Area of Sectors
Theory
A sector takes the same share of a circle’s area as its angle takes of a full turn: \(A = \dfrac{\theta}{360} \times \pi r^2\). With \(\theta = 60^\circ\) and \(r = 6\) cm, \(A = \dfrac{1}{6} \times 36\pi = 6\pi~\text{cm}^2\). This Year 8 maths page shows how.
A whole circle has \(360^\circ\) at the centre and area \(\pi r^2\), so a sector with angle \(\theta\) has \(A = \dfrac{\theta}{360} \times \pi r^2\). It is the arc formula with \(\pi r^2\) in place of \(2\pi r\).
Special sectors. A quadrant (\(90^\circ\)) is a quarter of the circle and a semicircle (\(180^\circ\)) is a half. Cut a circle into \(n\) equal slices and each has angle \(360 \div n\). The major sector uses \(360 - \theta\).
Composite shapes. Add the pieces (a square with a semicircle on one side) or subtract them (a square with a quadrant cut out). A ring-shaped sector is the big sector take away the small sector.
A segment is the sector take away the triangle made by the two radii and the chord.
Working backwards. Substitute what you know, then solve for \(\theta\), or for \(r\) (divide by \(\pi\), then take the square root).
Area of a sector:
How to find the area of a sector
- Write the fraction \(\dfrac{\theta}{360}\) for the shaded piece and simplify it.
- Multiply by \(\pi r^2\).
- Give an exact answer or round at the end, in square units.
| \(A\) | \(=\) | \(\dfrac{60}{360} \times \pi \times 6^2\) |
| \(\) | \(=\) | \(\dfrac{1}{6} \times 36\pi = 6\pi\) |
\(6\pi~\text{cm}^2\)
| \(24\pi\) | \(=\) | \(\dfrac{\theta}{360} \times 64\pi\) |
| \(\theta\) | \(=\) | \(\dfrac{24}{64} \times 360 = 135\) |
\(\theta = 135^\circ\)
(i) Find its area in terms of \(\pi\).
(ii) Find its area to \(1\) decimal place.
(iii) A sector of the same radius has area \(60\pi~\text{cm}^2\). Find its angle.
| \(A\) | \(=\) | \(\dfrac{1}{12} \times 144\pi = 12\pi\) |
| \(\) | \(=\) | \(37.69\ldots\) |
| \(\theta\) | \(=\) | \(\dfrac{60}{144} \times 360 = 150\) |
(i) \(12\pi~\text{cm}^2\)
(ii) \(37.7~\text{cm}^2\)
(iii) \(150^\circ\)
| \(\theta\) | \(=\) | \(360 - 95 = 265\) |
| \(A\) | \(=\) | \(\dfrac{265}{360} \times 144\pi = 106\pi\) |
| \(\) | \(=\) | \(333.0\ldots\) |
\(333~\text{m}^2\)
| \(A_1\) | \(=\) | \(1.2 \times 1.2 = 1.44\) |
| \(A_2\) | \(=\) | \(\dfrac{180}{360} \times \pi \times 0.6^2 = 0.565\ldots\) |
| \(A\) | \(=\) | \(2.005\ldots\) |
\(2.01~\text{m}^2\)
| \(A\) | \(=\) | \(\dfrac{200}{360} \times (25\pi - 9\pi)\) |
| \(\) | \(=\) | \(\dfrac{5}{9} \times 16\pi = 27.92\ldots\) |
\(27.9~\text{m}^2\): the big sector take away the small sector.
Common pitfalls
Frequently asked questions
What is the formula for the area of a sector?
\(A = \dfrac{\theta}{360} \times \pi r^2\).
What is the area of a quadrant?
A quarter of the circle: \(\dfrac{1}{4}\pi r^2\).
How do I find the area of a segment?
Sector area minus the area of the triangle made by the two radii and the chord.
How do I find the angle of a sector from its area?
Divide the sector area by \(\pi r^2\), then multiply by \(360\).
What is a major sector?
The larger piece of the circle, with angle \(360 - \theta\).