Circle Mensuration
Master circle mensuration in radians for Queensland Year 11 Mathematical Methods (QCAA). When the angle is measured in radians, the length of an arc is simply the radius times the angle.
You will learn to find arc length, work out the area of a sector as half the radius squared times the angle, and calculate the perimeter of a sector and the area of a segment — practical geometry in the QCAA course.
Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), circular measure uses radians to give neat formulas for a circle. The arc length of a sector is \(\ell=r\theta\) and the sector area is \(A=\dfrac12 r^2\theta\), where \(\theta\) is in radians. This page also covers the perimeter of a sector and the area of a segment.
A radian is the angle subtended at the centre of a circle by an arc equal in length to the radius. A full turn is \(2\pi\) radians \(=360^\circ\), so \(\pi\ \text{rad}=180^\circ\).
A sector is the pie-slice region between two radii and an arc. Its arc length is \(\ell=r\theta\) and its area is \(A=\dfrac12 r^2\theta\) — both requiring \(\theta\) in radians.
The perimeter of a sector is the arc plus the two radii, \(2r+\ell\). A segment is the region between a chord and its arc; its area is the sector area minus the triangle, \(\dfrac12 r^2(\theta-\sin\theta)\).
With \(\theta\) in radians:
How to solve a circle-mensuration problem
- Units: make sure \(\theta\) is in radians; convert from degrees with \(\pi\ \text{rad}=180^\circ\) if needed.
- Choose the formula: \(\ell=r\theta\) for arc length, \(A=\dfrac12 r^2\theta\) for sector area, \(2r+\ell\) for perimeter.
- Substitute and evaluate: put in \(r\) and \(\theta\); keep \(\pi\) exact where you can.
- Rearrange if solving backwards: from a given \(\ell\) or \(A\), make \(\theta\) or \(r\) the subject.
Formula — \(\ell=r\theta\) with \(\theta\) in radians:
| \(\ell\) | \(=\) | \(r\theta\) |
| \(=\) | \(6\times\dfrac{\pi}{3}\) | |
| \(=\) | \(2\pi\) |
Arc length \(=2\pi\) cm \(\approx 6.28\) cm.
Formula — \(A=\dfrac12 r^2\theta\):
| \(A\) | \(=\) | \(\dfrac12 r^2\theta\) |
| \(=\) | \(\dfrac12\times 8^2\times\dfrac{\pi}{4}\) | |
| \(=\) | \(\dfrac12\times 64\times\dfrac{\pi}{4}\) | |
| \(=\) | \(8\pi\) |
Area \(=8\pi\) cm\(^2\approx 25.1\) cm\(^2\).
Arc length first:
| \(\ell\) | \(=\) | \(r\theta\) |
| \(=\) | \(5\times 1.2\) | |
| \(=\) | \(6\) |
Perimeter — arc plus two radii:
| \(P\) | \(=\) | \(2r+\ell\) |
| \(=\) | \(2\times 5+6\) | |
| \(=\) | \(16\) |
Perimeter \(=16\) cm.
Segment \(=\) sector \(-\) triangle \(=\dfrac12 r^2(\theta-\sin\theta)\):
| \(A\) | \(=\) | \(\dfrac12 r^2(\theta-\sin\theta)\) |
| \(=\) | \(\dfrac12\times 10^2\left(\dfrac{\pi}{2}-\sin\dfrac{\pi}{2}\right)\) |
Evaluate \(\sin\dfrac{\pi}{2}=1\):
| \(=\) | \(50\left(\dfrac{\pi}{2}-1\right)\) | |
| \(=\) | \(25\pi-50\) |
Segment area \(=25\pi-50\) cm\(^2\approx 28.5\) cm\(^2\).
Common pitfalls
Frequently asked questions
What is the formula for arc length?
\(\ell=r\theta\), where \(r\) is the radius and \(\theta\) is the central angle in radians.
What is the formula for the area of a sector?
\(A=\dfrac12 r^2\theta\), with \(\theta\) in radians. Remember the factor of one half and to square the radius.
Do I have to use radians?
Yes for \(\ell=r\theta\) and \(A=\dfrac12 r^2\theta\). If the angle is in degrees, either convert with \(\pi\ \text{rad}=180^\circ\) or use \(\dfrac{\theta}{360}\times 2\pi r\) and \(\dfrac{\theta}{360}\times\pi r^2\).
How do I find the perimeter of a sector?
Add the arc length to the two straight radii: perimeter \(=2r+\ell=2r+r\theta\).
What is the area of a segment?
It is the sector minus the triangle: \(\dfrac12 r^2(\theta-\sin\theta)\).