SSC CGL 30 Aug 2015 Paper

∠BAC = 90°
From ΔABC,
BC = $\sqrt{\mathit{A}{\mathit{B}}^2+\mathit{A}{\mathit{C}}^2}$
= $\sqrt{6^2+8^2}$
= $\sqrt{36+64}$
= $\sqrt{100}$ = 10 cm
∴ Semi–perimeter of ΔABC = s
= $\frac{6+8+10}{2}$ = $\frac{24}{2}$ = 12 cm
Area of ΔABC = $\frac{1}{2}$ × AC × AB
= $\frac{1}{2}\times 8$ × 8 = 24 sq. cm.
∴ In–radius = $\frac{\mathit{\Delta }}{\mathit{s}}$ = $\frac{24}{12}$ = 2 cm