A brief note on Arcs, Sectors and Segments . Read it before a test. Try each question at the end before you open its answer.
Arcs and sectors Arc length: l = θ 360 × 2 π r l = \dfrac{\theta}{360} \times 2\pi r l = 360 θ × 2 π r , with θ \theta θ in degrees. In radians, l = r θ l = r\theta l = r θ . Perimeter of a sector: l + 2 r l + 2r l + 2 r . Area of a sector: A = θ 360 × π r 2 = 1 2 l r A = \dfrac{\theta}{360} \times \pi r^2 = \dfrac{1}{2}\,l\,r A = 360 θ × π r 2 = 2 1 l r . The minor and the major sector add up to the whole circle. The major sector has the angle 360 ∘ − θ 360^\circ - \theta 36 0 ∘ − θ . A wiper blade between the radii r r r and R R R sweeps θ 360 π ( R 2 − r 2 ) \dfrac{\theta}{360}\,\pi\,(R^2 - r^2) 360 θ π ( R 2 − r 2 ) . Segments Minor segment: the sector minus the triangle O A B OAB O A B : θ 360 π r 2 − 1 2 r 2 sin θ \dfrac{\theta}{360}\pi r^2 - \dfrac{1}{2}r^2\sin\theta 360 θ π r 2 − 2 1 r 2 sin θ . Major segment: the whole circle minus the minor segment. For θ = 90 ∘ \theta = 90^\circ θ = 9 0 ∘ the triangle is 1 2 r 2 \dfrac{1}{2}r^2 2 1 r 2 . For θ = 60 ∘ \theta = 60^\circ θ = 6 0 ∘ it is 3 4 r 2 \dfrac{\sqrt{3}}{4}r^2 4 3 r 2 . Use π = 22 7 \pi = \dfrac{22}{7} π = 7 22 only when told, and show the exact form with π \pi π . A worked example A chord subtends 60 ∘ 60^\circ 6 0 ∘ at the centre of a circle of radius 10 cm 10\ \text{cm} 10 cm . Find the area of the minor segment.
The sector is 1 6 π × 100 = 50 π 3 cm 2 \dfrac{1}{6}\pi \times 100 = \dfrac{50\pi}{3}\ \text{cm}^2 6 1 π × 100 = 3 50 π cm 2 . The triangle is equilateral: 3 4 × 100 = 25 3 cm 2 \dfrac{\sqrt{3}}{4} \times 100 = 25\sqrt{3}\ \text{cm}^2 4 3 × 100 = 25 3 cm 2 . The segment is ( 50 π 3 − 25 3 ) cm 2 \left(\dfrac{50\pi}{3} - 25\sqrt{3}\right)\ \text{cm}^2 ( 3 50 π − 25 3 ) cm 2 , which is 9.06 cm 2 9.06\ \text{cm}^2 9.06 cm 2 . Practice with new numbers
A program to try The program finds the arc, the sector and both segments for the worked example.
Run it. Does the minor segment agree with the worked example?
1. Change angle = 60 to 90. What is the minor segment now?
Mistakes to avoid Giving the sector as the segment. Using 2 π r 2\pi r 2 π r for an area. The area has r 2 r^2 r 2 . Forgetting the two radii in the perimeter of a sector. Using degrees in l = r θ l = r\theta l = r θ . That rule needs radians. Check yourself An arc makes 90 ∘ 90^\circ 9 0 ∘ at the centre of a circle of radius 14 cm 14\ \text{cm} 14 cm . Take π = 22 7 \pi = \dfrac{22}{7} π = 7 22 . How long is it? Stuck? Show a hint Show the answerA sector of a circle of radius 6 cm 6\ \text{cm} 6 cm has an angle of 60 ∘ 60^\circ 6 0 ∘ . Write its area in terms of π \pi π . Stuck? Show a hint Show the answerA sector has an arc of 20 cm 20\ \text{cm} 20 cm and a radius of 5 cm 5\ \text{cm} 5 cm . What is its area? Stuck? Show a hint Show the answerThe minor segment of a circle has an area of 12 cm 2 12\ \text{cm}^2 12 cm 2 and the circle has an area of 50 cm 2 50\ \text{cm}^2 50 cm 2 . What is the major segment? Stuck? Show a hint Show the answer