CAT 2023 Slot 2 Solved Paper

Since $\mathit{B}\mathit{C}$ is the diameter of the circle, which implies angle $\mathit{B}\mathit{A}\mathit{C}$ is 90 degrees. Let $\mathit{A}\mathit{B}=\mathit{a}\mathit{c}\mathit{m}$ , which implies $\mathit{A}\mathit{C}=$ $\mathrm{b}\mathrm{cm}$ . Hence, $\mathit{B}\mathit{C}=\sqrt{{\mathit{a}}^2+{\mathit{b}}^2}$ , which is diameter of the circle (2r).
Hence, $2\mathit{r}=\sqrt{{\mathit{a}}^2+{\mathit{b}}^2}$
$\Rightarrow 4{\mathit{r}}^2={\mathit{a}}^2+{\mathit{b}}^2$
The area of the triangle is $\frac{1}{2}\times \mathit{a}\times \mathit{b}$ , which can be written as
$\Rightarrow \frac{\mathit{a}\cdot \mathit{b}}{2\left({\mathit{a}}^2+{\mathit{b}}^2\right)}\times \left({\mathit{a}}^2+{\mathit{b}}^2\right)$
$\Rightarrow \frac{\mathit{a}\cdot \mathit{b}}{2\left({\mathit{a}}^2+{\mathit{b}}^2\right)}\times 4{\mathit{r}}^2$
$\Rightarrow \frac{\mathit{a}\cdot \mathit{b}}{{\mathit{a}}^2+{\mathit{b}}^2}\times 2{\mathit{r}}^2$
The correct option is B