A slice of a circle
Tonight's idea
Cut a circle from its centre, like a pizza. The slice is a sector. Its curved edge is an arc. The angle at the centre, \(\theta\) (theta), says how big the slice is.
A full turn is 360°. Turning a circle about its centre changes nothing, so every 1° slice is identical to every other, and 360 of them make the whole circle. Each has \(\tfrac{1}{360}\) of the circumference and \(\tfrac{1}{360}\) of the area. A slice of \(\theta\) degrees is \(\theta\) of these, so it has the fraction \(\tfrac{\theta}{360}\) of both.
Arc length \(= \dfrac{\theta}{360} \times 2\pi r\) Sector area \(= \dfrac{\theta}{360} \times \pi r^2\)
There is nothing new to memorise. Find the fraction first, then take that fraction of the circumference \(2\pi r\) or the area \(\pi r^2\). A 90° sector is a quarter of the circle; a 60° sector is a sixth. Leave \(\pi\) in the answer for an exact value, and turn it into a decimal only at the end.
Watch one: a 120° sector of radius 6 cm
Fraction: \(\tfrac{120}{360} = \tfrac13\).
Arc: \(\tfrac13 \times 2\pi \times 6 = 4\pi \approx 12.57\) cm.
Area: \(\tfrac13 \times \pi \times 6^2 = 12\pi \approx 37.70\) cm².
The perimeter of the slice is the arc plus two straight edges: \(4\pi + 12 \approx 24.57\) cm. Forgetting the two radii is the usual slip.
Where you met the whole circle: Roll a wheel once (grade 7) for \(2\pi r\), and Why a pizza is πr² (grade 6).
Your turn
- Find the arc length and the area of each sector, exactly and to two decimal places. (a) Radius 10 cm, angle 90°. (b) Radius 5 cm, angle 36°.
- The minute hand of a clock is 12 cm long. In 25 minutes, (a) what angle does it turn through, (b) how far does its tip travel, and (c) what area does it sweep over?
- Work backwards. (a) An arc of a circle of radius 12 cm is \(5\pi\) cm long. What is the angle at the centre? (b) A 60° sector has area \(6\pi\) cm². What is its radius?
- One slice is 40° of a pizza of radius 18 cm. Another is 90° of a pizza of radius 12 cm. (a) Which slice has more pizza? (b) Which has more crust along its curved edge? (c) Explain how both answers can be true.
Stretch: a sector as a triangle
(a) Show that sector area \(= \tfrac12 \times \text{arc length} \times r\), whatever the angle. Which triangle formula does that look like? (b) Find the angle of a sector whose arc is exactly as long as its radius.
Answers
- (a) A quarter: arc \(\tfrac14 \times 20\pi = 5\pi \approx 15.71\) cm; area \(\tfrac14 \times 100\pi = 25\pi \approx 78.54\) cm². (b) A tenth: arc \(\tfrac{1}{10} \times 10\pi = \pi \approx 3.14\) cm; area \(\tfrac{1}{10} \times 25\pi = 2.5\pi \approx 7.85\) cm².
- (a) 60 minutes is 360°, so each minute is 6° and 25 minutes is 150°. (b) The fraction is \(\tfrac{150}{360} = \tfrac{5}{12}\): \(\tfrac{5}{12} \times 24\pi = 10\pi \approx 31.4\) cm. (c) \(\tfrac{5}{12} \times 144\pi = 60\pi \approx 188.5\) cm².
- (a) The circumference is \(24\pi\), so the fraction is \(\tfrac{5\pi}{24\pi} = \tfrac{5}{24}\) and the angle is \(\tfrac{5}{24} \times 360 = 75^\circ\). (b) A sixth of the circle is \(6\pi\), so the circle is \(36\pi = \pi r^2\) and \(r = 6\) cm.
- (a) Neither: \(\tfrac19 \times 324\pi = 36\pi\) and \(\tfrac14 \times 144\pi = 36\pi\), about 113.1 cm² each. (b) The second: its arc is \(\tfrac14 \times 24\pi = 6\pi \approx 18.8\) cm against \(\tfrac19 \times 36\pi = 4\pi \approx 12.6\) cm. (c) Area depends on \(r^2\) but arc only on \(r\). The first slice is long and thin, the second short and wide.
Stretch. (a) \(\tfrac12 \times \left(\tfrac{\theta}{360} \times 2\pi r\right) \times r = \tfrac{\theta}{360} \times \pi r^2\), the sector area. It looks like \(\tfrac12 \times \text{base} \times \text{height}\), with the arc as base and the radius as height. Check with question 4: \(\tfrac12 \times 4\pi \times 18 = 36\pi\). (b) \(\tfrac{\theta}{360} \times 2\pi r = r\) gives \(\theta = \dfrac{360}{2\pi} = \dfrac{180}{\pi} \approx 57.3^\circ\), the same for every radius. This angle is called one radian.