Question

The least area of a circle circumscribing any right angled triangle of area $\frac{9}{\pi }$ is

Answer 1

If a circle is circumscibing a right-angled triangle then the hypotenuse must be a diameter.

Let the radius of the circle be $r.$
Area of the triangle $=\frac{1}{2}\times a\times b$
$\Rightarrow \frac{9}{\pi }=\frac{1}{2}\times a\times b\Rightarrow ab=\frac{18}{\pi }$
$a^2+b^2=4r^2$

$\text{AM}\ge \text{GM}$
$\frac{a^2+b^2}{2}\ge ab\Rightarrow 2r^2\ge \frac{18}{\pi }\Rightarrow \pi r^2\ge 9\Rightarrow {\text{Area}}_{min}=9$