Arc Lengths Problems
The perimeter of a shape with a curved edge is still the total distance around the outside. In NSW Year 8 maths (NESA outcome MA4‑LEN‑C‑01) a sector has a perimeter of its arc plus two radii, \(P = \frac{\theta}{360} \times 2\pi r + 2r\), with semicircles and quadrants as special cases.
You will find the perimeters of sectors, quadrants, the shape left when a slice is cut out, and composite figures made of straight sides, semicircles and arcs, leaving out any line drawn inside the shape. You will also work backwards from a perimeter to a radius or an angle, and cost timber edging for a sandpit or ribbon for a paper fan, rounding up to whole metres. It is part of Year 8 Area and Perimeter.
Theory
The perimeter of a shape with a curved edge is still the total distance around the outside. For a sector, \(P = \dfrac{\theta}{360} \times 2\pi r + 2r\): with \(r = 6\) cm and \(\theta = 120^\circ\), \(P = 4\pi + 12 \approx 24.57\) cm. This Year 8 maths page shows how.
A sector is bounded by an arc and two radii: \(P = \dfrac{\theta}{360} \times 2\pi r + 2r\).
A semicircle has \(P = \pi r + 2r\) (the same as \(\dfrac{\pi d}{2} + d\)); a quadrant has \(P = \dfrac{\pi r}{2} + 2r\). If a slice of \(\theta^\circ\) is cut away, the shape left has angle \(360 - \theta\).
Composite shapes. Add only the edges on the outside. A line inside the shape is not counted, but a curve that bulges inwards is. Two semicircles on the same diameter make one whole circle.
Missing straight pieces come from subtraction: the flat end of a half ring is the outer radius minus the inner radius.
Working backwards. Write \(P\) in terms of \(r\), take out \(r\) as a factor, then divide: \(P = \pi r + 2r = r(\pi + 2)\).
Perimeters with curves:
| Shape | Perimeter |
|---|---|
| sector | \(\dfrac{\theta}{360} \times 2\pi r + 2r\) |
| semicircle | \(\pi r + 2r\) |
| quadrant | \(\dfrac{\pi r}{2} + 2r\) |
| whole circle | \(2\pi r\) (nothing added) |
How to find a curved perimeter
- List every edge on the outside.
- Work out each arc with \(\dfrac{\theta}{360} \times 2\pi r\) and each straight edge.
- Add them, then round (or round up when buying).
Each curve is half a circumference, \(\dfrac{1}{2} \times 2\pi r = \pi r\). The big one has \(r = 4\) and the small one \(r = 2\); the straight edge \(AB\) is \(4\) cm.
| \(\text{big curve}\) | \(=\) | \(\pi \times 4 = 4\pi\) |
| \(\text{small curve}\) | \(=\) | \(\pi \times 2 = 2\pi\) |
| \(P\) | \(=\) | \(4 + 4\pi + 2\pi\) |
| \(\) | \(=\) | \(4 + 6\pi\) |
\((4 + 6\pi)\) cm
| \(\text{big curve}\) | \(=\) | \(\pi \times 3.5 = 3.5\pi\) |
| \(\text{each arc}\) | \(=\) | \(\dfrac{1}{3} \times 5\pi = \dfrac{5\pi}{3}\) |
| \(P\) | \(=\) | \(\dfrac{7\pi}{2} + \dfrac{10\pi}{3}\) |
| \(\) | \(=\) | \(\dfrac{41\pi}{6}\) |
\(\dfrac{41\pi}{6}\) cm
(i) exactly
(ii) to \(2\) decimal places
| \(l\) | \(=\) | \(\dfrac{1}{3} \times 12\pi = 4\pi\) |
| \(P\) | \(=\) | \(4\pi + 12\) |
| \(\) | \(=\) | \(24.566\ldots\) |
(i) \((4\pi + 12)\) cm
(ii) \(24.57\) cm
| \(\text{curve}\) | \(=\) | \(\dfrac{1}{2} \times \pi \times 2.6 = 1.3\pi\) |
| \(P\) | \(=\) | \(1.3\pi + 2.6 = 6.684\ldots\) |
\(6.68\) m
| \(\text{curves}\) | \(=\) | \(\pi \times 9 = 9\pi\) |
| \(P\) | \(=\) | \(20 + 20 + 9\pi = 68.27\ldots\) |
The two semicircles make one circle of diameter \(9\) m, so the track is \(68.3\) m.
| \(P\) | \(=\) | \(\pi r + 2r = r(\pi + 2)\) |
| \(r\) | \(=\) | \(25 \div (\pi + 2) = 4.86\ldots\) |
\(4.9\) m
Common pitfalls
Frequently asked questions
How do I find the perimeter of a sector?
Find the arc with \(\dfrac{\theta}{360} \times 2\pi r\), then add two radii.
What is the perimeter of a quadrant?
\(\dfrac{\pi r}{2} + 2r\).
Do lines inside a shape count?
No. Only the edges on the outside make up the perimeter.
How do I find the radius from a perimeter?
Write the perimeter in terms of \(r\), take \(r\) out as a factor, then divide.
What is the perimeter of a semicircle with radius r?
\(\pi r + 2r\).