Area of a Segment of a Circle

A segment is the sliver between a chord and its arc. Its area is the classic “combined shape” subtraction: sector minus triangle. The triangle uses the $\frac12 ab\sin C$ rule with both sides equal to the radius, giving the compact formula $\frac12 r^2(\theta - \sin\theta)$.

A chord of a circle of radius $6$ cm subtends an angle of $2$ radians at the centre. Find the area of the minor segment cut off by the chord. [5]

The working

Step 1, sector area with $r = 6$, $\theta = 2$: $A_{\text{sector}} = \frac{1}{2}r^2\theta = \frac{1}{2}(36)(2) = 36 \text{ cm}^2 \quad \text{(M1)}$

Step 2, triangle area (two radii with included angle $\theta$), using $\frac12 r^2\sin\theta$: $A_{\text{triangle}} = \frac{1}{2}(6)^2 \sin 2 = 18\sin 2 \quad \text{(M1)}$

With the calculator in radian mode, $\sin 2 = 0.9093$, so $A_{\text{triangle}} = 18(0.9093) = 16.37 \text{ cm}^2$. (A1)

Step 3, segment = sector − triangle: $A_{\text{segment}} = 36 - 16.37 = 19.6 \text{ cm}^2 \ (3 \text{ s.f.}) \quad \text{(M1, A1)}$

Equivalently, $\frac12 r^2(\theta - \sin\theta) = \frac12(36)(2 - \sin 2) = 18(2 - 0.9093) = 19.6$ cm².

Where the marks are won and lost

• Radian mode is essential: $\sin 2$ means the sine of $2$ radians ($\approx 0.909$), not $2^\circ$. A calculator left in degrees gives $\sin 2^\circ = 0.035$ and wrecks the answer.
• The triangle area is $\frac12 r^2\sin\theta$, using the included angle. Using $\frac12 \times \text{base} \times \text{height}$ needs extra work you do not have here.
• Segment $=$ sector $-$ triangle for the minor segment. The major segment would be the circle’s area minus this.

Common mistakes

• Calculator in degree mode for $\sin\theta$.
• Adding the triangle instead of subtracting it.
• Using $\frac12 r^2\theta$ for the triangle (that is the sector, the triangle uses $\sin\theta$).

Full method: Problem-Solving with Circular Measure notes. Topic home: Circular Measure pillar.