C2: Arcs, Sectors and Segments

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Presentation transcript:

C2: Arcs, Sectors and Segments Learning Objective: to use radian measure to calculate an arc length, area of a sector and segment of a circle

Starter: 360o = 180o = 90o = 270o = 45o = 60o = 120o = 30o =

Using radians to measure arc length Suppose an arc AB of a circle of radius r subtends an angle of θ radians at the centre. If the angle at the centre is 1 radian then the length of the arc is r. r θ O A B If the angle at the centre is 2 radians then the length of the arc is 2r. If the angle at the centre is 0.3 radians then the length of the arc is 0.3r. In general: Point out that the size of the angle subtended by an arc is directly proportional to the length of the arc. Length of arc AB = θr where θ is measured in radians. When θ is measured in degrees the length of AB is

Finding the area of a sector We can also find the area of a sector using radians. r θ O A B Again suppose an arc AB subtends an angle of θ radians at the centre O of a circle. The angle at the centre of a full circle is 2π radians. So the area of the sector AOB is of the area of the full circle.  Area of sector AOB = If necessary remind students that the area of a full circle is πr2. In general: Area of sector AOB = r2θ where θ is measured in radians. When θ is measured in degrees the area of AOB is

Finding arc length and sector area A chord AB subtends an angle of radians at the centre O of a circle of radius 9 cm. Find in terms of π: a) the length of the arc AB. b) the area of the sector AOB. a) length of arc AB = θr 9 cm O A B = 6π cm b) area of sector AOB = r2θ = 27π cm2

Task 1: An arc AB of a circle, centre O and radius r cm, subtends an angle θ radians at O. Find the length of the arc AB and the area of the sector AOB when r = 8cm, θ = 0.85 r = 3.5cm, θ = 0.15 r = 6cm, θ = 3/8 π A sector of a circle of radius r cm contains an angle of 1.2 radians. Given that the sector has the same perimeter as a square of area 36 cm2 , find the value of r. Hence calculate the area of the sector. The area of a sector of a circle of radius 12cm is 100 cm2. Find the perimeter of the sector.

Finding the area of a segment The formula for the area of a sector can be combined with the formula for the area of a triangle to find the area of a segment. For example: A chord AB divides a circle of radius 5 cm into two segments. If AB subtends an angle of 45° at the centre of the circle, find the area of the minor segment to 3 significant figures. 5 cm 45° O A B The dots at the end of the number indicate that it has not been rounded but kept in the calculator. This is to avoid rounding errors. Occasionally the solution is required to be written exactly in which case it is left in terms of π. Let’s call the area of sector AOB AS and the area of triangle AOB AT.

Finding the area of a segment Now: Area of the minor segment = AS – AT = 9.8174… – 8.8388… = 0.979 cm2 (to 3 sig. figs.) This formula does not have to be leant since it can be derived from the formulae for the area of a sector and the area of a triangle. Students sometimes confuse sectors and segments. Remember, a sector is like a slice of cake (both contain the letter c). In general, the area of a segment of a circle of radius r is: where θ is measured in radians.

Task 2: Calculate the area of the shaded segment A B Calculate the area of the segment when r = 12cm and θ = π / 4 8 cm 30° O A B

Examination-style question In the following diagram AC is an arc of a circle with centre O and radius 10 cm and BD is an arc of a circle with centre O and radius 6 cm. = θ radians. O θ A B C D 6 cm 10 cm Find an expression for the area of the shaded region in terms of θ. Given that the shaded region is 25.6 cm2 find the value of θ. Calculate the perimeter of the shaded region.

Examination-style question a) Area of sector AOC = × 102 × θ = 50θ Area of sector BOD = × 62 × θ = 18θ Area of shaded region = 50θ – 18θ = 32θ b) 32θ = 25.6 θ = 25.6 ÷ 32 θ = 0.8 radians c) Perimeter of the shaded region = length of arc AC + length of arc BD + AB + CD = (10 × 0.8) + (6 × 0.8) + 8 = 20.8 cm