8.02 Circumference and area of circular shapes
Circumference of a circle
The circumference of a circle is the distance around the edge of a circle. In other words, 'circumference' is a specific term for the perimeter of the circle.
$C=2\pi r$C=2πr
where $C$C is the circumference, and $r$r is the radius.
For example, if the radius of a circle is $8$8 cm, the circumference, $C=2\pi r=2\times\pi\times8=50.3$C=2πr=2×π×8=50.3 cm (rounded to one decimal place).
Arc of a circle
An arc of a circle is a section of the edge of a circle.
The curved side of a sector of a circle is also an arc of the circle.
Arc length
As the arc of a circle is a fraction of the edge of a circle, we can calculate its arc length $s$s using a variation of the circumference of a circle formula.
Semicircular arc
Imagine first that the arc length that we want is half of the edge of the circle, then we could take half the circumference and get $s=\frac{1}{2}\times2\pi r$s=12×2πr
Quarter circle arc
What if we wanted a quarter of the edge? Then the arc length would be $s=\frac{1}{4}\times2\pi r$s=14×2πr.
What if we want a different fraction? Particularly, the fraction created by using an angle $\theta$θ. Then we would use $s=\frac{\theta}{360}2\pi r$s=θ3602πr.
If $\theta$θ is the angle at the centre of the circle, measured in degrees and subtended by an arc, then the arc length can be calculated using the formula:
| $s$s | $=$= | $\frac{\theta}{360}\times2\pi r$θ360×2πr |
| $=$= | $r\times\frac{\pi}{180}\times\theta$r×π180×θ |
The following applet helps with a visual connection between the circle, the arc and the formula.
Practice questions
Question 1
Find the length of the arc in the figure correct to one decimal place.
Question 2
Find the length of the arc in the figure correct to one decimal place.
Question 3
If the arc formed by two points on a sphere with a radius of $2$2m subtends an angle of $37$37° at the centre, find the length of the arc correct to two decimal places.
Area of a circle
The area of a full circle, measured in square units, can be found using the following formula:
$\text{Area of a circle}=\pi r^2$Area of a circle=πr2
Sectors
Given that the area of a circle is $\pi r^2$πr2 the area of a sector is some fraction of that full area $\frac{\theta}{360}\times\pi r^2$θ360×πr2.
There is a minor sector and a major sector associated with any given angle at the centre. The area of the corresponding sector can be found by replacing $\theta$θ with $360-\theta$360−θ in the formula for the area of a sector, to become $\frac{360-\theta}{360}\times\pi r^2$360−θ360×πr2. Notice this simplifies to $\pi r^2-\frac{\theta}{360}\times\pi r^2$πr2−θ360×πr2, i.e. subtracting the area of the original sector from the area of the whole circle.
$Area=\pi r^2$Area=πr2
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| $Area=\frac{\theta}{360}\times\pi r^2$Area=θ360×πr2 | $Area=\frac{360-\theta}{360}\times\pi r^2$Area=360−θ360×πr2 |
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| $Area=\frac{360-\theta}{360}\times\pi r^2$Area=360−θ360×πr2 | $Area=\frac{\theta}{360}\times\pi r^2$Area=θ360×πr2 |
Segment
A segment is the region bounded by a chord and the arc subtended by the chord (a chord is an interval connecting two points on a circle).
The area of a minor segment is the area of a sector minus the area of a triangle.
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| $Area=\frac{\theta}{360}\times\pi r^2$Area=θ360×πr2 | $\frac{1}{2}r^2\sin\theta$12r2sinθ | $Area=\frac{\theta}{360}\times\pi r^2-\frac{1}{2}r^2\sin\theta$Area=θ360×πr2−12r2sinθ |
Here we obtained the area of an isoceles triangle $\frac{1}{2}r^2\sin\theta$12r2sinθ from the formula $Area=\frac{1}{2}ab\sin C$Area=12absinC.
For a segment which subtends an angle $\theta$θ at the centre of a circle:
$\text{Area of a segment}$Area of a segment$=$=$\frac{\theta}{360}\times\pi r^2-\frac{1}{2}r^2\sin\theta$θ360×πr2−12r2sinθ
Area of a major segment
To find the area of a major segment, we can just take away its corresponding minor segment from the area of the whole circle.
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| $\pi r^2$πr2 | $\frac{\theta}{360}\times\pi r^2-\frac{1}{2}r^2\sin\theta$θ360×πr2−12r2sinθ | $Area=\pi r^2\left(\frac{360-\theta}{360}\right)+\frac{1}{2}r^2\sin\theta$Area=πr2(360−θ360)+12r2sinθ |
For the major segment, formed from the minor segment which subtends an angle $\theta$θ at the centre of a circle:
$\text{Area of major segment}$Area of major segment$=$=$\frac{360-\theta}{360}\times\pi r^2+\frac{1}{2}r^2\sin\theta$360−θ360×πr2+12r2sinθ
Practice questions
Question 4
The diagram shows a circle with radius $8$8 units, and chord $AB$AB subtending an angle of $\frac{\pi}{3}$π3 radians at the centre.
Find the exact area of the minor segment cut off by chord $AB$AB. Fully simplify your answer.
Question 5
In the diagram, $O$O is the centre of the circle, and sector $OAB$OAB takes up $\frac{4}{9}$49 of the circle.
Find the area of the minor segment cut off by chord $AB$AB correct to one decimal place.
Question 6
Consider a circle with centre $O$O and a chord $AB$AB subtended by an angle of $\theta$θ radians at the centre. The radius is $30$30 cm and the area of sector $OAB$OAB is $75\pi$75π cm2.
Solve for $\theta$θ, the angle at the centre.
Find the area of the minor segment cut off by chord $AB$AB.
Find the area of the major segment cut off by chord $AB$AB.
Determine the exact ratio of the area of the major segment to the area of the minor segment.