In a right-angled triangle $A B C$ of maximum area is inscribed within a circle of radius $R$ , then (where $△$ is area and $s$ is semi-perimeter of $△ A B C$ )

△ = 2 R^{2}

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\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} = \frac{\sqrt{2} + 1}{R}

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r = (\sqrt{2} - 1) R

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s = (\sqrt{2} + 2) R

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Solution

The correct option is B $\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} = \frac{\sqrt{2} + 1}{R}$
$∵$ Area of $△ A B C$ is maximum
$\Rightarrow △ A B C$ is isosceles.
$\Rightarrow A B = A C = R \sqrt{2}$
$∴ S = \frac{R \sqrt{2} + R \sqrt{2} + 2 R}{2}$
$= \frac{2 R (\sqrt{2} + 1)}{2} = R (\sqrt{2} + 1)$
$∴ △ = \frac{1}{2} .2 R . R = R^{2}$
$\Rightarrow r = \frac{△}{s} = \frac{R^{2}}{R (\sqrt{2} + 1)} = \frac{R}{\sqrt{2} + 1}$
$\Rightarrow \frac{1}{r} = \frac{\sqrt{2} + 1}{R}$
Hence $\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}} = \frac{\sqrt{2} + 1}{R}$

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The correct option is B 1r1+1r2+1r3=√2+1R∵Area of △ABC is maximum ⇒△ABC is isosceles.⇒AB=AC=R√2∴S=R√2+R√2+2R2 =2R(√2+1)2=R(√2+1)∴△=12.2R.R=R2⇒r=△s=R2R(√2+1)=R√2+1⇒1r=√2+1RHence 1r1+1r2+1r3=√2+1R

In a right-angled triangle $A B C$ of maximum area is inscribed within a circle of radius $R$ , then (where $△$ is area and $s$ is semi-perimeter of $△ A B C$ )