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Summary

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):

$Area = \dfrac {\theta}{360 \degree} \times \pi r^2$

Converting this formula such that radians is used gives:

$\begin{aligned}Area &= \dfrac {\theta}{2 \pi} \times \pi r^2 \\\\& = \dfrac {\theta}{2 \cancel\pi} \times \cancel\pi r^2\end {aligned}$

This gives the formula for the area of a sector as follows:

\boxed{A = \dfrac 1 2 r^2 \theta}

$A$ - area of the sector
$r$ - radius
$\theta$ - angle in radians

Maths; Radians; KS5 Year 13; Areas of sectors and segments

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:

$\begin{aligned}A &= \dfrac 1 2 r_L^2 \theta - \dfrac 1 2 r_S^2 \theta \\\\&= \dfrac 1 2 \times 4^2 \times \dfrac {\pi} {3} - \dfrac 1 2 \times 2^2 \times \dfrac {\pi} {3} \\\\&= \underline{ 2 \pi \ cm^2}\end{aligned}$

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 $A$ $=$ area of sector $-$ area of triangle $OPQ$ For the area of the triangle $OPQ$ , use $A = \dfrac 1 2 ab \sin C$ : $A = \dfrac 1 2 r^2 \theta - \dfrac 1 2 r^2 \sin \theta$ Factorising gives:
$A$ - area of the segment $r$ - radius $\theta$ - angle in radians	$\boxed{A = \dfrac 1 2 r^2 (\theta - \sin\theta)}$

Example 2

Two identical circles of radius $10 \ cm$ overlap each other so that their centres are $10 \sqrt3 \ cm$ 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 $\angle AOB$ . Use triangle $AOB$ :

Use trigonometry to find angle $\theta$ 

$\begin {aligned}\cos \Big (\dfrac {\theta}{2}\Big ) &= \dfrac {5\sqrt3}{10} \\\\\dfrac {\theta}{2} &= \cos^{-1}\Big ( \dfrac {5\sqrt3}{10}\Big ) \\\\\dfrac {\theta}{2} &= \dfrac {\pi}{6} \ rads \\\\\theta &= \dfrac {\pi}{3} \ rads \end{aligned}$

The overlapping area is the area of two segments:

$\begin {aligned}Area &= \dfrac 1 2 r^2 (\theta - \sin (\theta)) \times 2 \\ \\&= \dfrac 1 2 \times 10^2 \times \Big( \dfrac{\pi}{3} - \sin \big(\dfrac{\pi}{3}\big) \Big) \times 2 \\ \\&=\underline{18.1 \ cm^2 \ (3 \ s.f.)}\end {aligned}$

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