A circle of radius of 12 cm. Two radii are drawn such that the length of the chord connecting their endpoints is 12 cm. What is the area of the minor segment formed?

Correct — B. The chord equals the radius, so the triangle of two radii and the chord is equilateral and the central angle is 60°.

Sector = (60⁄360) × π × 12²
= (1⁄6) × 144π = 24π

Triangle = (√3⁄4) × 12²
= (√3⁄4) × 144 = 36√3

Minor segment = sector − triangle
= (24π − 36√3) sq. cm → option (b)

Why the others are wrong

• (a)
  (12π−18√3) sq. cm — 12π − 18√3 is exactly half the segment: both the sector and the triangle are halved. A 60° sector of a 144π circle is 24π, not 12π.
• (c)
  (12π−36√3) sq. cm — 12π − 36√3 ≈ 37.70 − 62.35 is negative, so it cannot be an area. It pairs half the sector with the full triangle.
• (d)
  (24π−18√3) sq. cm — 24π − 18√3 has the right sector but half the triangle. An equilateral triangle of side 12 cm has area (√3⁄4) × 144 = 36√3, not 18√3.

A chord splits a circle into two segments. The minor segment is the sector on the smaller arc minus the triangle formed by the chord and the two radii.

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

Here θ is not given. It comes from the lengths: a chord equal to the radius makes an equilateral triangle, so θ = 60°.

Key facts

• Area of a minor segment = (θ⁄360°) × πr² − ½ r² sin θ.
• A chord equal to the radius subtends 60° at the centre.
• An equilateral triangle of side a has area (√3⁄4)a².
• A 60° sector is one-sixth of the circle: 144π ÷ 6 = 24π.

Common traps

• Stopping at the sector area, 24π, and forgetting to subtract the triangle.
• Taking the triangle as ½ × 12 × 12 = 72, which treats the 60° angle as a right angle.
• Looking for a stated angle: the stem gives only lengths, and chord = radius is what fixes 60°.

No directly related past PYQ was found.