Areas of sectors and segments In a nutshell The formulae for calculating areas of sectors and segments is simplified now that the angle can be measured in radians.
Area of a sector Previously, you have learnt that the area of a sector can be calculated using the formula (using an angle measured in degrees):
A r e a = θ 360 ° × π r 2 Area = \dfrac {\theta}{360 \degree} \times \pi r^2 A re a = 360° θ × π r 2
Converting this formula such that radians is used gives:
A r e a = θ 2 π × π r 2 = θ 2 π × π r 2 \begin{aligned}Area &= \dfrac {\theta}{2 \pi} \times \pi r^2 \\\\& = \dfrac {\theta}{2 \cancel\pi} \times \cancel\pi r^2\end {aligned} A re a = 2 π θ × π r 2 = 2 π θ × π r 2
This gives the formula for the area of a sector as follows:
A = 1 2 r 2 θ \boxed{A = \dfrac 1 2 r^2 \theta} A = 2 1 r 2 θ
A A A - area of the sector r r r - radius θ \theta θ - angle in radians
Example 1 Calculate the exact area of the shaded region shown:
The area of the shaded region can be found by subtracting the area of the smaller sector from the area of the larger sector:
A = 1 2 r L 2 θ − 1 2 r S 2 θ = 1 2 × 4 2 × π 3 − 1 2 × 2 2 × π 3 = 2 π c m 2 ‾ \begin{aligned}A &= \dfrac 1 2 r_L^2 \theta - \dfrac 1 2 r_S^2 \theta \\\\&= \dfrac 1 2 \times 4^2 \times \dfrac {\pi} {3} - \dfrac 1 2 \times 2^2 \times \dfrac {\pi} {3} \\\\&= \underline{ 2 \pi \ cm^2}\end{aligned} A = 2 1 r L 2 θ − 2 1 r S 2 θ = 2 1 × 4 2 × 3 π − 2 1 × 2 2 × 3 π = 2 π c m 2
Area of a segment The area of a segment can be found by taking the area of a sector and subtracting the area of a triangle:
Area
A A A = = = area of sector
− - − area of triangle
O P Q OPQ OPQ
For the area of the triangle
O P Q OPQ OPQ ,
use
A = 1 2 a b sin C A = \dfrac 1 2 ab \sin C A = 2 1 ab sin C :
A = 1 2 r 2 θ − 1 2 r 2 sin θ A = \dfrac 1 2 r^2 \theta - \dfrac 1 2 r^2 \sin \theta A = 2 1 r 2 θ − 2 1 r 2 sin θ
Factorising gives:
A A A - area of the segmentr r r - radiusθ \theta θ - angle in radians
A = 1 2 r 2 ( θ − sin θ ) \boxed{A = \dfrac 1 2 r^2 (\theta - \sin\theta)} A = 2 1 r 2 ( θ − sin θ )
Example 2 Two identical circles of radius 10 c m 10 \ cm 10 c m overlap each other so that their centres are 10 3 c m 10 \sqrt3 \ cm 10 3 c m apart. Calculate the area of the overlap to three significant figures.
Draw a sketch of the circles:
The area of the overlap is the area of one of the segments multiplied by two. It is therefore necessary to find the angle ∠ A O B \angle AOB ∠ A OB . Use triangle A O B AOB A OB :
Use trigonometry to find angle θ \theta θ
cos ( θ 2 ) = 5 3 10 θ 2 = cos − 1 ( 5 3 10 ) θ 2 = π 6 r a d s θ = π 3 r a d s \begin {aligned}\cos \Big (\dfrac {\theta}{2}\Big ) &= \dfrac {5\sqrt3}{10} \\\\\dfrac {\theta}{2} &= \cos^{-1}\Big ( \dfrac {5\sqrt3}{10}\Big ) \\\\\dfrac {\theta}{2} &= \dfrac {\pi}{6} \ rads \\\\\theta &= \dfrac {\pi}{3} \ rads \end{aligned} cos ( 2 θ ) 2 θ 2 θ θ = 10 5 3 = cos − 1 ( 10 5 3 ) = 6 π r a d s = 3 π r a d s
The overlapping area is the area of two segments:
A r e a = 1 2 r 2 ( θ − sin ( θ ) ) × 2 = 1 2 × 1 0 2 × ( π 3 − sin ( π 3 ) ) × 2 = 18.1 c m 2 ( 3 s . f . ) ‾ \begin {aligned}Area &= \dfrac 1 2 r^2 (\theta - \sin (\theta)) \times 2 \\ \\&= \dfrac 1 2 \times 10^2 \times \Big( \dfrac{\pi}{3} - \sin \big(\dfrac{\pi}{3}\big) \Big) \times 2 \\ \\&=\underline{18.1 \ cm^2 \ (3 \ s.f.)}\end {aligned} A re a = 2 1 r 2 ( θ − sin ( θ )) × 2 = 2 1 × 1 0 2 × ( 3 π − sin ( 3 π ) ) × 2 = 18.1 c m 2 ( 3 s . f . )