Question

If a right angled triangle ABC of maximum area $\Delta$ is inscribed in a circle of radius R, then

• A. $\Delta =2R^2$
• B. $r=(\sqrt{2}-1)/R$
• C. $\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{\sqrt{2}-1}{R}$
• D. None of these

Answer 1

Answer: (D)

None of these

For a right angled inscribed in a circle of radius $R$, the length of the hypotenuse is $2R$
Therefore the $R$ is maximum when the triangle is isosceles with each side $=\sqrt{2}R$
$\therefore s=\frac{1}{2}(2\sqrt{2}+2)R=(\sqrt{2}+1)R$
$\Delta =\frac{1}{2}\left(\sqrt{2}R\right).\sqrt{2}R=R^2\Rightarrow \frac{1}{r}=\frac{\sqrt{2}+1}{R}$
$\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{s-a}{\Delta }+\frac{s-b}{\Delta }+\frac{s-c}{\Delta }$
$=\frac{s}{\Delta }=\frac{1}{r}=\frac{\sqrt{2}+1}{R}$