SSC CGL 2025 PYQ — Question 51

The correct answer is option 1 i.e. ${32\pi \over 3} - 16\sqrt{3}$.

Radius (r) = 8, angle = 60°

Area of a sector = $60\over360$ × π × 8² = $1\over6$ × π × 64 = $32π\over3$

Area of triangle = $1\over2$r² sin60° = $1\over2$ × 64 × $√3\over2$ = 16√3

Area of segment = sector − triangle = ${32\pi \over 3} - 16\sqrt{3}$ cm²