A circle touches two of the smaller sides of a $Δ A B C (a < b < c)$ and has its center on the greatest side. Then 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 C $\frac{2 Δ}{a + b}$

Let O be the center and r be the radius of the circle.
Then
area $Δ B O C$ $+$ area $Δ A O C$ $=$ area $Δ A B C$
$\Rightarrow (a + b) \frac{r}{2} = \sqrt{s (s - a) (s - b) (s - c)}$
$\Rightarrow r = \frac{2 \sqrt{s (s - a) (s - b) (s - c)}}{a + b} = \frac{2 Δ}{a + b}$

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

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