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Question: If a right-angled DABC of maximum area is inscribed within a circle of radius R, then-...

If a right-angled DABC of maximum area is inscribed within a circle of radius R, then-

A

D = 2R2

B

1r1\frac{1}{r_{1}}+ 1r2\frac{1}{r_{2}}+ 1r3\frac{1}{r_{3}}= 2+1R\frac{\sqrt{2} + 1}{R}

C

r = (2\sqrt{2}– 1) R

D

s = (1 + 2\sqrt{2})R

Answer

1r1\frac{1}{r_{1}}+ 1r2\frac{1}{r_{2}}+ 1r3\frac{1}{r_{3}}= 2+1R\frac{\sqrt{2} + 1}{R}

Explanation

Solution

For a right-angled triangle inscribed in a circle of radius

R, the length of the hypotenuse is 2R. \ the area is maximum

when the triangle is isosceles with each side = 2\sqrt{2}R.

\ s = 12\frac{1}{2} (22\sqrt{2}+ 2) R = (2\sqrt{2}+ 1)R

\ D =12\frac{1}{2} 2\sqrt{2}R. 2\sqrt{2}R = R2 ή 1r\frac{1}{r}= (2+1)R\frac{(\sqrt{2} + 1)}{R}

1r1\frac{1}{r_{1}}+ 1r2\frac{1}{r_{2}}+ 1r3\frac{1}{r_{3}}= s−aΔ\frac{s - a}{\Delta}+ s−bΔ\frac{s - b}{\Delta}+ s−cΔ\frac{s - c}{\Delta}

= sΔ\frac{s}{\Delta}=1r\frac{1}{r}= 2+1R\frac{\sqrt{2} + 1}{R}