Perimeters and Areas of Composite Shapes
A composite shape is made from simpler shapes, so its area is found by adding or subtracting the parts. This NSW Year 9 Advanced maths subtopic builds on Stage 4 area work to solve practical composite area problems (NESA outcome MA5‑ARE‑C‑01).
You will find shaded areas and perimeters of composite shapes, including ones with semicircles and a square drawn inside another square. You will cost a path around a pool and turf for a lawn, find a missing width, write an algebraic area for an L-shaped garden and solve for \(x\), and find the width of a frame whose area equals the photo's. It is part of Year 9 Advanced Surface Area and Volume.
Every question with a fully worked solution.
- Perimeters and Areas of Composite Shapes - Video - area perimeter composite figures Watch
Theory
A composite shape is built from simpler shapes. Find its area by splitting it into pieces and adding, or by taking a whole shape and subtracting the part that is missing. Find its perimeter by walking around the outside edge. This Year 9 Advanced maths page revises the earlier area work and takes it further, with curved edges, slanted sides and unknown lengths.
A composite shape is made of two or more simple shapes: rectangles, triangles, parallelograms, trapeziums, circles and parts of circles.
Its area is found by addition (split into pieces and add) or subtraction (whole shape minus the hole or cut-out). A shaded region is usually 'whole minus white'.
Its perimeter is the total length of the outside edge only. Lines drawn inside the shape to split it are not part of the perimeter. A cut-out on the edge can change the perimeter, even though it reduces the area.
Missing sides come from the sides you know: in a shape with only right angles, the lengths on one side add up to the length opposite. A slanted side is often the hypotenuse of a right-angled triangle, found with Pythagoras' theorem.
A curved edge is part of a circle: a semicircle is half, a quadrant is a quarter.
Area and perimeter of the building blocks:
| Shape | Area | Perimeter |
|---|---|---|
| Rectangle | \(A = lw\) | \(P = 2(l + w)\) |
| Triangle | \(A = \dfrac{1}{2}bh\) | add the three sides |
| Parallelogram | \(A = bh\) | \(P = 2(a + b)\) |
| Trapezium | \(A = \dfrac{h}{2}(a + b)\) | add the four sides |
| Circle | \(A = \pi r^2\) | \(C = 2\pi r = \pi d\) |
| Semicircle / quadrant | \(\dfrac{1}{2}\pi r^2\) / \(\dfrac{1}{4}\pi r^2\) | curved part \(\pi r\) / \(\dfrac{1}{2}\pi r\), plus straight edges |
How to find a composite area
- Sketch the shape and mark every length you know.
- Find missing lengths from the sides you know (opposite sides balance), or with Pythagoras' theorem for a slanted side.
- Choose add or subtract: split along a line into simple shapes, or start from a whole shape and take away the missing part.
- Calculate each part with its formula, then add or subtract. Keep \(\pi\) in the calculator until the end.
- Answer with square units, rounded as asked.
Perimeter and harder problems
- Perimeter: start at a corner and walk around the outside, adding each edge once. Use only the curved part of a circle, never its diameter, when the diameter is inside the shape.
- Two triangles on one base: a shape split by a line into two triangles can use that line as the base of both.
- Shapes inside shapes: a square joining the midpoints of a larger square's sides cuts off four equal right-angled triangles.
- Working backwards: if the area is given, write it with the unknown length, form an equation and solve. A quadratic may appear; reject a negative length.
The two top pieces are \(\dfrac{14 - 5}{2} = 4.5\) m each. Walk around the edge and add every side.
| \(P\) | \(=\) | \(14 + 10 + 4.5 + 4 + 5 + 4 + 4.5 + 10\) |
| \(\) | \(=\) | \(\mathbf{56}~\text{m}\) |
The notch removes \(5\) m of edge but adds two \(4\) m sides, so the perimeter is the rectangle's \(48\) m plus \(8\) m.
| \(A\) | \(=\) | \(14 \times 10 - 5 \times 4\) |
| \(\) | \(=\) | \(140 - 20 = \mathbf{120}~\text{m}^2\) |
| \(A\) | \(=\) | \(\dfrac{8}{2}(14 + 20) = \mathbf{136}~\text{m}^2\) |
The slanted side is the hypotenuse of a right-angled triangle with legs \(20 - 14 = 6\) m and \(8\) m.
| \(s\) | \(=\) | \(\sqrt{6^2 + 8^2} = \sqrt{100} = 10~\text{m}\) |
| \(P\) | \(=\) | \(14 + 20 + 8 + 10 = \mathbf{52}~\text{m}\) |
| \(\text{cost}\) | \(=\) | \(52 \times 32.50 = \mathbf{\$1690.00}\) |
Area: the rectangle minus a quarter of a circle.
| \(A\) | \(=\) | \(12 \times 8 - \dfrac{1}{4}\pi \times 3^2\) |
| \(\) | \(=\) | \(96 - 7.0685\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{88.93}~\text{m}^2\) |
Perimeter: two full sides, two shortened sides, and the curved edge (a quarter of a circumference).
| \(P\) | \(=\) | \(12 + 8 + 9 + 5 + \dfrac{1}{4} \times 2\pi \times 3\) |
| \(\) | \(=\) | \(34 + 4.7123\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{38.71}~\text{m}\) |
(i) Show that \(x^2 + 16x - 36 = 0\).
(ii) Find the width of the path.
(i) The garden and path together make a rectangle \((10 + x)\) by \((6 + x)\). The path is that rectangle minus the garden.
| \((10 + x)(6 + x) - 60\) | \(=\) | \(36\) |
| \(60 + 16x + x^2 - 60\) | \(=\) | \(36\) |
| \(x^2 + 16x - 36\) | \(=\) | \(0\) |
(ii) Factorise and reject the negative width.
| \((x + 18)(x - 2)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\mathbf{2}~\text{m} \quad (x = -18 \text{ rejected})\) |
Check: \(12 \times 8 - 60 = 36\) \(\checkmark\)
Common pitfalls
Frequently asked questions
How do you find the area of a composite shape?
Split it into simple shapes such as rectangles, triangles and semicircles, find each area and add them. Or take a larger simple shape and subtract the part that is missing.
How do you find the perimeter of a composite shape?
Add the lengths of every edge around the outside of the shape. Work out any missing lengths first, and for curved edges use the right fraction of the circumference.
How do you find the shaded area?
Find the area of the whole shape and subtract the area of the white or unshaded parts. Sometimes it is quicker to add the shaded pieces directly.
How do you find a missing side in an L-shape?
Use the sides opposite it. In a shape with only right angles, the horizontal lengths on one side add up to the horizontal length on the other side, and the same for vertical lengths.
What is the perimeter of a semicircle?
The curved part is half the circumference, \(\pi r\). If the straight diameter is part of the outside edge, add it as well, giving \(\pi r + 2r\).
How do you find the area of a path around a rectangle?
Find the area of the outer rectangle, including the path, and subtract the area of the inner rectangle. Remember the path adds its width to both sides of each dimension.