A sector of a circle with radius 6 cm has a central angle of 60°. What is the area of the corresponding segment?

Correct — B. A segment is the sector minus the triangle that the chord forms with the two radii.

Sector: (60⁄360) × π × 6²
= (1⁄6) × 36π = 6π

Triangle: two 6 cm radii with 60° between them, so it is equilateral
Area = (√3⁄4) × 6² = (√3⁄4) × 36 = 9√3

Segment = sector − triangle = (6π − 9√3) sq. cm → option (b)

Why the others are wrong

• (a)
  (3π−9√3) sq. cm — 3π is half the sector: (60⁄360) × 36π = 6π. And 3π ≈ 9.42 is less than 9√3 ≈ 15.59, so this expression is negative, which no area can be.
• (c)
  (6π−18√3) sq. cm — 18√3 is twice the triangle's area, (√3⁄4) × 36 = 9√3. The result 6π − 18√3 ≈ 18.85 − 31.18 is negative, so it cannot be an area.
• (d)
  (12π−18√3) sq. cm — 12π − 18√3 is exactly twice the segment, 2 × (6π − 9√3). A 60° sector is one-sixth of 36π, which is 6π, not 12π.

A chord splits a circle into two segments. The minor one, on the side of the smaller arc, is what is left of the sector once the triangle of two radii and the chord is removed.

Area of minor segment = (θ⁄360°) × πr² − ½ r² sin θ.

At 60° the triangle is equilateral: the angles opposite the equal radii are equal and share the remaining 120°. Its area is (√3⁄4)r².

Key facts

• Area of a minor segment = (θ⁄360°) × πr² − ½ r² sin θ.
• A 60° central angle makes the triangle of two radii and the chord equilateral, with area (√3⁄4)r².
• A 60° sector is one-sixth of the circle: here 36π ÷ 6 = 6π.
• The major segment is the rest of the circle: πr² minus the minor segment.

Common traps

• Stopping at the sector area, 6π, and forgetting to subtract the triangle.
• Taking the triangle as ½ × 6 × 6 = 18, which treats the 60° angle as a right angle.
• Picking an option without checking its sign: 3π − 9√3 and 6π − 18√3 are both negative.

No directly related past PYQ was found.