Area
Theory
The area of a shape is the space inside it, measured in square units. This Year 9 maths page revises the Year 8 formulas for triangles, quadrilaterals, circles and sectors, and uses them for shaded regions and to work back from an area to a missing length.
The area is the amount of flat space inside a shape. It is measured in square units: \(\text{mm}^2\), \(\text{cm}^2\), \(\text{m}^2\) or hectares.
A triangle, parallelogram or trapezium uses its perpendicular height: the distance measured at right angles to the base. Look for the right-angle mark. It is not the sloping side. In an obtuse triangle the height can fall outside the triangle, meeting the base line extended.
A trapezium has one pair of parallel sides, \(a\) and \(b\). A rhombus and a kite have diagonals that cross at right angles, so their area comes from the two whole diagonals \(x\) and \(y\).
A circle of radius \(r\) has area \(\pi r^2\). A sector with angle \(\theta\) at the centre is the fraction \(\dfrac{\theta}{360}\) of the circle.
A shaded region is often a whole shape with a part removed: find the whole area and subtract the white part.
| Shape | Area |
|---|---|
| Rectangle | \(A = lw\) |
| Square | \(A = s^2\) |
| Triangle | \(A = \dfrac{1}{2}bh\) |
| Parallelogram | \(A = bh\) |
| Trapezium | \(A = \dfrac{1}{2}(a + b)h\) |
| Rhombus or kite | \(A = \dfrac{1}{2}xy\) |
| Circle | \(A = \pi r^2\) |
| Sector | \(A = \dfrac{\theta}{360} \times \pi r^2\) |
Working backwards, rearrange the formula:
How to find an area
- Name the shape and choose its formula.
- Check each measurement: the height must be perpendicular, diagonals must be whole, and a circle needs its radius (halve a diameter).
- Change all lengths to the same unit.
- Substitute and calculate. Give the answer in square units, rounded as asked.
Missing lengths and shaded regions
- Missing length: substitute the area and the lengths you know, then solve. For a circle, divide by \(\pi\) and take the square root.
- Shaded region: whole shape minus the white part, or add the shaded pieces.
- Changing one length: if one length of a rectangle or triangle is halved and the other stays the same, the area is halved too.
Use the base and the perpendicular height. The \(6.5\) cm sloping side is not needed.
| \(A\) | \(=\) | \(bh\) |
| \(\) | \(=\) | \(14 \times 6\) |
| \(\) | \(=\) | \(\mathbf{84} \text{ cm}^2\) |
Using \(6.5\) would give \(91 \text{ cm}^2\), which is too big. A rhombus is a parallelogram too, so \(A = bh\) works for it as well.
\(a\) and \(b\) are the two parallel sides. \(h\) is the distance between them.
| \(A\) | \(=\) | \(\dfrac{1}{2}(a + b)h\) |
| \(\) | \(=\) | \(\dfrac{1}{2} \times (7 + 11) \times 6\) |
| \(\) | \(=\) | \(\dfrac{1}{2} \times 18 \times 6\) |
| \(\) | \(=\) | \(\mathbf{54} \text{ m}^2\) |
Add the parallel sides first. The sloping sides play no part.
(ii) A sector has a radius of \(10\) cm and an angle of \(45^\circ\). Find its area.
Give both answers correct to two decimal places.
(i) The formula needs the radius, so halve the diameter first: \(r = 11\) cm.
| \(A\) | \(=\) | \(\pi r^2 = \pi \times 11^2\) |
| \(\) | \(=\) | \(121\pi = 380.132\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{380.13} \text{ cm}^2\) |
(ii) The sector is \(\dfrac{45}{360} = \dfrac{1}{8}\) of the circle.
| \(A\) | \(=\) | \(\dfrac{1}{8} \times \pi \times 10^2\) |
| \(\) | \(=\) | \(39.269\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{39.27} \text{ cm}^2\) |
Using \(22\) as the radius would give an area four times too big.
(ii) A circle has an area of \(50 \text{ cm}^2\). Find its radius, correct to two decimal places.
(i) Substitute what you know into \(A = \dfrac{1}{2}bh\), then solve.
| \(\dfrac{1}{2} \times 7.5 \times h\) | \(=\) | \(30\) |
| \(3.75h\) | \(=\) | \(30\) |
| \(h\) | \(=\) | \(30 \div 3.75 = \mathbf{8} \text{ m}\) |
(ii) Substitute into \(A = \pi r^2\) and undo the steps: divide by \(\pi\), then take the square root.
| \(\pi r^2\) | \(=\) | \(50\) |
| \(r^2\) | \(=\) | \(\dfrac{50}{\pi} = 15.915\ldots\) |
| \(r\) | \(=\) | \(\sqrt{15.915\ldots} = 3.989\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{3.99} \text{ cm}\) |
Check (i): \(\dfrac{1}{2} \times 7.5 \times 8 = 30\) \(\checkmark\)
Common pitfalls
Frequently asked questions
What is the formula for the area of a triangle?
\(A = \dfrac{1}{2}bh\), where \(b\) is the base and \(h\) is the perpendicular height to that base.
What is the formula for the area of a trapezium?
\(A = \dfrac{1}{2}(a + b)h\), where \(a\) and \(b\) are the parallel sides and \(h\) is the perpendicular height between them.
How do you find the area of a kite or rhombus?
Multiply the two diagonals and halve the result: \(A = \dfrac{1}{2}xy\). Diagonals of \(16\) cm and \(9\) cm give \(72 \text{ cm}^2\).
How do you find the area of a circle from the diameter?
Halve the diameter to get the radius, square it, then multiply by \(\pi\). For \(d = 22\) cm, \(A = 121\pi \approx 380.13 \text{ cm}^2\).
How do you find the area of a sector?
Multiply the area of the whole circle by \(\dfrac{\theta}{360}\), where \(\theta\) is the angle at the centre.
How do you find the radius from the area of a circle?
Divide the area by \(\pi\), then take the square root: \(r = \sqrt{\dfrac{A}{\pi}}\).