Areas of Quadrilaterals
Each special quadrilateral has an area formula, such as \(A = \dfrac{1}{2}(a + b)h\) for a trapezium and \(A = \dfrac{1}{2}xy\) for a rhombus or kite with diagonals \(x\) and \(y\). This NSW Year 9 Advanced maths subtopic revises these Stage 4 formulas (NESA outcome MA4‑ARE‑C‑01).
You will find areas of parallelograms, trapeziums, kites and rhombuses, and shaded areas made from them. You will work back to a missing base, height, diagonal or parallel side, cost a kite-shaped sail, find a rhombus diagonal with Pythagoras' theorem, solve a quadratic for a trapezium's sides, and prove the diagonal formula for any quadrilateral with perpendicular diagonals. It is part of Year 9 Advanced Surface Area and Volume.
Every question with a fully worked solution.
- Areas of Quadrilaterals - Video - Areas of Triangles and Quadrilaterals Watch
Theory
Every quadrilateral area formula comes from a rectangle or a triangle. Parallelograms and trapeziums use the perpendicular height; rhombuses and kites use their diagonals. This Year 9 Advanced maths page revises the earlier formulas and uses them backwards, with Pythagoras' theorem and with algebra.
A parallelogram has two pairs of parallel sides. Cutting a triangle off one end and moving it to the other makes a rectangle, so \(A = bh\), where \(h\) is the perpendicular height.
A trapezium has one pair of parallel sides, \(a\) and \(b\). Two copies make a parallelogram with base \(a + b\), so \(A = \dfrac{h}{2}(a + b)\). A trapezium can be drawn on its side: the parallel sides are the ones marked parallel, or both at right angles to the same line.
A rhombus has four equal sides; a kite has two pairs of equal adjacent sides. In both, the diagonals cross at right angles, so \(A = \dfrac{1}{2}xy\). In a rhombus the diagonals bisect each other; in a kite only the short one is bisected.
A square is a special rhombus, so \(\dfrac{1}{2}xy\) works for it too. Any quadrilateral whose diagonals cross at right angles has area \(\dfrac{1}{2}xy\).
| Quadrilateral | Area | What the letters mean |
|---|---|---|
| Rectangle | \(A = lb\) | length and breadth |
| Square | \(A = s^2\) | side length |
| Parallelogram | \(A = bh\) | base and perpendicular height |
| Trapezium | \(A = \dfrac{h}{2}(a + b)\) | parallel sides \(a\), \(b\); perpendicular height \(h\) |
| Rhombus | \(A = \dfrac{1}{2}xy\) or \(A = bh\) | diagonals \(x\), \(y\) |
| Kite | \(A = \dfrac{1}{2}xy\) | diagonals \(x\), \(y\) |
How to find the area of a quadrilateral
- Identify the shape from its parallel sides, equal sides or right angles.
- Choose the formula and mark the lengths it needs: the perpendicular height, both parallel sides, or both full diagonals.
- Convert lengths to the same unit, and double any half-diagonals.
- Substitute and calculate, with square units.
Harder problems
- Missing length: substitute the area and the known lengths, then solve the equation. For a trapezium, undo the halving, then the multiplying, then the adding.
- Missing diagonal or side: the diagonals of a rhombus or kite make right-angled triangles, so use Pythagoras' theorem.
- Shaded regions: find the whole shape and subtract the unshaded triangle or rectangle.
- Cost: area \(\times\) price per square unit, in matching units.
- Algebra: write the lengths with a pronumeral, form the area equation and solve; a quadratic may need factorising, and a negative length is rejected.
(i) the area
(ii) the perimeter of the rhombus.
(i) The diagonals of a rhombus bisect each other, so each full diagonal is double the given half.
| \(AC\) | \(=\) | \(2 \times 9 = 18~\text{cm}\) |
| \(BD\) | \(=\) | \(2 \times 12 = 24~\text{cm}\) |
| \(A\) | \(=\) | \(\dfrac{1}{2} \times 18 \times 24 = \mathbf{216}~\text{cm}^2\) |
(ii) Triangle \(AOB\) is right-angled at \(O\), so Pythagoras' theorem gives the side.
| \(AB\) | \(=\) | \(\sqrt{9^2 + 12^2} = \sqrt{225} = 15~\text{cm}\) |
| \(P\) | \(=\) | \(4 \times 15 = \mathbf{60}~\text{cm}\) |
Substitute everything known, then undo the operations one at a time. The unknown is inside the bracket.
| \(\dfrac{12}{2}(9 + b)\) | \(=\) | \(168\) |
| \(6(9 + b)\) | \(=\) | \(168\) |
| \(9 + b\) | \(=\) | \(28\) |
| \(b\) | \(=\) | \(\mathbf{19}~\text{m}\) |
Check: \(\dfrac{12}{2}(9 + 19) = 6 \times 28 = 168\) \(\checkmark\)
Use the perpendicular height, not the slanted side, and put both lengths in metres.
| \(350~\text{cm}\) | \(=\) | \(3.5~\text{m}\) |
| \(A\) | \(=\) | \(bh = 6.4 \times 3.5 = 22.4~\text{m}^2\) |
| \(\text{cost}\) | \(=\) | \(22.4 \times 48 = \mathbf{\$1075.20}\) |
The \(4\) m slanted side is extra information: it would be used for the perimeter, not the area.
(i) Find the diagonals.
(ii) The short diagonal cuts the long one into parts of \(3\) cm and \(9\) cm. Find the perimeter in simplest surd form and correct to \(2\) decimal places.
(i) Let the short diagonal be \(x\) cm, so the long one is \((x + 6)\) cm.
| \(\dfrac{1}{2}x(x + 6)\) | \(=\) | \(36\) |
| \(x^2 + 6x - 72\) | \(=\) | \(0\) |
| \((x + 12)(x - 6)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(6 \quad (x = -12 \text{ rejected})\) |
The diagonals are 6 cm and 12 cm.
(ii) The long diagonal bisects the short one, so each half is \(3\) cm. Each side is the hypotenuse of a right-angled triangle.
| \(\text{short side}\) | \(=\) | \(\sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}\) |
| \(\text{long side}\) | \(=\) | \(\sqrt{3^2 + 9^2} = \sqrt{90} = 3\sqrt{10}\) |
| \(P\) | \(=\) | \(2(3\sqrt{2} + 3\sqrt{10})\) |
| \(\) | \(=\) | \(\boldsymbol{6\sqrt{2} + 6\sqrt{10}} \approx \mathbf{27.46}~\text{cm}\) |
Common pitfalls
Frequently asked questions
What is the formula for the area of a trapezium?
\(A = \dfrac{h}{2}(a + b)\): half the height times the sum of the parallel sides. The height is the perpendicular distance between the parallel sides.
How do you find the area of a kite?
Multiply the lengths of the two diagonals and halve the result: \(A = \dfrac{1}{2}xy\). This works because the diagonals cross at right angles.
How do you find the area of a rhombus?
Use half the product of the diagonals, \(\dfrac{1}{2}xy\), or base times perpendicular height, \(bh\). Both give the same answer.
What is the area of a parallelogram?
Base times perpendicular height, \(A = bh\). The slanted side is not used for the area.
How do you find the height of a trapezium from its area?
Substitute the area and the two parallel sides into \(A = \dfrac{h}{2}(a + b)\), then solve for \(h\): multiply both sides by \(2\) and divide by the sum of the parallel sides.
Can you use half the product of the diagonals for a square?
Yes. A square is a special rhombus, so its area is half the product of its diagonals. Because both diagonals are equal, that is \(\dfrac{1}{2}d^2\).