JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Properties of Triangle

Since $\mathrm{AC}$ is perpendicular to $\mathrm{AB}$, $△\mathrm{ABC}$ is right-angled at $\mathrm{A}$. The circumcentre of a right triangle is the midpoint of the hypotenuse $\mathrm{BC}$, so $\left(\frac{5\alpha }{8},2\right)=\left(5,\frac{\alpha }{4}\right)$. This gives $\frac{5\alpha }{8}=5$ and $\frac{\alpha }{4}=2$, both yielding $\alpha =8$. Thus $\mathrm{A}(8,-2)$, $\mathrm{B}(8,6)$, $\mathrm{C}(2,-2)$. We have $\mathrm{AB}=8$, $\mathrm{AC}=6$, $\mathrm{BC}=10$ (by Pythagoras). Area $=\frac{1}{2}(6)(8)=24$. Perimeter $=6+8+10=24$ (not 25). Circumradius $=\frac{\mathrm{BC}}{2}=5$. Inradius $=\frac{\text{Area}}{s}=\frac{24}{12}=2$. So option (B) is NOT correct.