side. The radius of the circle is