Area of Triangles and Quadrilaterals
Area of Triangles and Quadrilaterals in NSW Year 10 Core maths revises the Stage 4 area formulas for triangles, parallelograms, trapeziums, rhombuses and kites (NESA outcome MA4‑ARE‑C‑01), then applies them to composite shapes and practical problems at Stage 5 (MA5‑ARE‑C‑01).
You will find the areas of shaded regions by adding and subtracting simpler shapes, use Pythagoras' theorem to find a missing height or diagonal before finding an area, and work backwards from an area to a base, a height or a pair of parallel sides. You will also cost turf for a trapezium-shaped backyard, find the area left in a paddock after a dam is fenced off, and find the area of a shed's end wall with a window. It is part of Year 10 Core Measurement.
Every question with a fully worked solution.
- Area of Triangles and Quadrilaterals - Video - Finding the Area of Rectangles, Parallelograms, Triangles and Trapeziums Watch
Theory
The area of a triangle or quadrilateral comes from a formula that uses the perpendicular height or, for a rhombus or kite, the diagonals. This Year 10 maths page revises the earlier area formulas and takes them further: composite shapes, finding a missing height with Pythagoras' theorem, and working backwards from a known area.
Area is the amount of flat space inside a shape, measured in square units such as \(\text{cm}^2\), \(\text{m}^2\) or hectares.
The perpendicular height is the distance between the base and the opposite vertex or side, measured at right angles to the base. It is often drawn dashed. A sloping side is not the height.
A parallelogram has two pairs of parallel sides. A trapezium has one pair of parallel sides, called \(a\) and \(b\). A rhombus has four equal sides and a kite has two pairs of equal adjacent sides; in both, the diagonals cross at right angles.
A composite shape is made of simpler shapes. Add the areas of the parts, or find a larger shape and subtract the part that is missing.
The Year 10 questions often hide the height, so you need Pythagoras' theorem to find it first.
| Shape | Area formula | Lengths used |
|---|---|---|
| Rectangle | \(A = lb\) | length, breadth |
| Square | \(A = s^{2}\) | side |
| Triangle | \(A = \dfrac{1}{2}bh\) | base, perpendicular height |
| Parallelogram | \(A = bh\) | base, perpendicular height |
| Trapezium | \(A = \dfrac{1}{2}(a + b)h\) | parallel sides, perpendicular height |
| Rhombus or kite | \(A = \dfrac{1}{2}xy\) | diagonals |
How to find an area
- Name the shape and write its formula.
- Pick the lengths the formula needs: the perpendicular height, both parallel sides, or both diagonals. Ignore sloping sides.
- Find any missing length first, often with Pythagoras' theorem.
- Substitute, calculate and write square units.
Composite shapes and other problems
- Add the parts, or subtract the missing part from a bigger shape. Work out any side not labelled from the sides that are.
- A shape inside a shape: the shaded area is the outer area minus the inner area.
- Cost: multiply the area by the price per square metre. For example, \(40\text{ m}^2\) of paving at $55 per square metre costs $2200.
- Working backwards: put the known area into the formula with a pronumeral for the unknown, then solve the equation. A length must be positive.
(i) A parallelogram has a base of \(11\) cm, sloping sides of \(7\) cm and a perpendicular height of \(5\) cm. Find its area.
(ii) A kite has diagonals of \(18\) cm and \(11\) cm. Find its area.
(i) Use the base and the perpendicular height. The \(7\) cm sloping side is not needed.
| \(A\) | \(=\) | \(bh\) |
| \(\) | \(=\) | \(11 \times 5\) |
| \(\) | \(=\) | \(\mathbf{55}\text{ cm}^2\) |
(ii) The diagonals of a kite cross at right angles.
| \(A\) | \(=\) | \(\dfrac{1}{2}xy\) |
| \(\) | \(=\) | \(\dfrac{1}{2} \times 18 \times 11\) |
| \(\) | \(=\) | \(\mathbf{99}\text{ cm}^2\) |
Find the whole rectangle, then subtract the triangle. The two shorter sides of a right-angled triangle are its base and height.
| \(\text{rectangle}\) | \(=\) | \(12 \times 7 = 84\text{ m}^2\) |
| \(\text{triangle}\) | \(=\) | \(\dfrac{1}{2} \times 4 \times 3 = 6\text{ m}^2\) |
| \(\text{lawn}\) | \(=\) | \(84 - 6\) |
| \(\) | \(=\) | \(\mathbf{78}\text{ m}^2\) |
(i) its perpendicular height \(h\)
(ii) its area.
The bottom is \(26 - 10 = 16\) cm longer than the top, shared equally between the two ends: \(8\) cm each.
(i) The sloping side is the hypotenuse.
| \(h^{2} + 8^{2}\) | \(=\) | \(17^{2}\) |
| \(h^{2}\) | \(=\) | \(289 - 64 = 225\) |
| \(h\) | \(=\) | \(\mathbf{15}\text{ cm}\) |
(ii)
| \(A\) | \(=\) | \(\dfrac{1}{2}(10 + 26) \times 15\) |
| \(\) | \(=\) | \(\mathbf{270}\text{ cm}^2\) |
(i) the other diagonal
(ii) the side length \(s\) of the rhombus.
(i) Substitute into the area formula and solve.
| \(\dfrac{1}{2} \times 12 \times y\) | \(=\) | \(96\) |
| \(6y\) | \(=\) | \(96\) |
| \(y\) | \(=\) | \(\mathbf{16}\text{ cm}\) |
(ii) The diagonals of a rhombus cut each other in half at right angles, giving a right-angled triangle with sides \(6\) and \(8\).
| \(s^{2}\) | \(=\) | \(6^{2} + 8^{2} = 100\) |
| \(s\) | \(=\) | \(\mathbf{10}\text{ cm}\) |
Common pitfalls
Frequently asked questions
How do you find the area of a trapezium?
Add the two parallel sides, multiply by the perpendicular height, then halve: \(A = \dfrac{1}{2}(a + b)h\). For parallel sides \(6\) m and \(10\) m, \(5\) m apart, \(A = \dfrac{1}{2} \times 16 \times 5 = 40\text{ m}^2\).
What is the formula for the area of a rhombus?
Multiply the diagonals and halve: \(A = \dfrac{1}{2}xy\). You can also use base \(\times\) perpendicular height, because a rhombus is a parallelogram.
What is the perpendicular height of a triangle?
It is the distance from the base to the opposite vertex, measured at right angles to the base. In an obtuse triangle it can fall outside the triangle, meeting the base extended.
How do you find the area of a composite shape?
Split the shape into rectangles, triangles and other simple shapes and add their areas, or find the area of a larger shape and subtract the part that is missing.
How do you find the height of an isosceles triangle?
Draw the height from the top vertex. It cuts the base in half and makes two right-angled triangles. Use Pythagoras' theorem with half the base and an equal side as the hypotenuse.