Perimeter of a Segment

A segment’s perimeter is its arc plus its chord. The arc is $r\theta$; the chord comes from the cosine rule on the isosceles triangle of two radii and the chord. Combining the two is the whole task.

A chord of a circle of radius $8$ cm subtends an angle of $1.5$ radians at the centre. Find the perimeter of the minor segment. [5]

The working

Step 1, arc length with $r = 8$, $\theta = 1.5$: $\text{arc} = r\theta = 8 \times 1.5 = 12 \text{ cm} \quad \text{(M1)}$

Step 2, chord length by the cosine rule on the triangle with two sides $r = 8$ and included angle $1.5$ rad: $\text{chord}^2 = 8^2 + 8^2 - 2(8)(8)\cos 1.5 = 128 - 128\cos 1.5 \quad \text{(M1)}$

With the calculator in radian mode, $\cos 1.5 = 0.0707$: $\text{chord}^2 = 128 - 128(0.0707) = 128 - 9.05 = 118.95 \;\Rightarrow\; \text{chord} = 10.9 \text{ cm} \quad \text{(A1)}$

Step 3, perimeter = arc + chord: $P = 12 + 10.9 = 22.9 \text{ cm} \ (3 \text{ s.f.}) \quad \text{(M1, A1)}$

Where the marks are won and lost

• The perimeter is arc plus chord, not arc plus two radii (that’s a sector). A segment is bounded by the arc and the chord.
• Radian mode for $\cos 1.5$: it’s $1.5$ radians ($\approx 0.071$), not $1.5^\circ$. Degree mode wrecks the chord.
• Use the cosine rule with the included angle $\theta$ between the two radii; both sides are $r$.

Common mistakes

• Adding two radii instead of the chord (confusing segment with sector).
• Calculator in degree mode for $\cos 1.5$.
• Forgetting to square-root to get the chord from chord².

Full method: Problem-Solving with Circular Measure notes. See also Area of a Segment. Topic home: Circular Measure pillar.