What is the area of the segment formed by a chord in a circle of radius 8 cm, if the angle subtended at the center is 60°?

Correct — A. Segment = sector − triangle, with r = 8 and θ = 60°.

Sector: (60⁄360) × π × 8² = 64π⁄6 = 32π⁄3
Triangle: ½ × 8 × 8 × sin 60° = 32 × √3⁄2 = 16√3

Segment = sector − triangle
= 32π⁄3 − 16√3 (≈ 5.8 cm²) → option (a)

Why the others are wrong

• (b)
  64π.3−16√3 — Printed as '64π.3−16√3'. Read as 64π⁄3, the sector is a third of the circle (120°), double the true 32π⁄3. Read as 64π × 3, it exceeds the whole circle's 64π.
• (c)
  32π⁄3 − 8√3 — 8√3 is half the triangle. OAB is equilateral with side 8, so its area is (√3⁄4) × 8² = 16√3. Subtracting 8√3 leaves about 19.7 cm², not the true 5.8.
• (d)
  64π⁄3 − 32√3 — Every term is doubled: 64π⁄3 − 32√3 = 2 × (32π⁄3 − 16√3). It drops the ½ from both ½r²θ for the sector (θ in radians) and ½r² sin θ for the triangle.

A minor segment is the sector minus the triangle cut off by the same two radii.

With θ in radians both pieces carry ½r², so segment = ½r²(θ − sin θ). Here ½ × 64 × (π⁄3 − √3⁄2) = 32π⁄3 − 16√3.

A 60° central angle makes triangle OAB equilateral with side 8, which is why its area is (√3⁄4) × 64 = 16√3.

Key facts

• Area of a sector = (θ⁄360) × πr².
• Area of the triangle formed by two radii = ½r² sin θ.
• Minor segment = sector − triangle = ½r²(θ − sin θ), with θ in radians.
• An equilateral triangle of side a has area (√3⁄4)a².

Common traps

• Halving the triangle: the equilateral triangle of side 8 has area 16√3, not 8√3.
• Using θ⁄180 in place of θ⁄360 for the sector, which doubles it to 64π⁄3.

No directly related past PYQ was found.