Areas of Quadrilaterals
The areas of parallelograms, trapeziums, rhombuses and kites all come from rectangles and triangles. This NSW Year 9 Core maths subtopic revises those area formulas from Stage 4 (NESA outcome MA4‑ARE‑C‑01) and adds questions that run the formulas backwards.
You will use \(A = bh\) for a parallelogram, \(A = \frac{1}{2}h(a + b)\) for a trapezium and \(A = \frac{1}{2}xy\) for a rhombus or kite, where \(x\) and \(y\) are the diagonals. You will read the perpendicular height, not the sloping side, off a diagram, convert units before calculating, and rearrange each formula to find a missing base, height or diagonal from a given area. Real problems include costing the cloth for a kite-shaped sail. It is part of Year 9 Core Surface Area and Volume.
Every question with a fully worked solution.
- Areas of Quadrilaterals - Video - Areas of Triangles and Quadrilaterals Watch
Theory
The area of a quadrilateral comes from a short formula, as long as you use a perpendicular measurement. This Year 9 maths page revises the formulas for the parallelogram, trapezium, rhombus and kite, shows where they come from, and works backwards from an area to a missing length.
A parallelogram has two pairs of parallel sides. A trapezium has exactly one pair of parallel sides. A rhombus is a parallelogram with all four sides equal. A kite has two pairs of equal sides next to each other.
The perpendicular height is the distance between parallel sides, measured at right angles to them. Look for the right-angle mark. It is not the sloping side.
The diagonals join opposite corners. In a rhombus and a kite they cross at right angles, which is why a diagonal formula works for them.
The diagonals of a rhombus cut each other in half. If a diagram shows only half a diagonal, double it.
Here \(b\) is the base, \(h\) the perpendicular height, \(a\) and \(b\) in the trapezium formula the two parallel sides, and \(x\) and \(y\) the two whole diagonals. The trapezium formula is also written \(A = \dfrac{h}{2}(a + b)\).
Working backwards:
How to find the area of a quadrilateral
- Name the shape: parallelogram, trapezium, rhombus or kite.
- Pick the formula that matches the measurements you are given: base and height, or diagonals.
- Check each measurement: the height must be perpendicular, and each diagonal must be whole. Change to the same units.
- Substitute and work it out. For a missing length, substitute the area and solve.
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.
The long diagonal is shown in two parts. Add them to get the whole diagonal.
| \(x\) | \(=\) | \(5 + 11 = 16 \text{ cm}\) |
| \(y\) | \(=\) | \(9 \text{ cm}\) |
| \(A\) | \(=\) | \(\dfrac{1}{2}xy\) |
| \(\) | \(=\) | \(\dfrac{1}{2} \times 16 \times 9\) |
| \(\) | \(=\) | \(\mathbf{72} \text{ cm}^2\) |
The same formula works for a rhombus.
(i) A parallelogram has an area of \(91 \text{ cm}^2\) and a perpendicular height of \(7\) cm. Find its base.
(ii) A trapezium has an area of \(75 \text{ m}^2\) and a height of \(5\) m. One parallel side is \(10\) m. Find the other.
(i) Substitute what you know, then divide.
| \(91\) | \(=\) | \(b \times 7\) |
| \(b\) | \(=\) | \(91 \div 7 = \mathbf{13} \text{ cm}\) |
(ii) Substitute, then undo one step at a time.
| \(75\) | \(=\) | \(\dfrac{1}{2} \times (10 + b) \times 5\) |
| \(75\) | \(=\) | \(2.5 \times (10 + b)\) |
| \(30\) | \(=\) | \(10 + b\) |
| \(b\) | \(=\) | \(\mathbf{20} \text{ m}\) |
Check: \(\dfrac{1}{2} \times (10 + 20) \times 5 = 75\).
Common pitfalls
Frequently asked questions
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.
How do you find the area of a kite?
Multiply the two diagonals and halve the result: \(A = \dfrac{1}{2}xy\). Diagonals of \(16\) cm and \(9\) cm give \(72 \text{ cm}^2\).
Is the area of a rhombus the same as a parallelogram?
Yes. A rhombus is a parallelogram, so \(A = bh\) works. If the diagonals are given instead, use \(A = \dfrac{1}{2}xy\).
Why do you use the perpendicular height?
A parallelogram rearranges into a rectangle whose height is the perpendicular height, not the sloping side. The sloping side is longer and gives too big an area.
How do you find the base of a parallelogram from its area?
Divide the area by the perpendicular height: \(b = \dfrac{A}{h}\). An area of \(91 \text{ cm}^2\) with height \(7\) cm gives a base of \(13\) cm.