Mathematics - Properties of Triangle Question with Solution | TestHub

$\frac{a+b}{\sqrt{ab}}=\frac{C}{R}\Rightarrow \frac{a+b}{2\sqrt{ab}}=\frac{C}{2R}=\sin C$

$${\cos }^{2}C=1-\frac{(a+b{)}^2}{4ab}=\frac{-(a-b{)}^2}{4ab}$$

But ${\cos }^{2}C\ge 0\Rightarrow a=b\Rightarrow c=\frac{\pi }{2}$

$$r=\frac{\Delta }{S}=\frac{\frac{1}{2}ab}{\frac{a+b+c}{2}}=\frac{a^2}{2a+\sqrt{2}a}=\frac{a}{2+\sqrt{2}}=\frac{2+\sqrt{2}}{2+\sqrt{2}}=1$$