Calculate arc length and sector area as fractions of a circle, and find the perimeter and area of segments

What this dot point is asking

SEAB wants you to find the length of an arc and the area of a sector by treating each as a fraction of the whole circle determined by the central angle, and to find the perimeter of a sector and the area of a segment. These build directly on the circle area and circumference formulas.

The answer

A sector as a fraction of a circle

A sector is a slice of a circle bounded by two radii and an arc. The fraction of the circle it covers is the central angle over $360^\circ$:

Everything about a sector follows from this fraction applied to the whole circle.

Arc length

The arc is that fraction of the full circumference:

Sector area

The sector area is the same fraction of the full circle's area:

Perimeter of a sector and area of a segment

The perimeter of a sector is the arc length plus the two bounding radii, so add $2r$ to the arc. A segment is the region between a chord and its arc; its area is the sector area minus the area of the triangle formed by the two radii and the chord.

Finding the segment area in full

A segment is the region between a chord and its arc, and its area is the sector area minus the triangle formed by the two radii and the chord. The triangle is found with the trigonometric area rule, $\tfrac{1}{2}r^2\sin\theta$, using the same central angle. So the segment area is $\tfrac{\theta}{360^\circ} \times \pi r^2 - \tfrac{1}{2}r^2\sin\theta$. For a sector of radius $10$ and angle $90^\circ$, the sector area is a quarter circle and the triangle is $\tfrac{1}{2}(10)^2\sin 90^\circ = 50$, so the segment is the difference. Combining the sector formula with the triangle area rule is the standard route to a segment, and remembering to subtract the triangle is where marks are commonly lost.

Working backwards to the angle or radius

Both the arc-length and sector-area formulas rearrange, so a question can give the arc or area and ask for the central angle or the radius. From arc length $= \tfrac{\theta}{360^\circ} \times 2\pi r$, you can solve for $\theta$ if the arc and radius are known, or for $r$ if the arc and angle are known. For instance, an arc of $\tfrac{1}{4}$ of the circumference must subtend $90^\circ$, since $\tfrac{\theta}{360} = \tfrac{1}{4}$. Recognising that the same formula solves for whichever of arc, angle, and radius is unknown, once the other two are given, is the flexibility these questions test.

Examples in context

Example 1. A pizza slice. A slice of a circular pizza is a sector; its curved crust is the arc and the cheese area is the sector area. Sharing a pizza into equal slices divides $360^\circ$ by the number of slices to find each central angle.

Example 2. A windscreen wiper. A wiper sweeping through an angle clears a sector-shaped region of the windscreen. The area cleared is a sector area, and the length the wiper tip travels is the arc length.

Try this

Q1. A sector has radius $12\ \text{cm}$ and angle $90^\circ$. State the fraction of the circle it covers. [1 mark]

• Cue. $\dfrac{90}{360} = \dfrac{1}{4}$.

Q2. Find the arc length of a sector with radius $5\ \text{cm}$ and angle $72^\circ$, taking $\pi = 3.142$. [2 marks]

• Cue. $\dfrac{72}{360} \times 2 \times 3.142 \times 5 = \dfrac{1}{5} \times 31.42 = 6.284\ \text{cm}$.

Q3. Find the area of a sector with radius $8\ \text{cm}$ and angle $45^\circ$, taking $\pi = 3.142$. [2 marks]

• Cue. $\dfrac{45}{360} \times 3.142 \times 64 = \dfrac{1}{8} \times 201.088 = 25.136\ \text{cm}^2$.