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Year 8 Maths Area and Perimeter

Composite Areas and Problems on Areas

20 practice questions 2 video lessons Theory + worked examples

A composite shape is made from simpler shapes, so its area is found by adding the parts or by taking a missing piece away from a larger shape. This NSW Year 8 maths subtopic (NESA outcome MA4‑ARE‑C‑01) combines rectangles, triangles, parallelograms, trapeziums and rhombuses.

You will split shapes along a dashed line, work out missing lengths first, subtract one or more holes, and handle two paths that cross by taking the shared square away once. You will also cost a pebble path around a pool, count the floor tiles for an L-shaped laundry, find the lawn left around a shed slab and work backwards from the cost of turf to a missing length. It is part of Year 8 Area and Perimeter.

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Theory

A composite shape is built from simple shapes. Add the parts (\(A = A_1 + A_2\)), or start with a bigger shape and take away what is missing (\(A = A_1 - A_2\)). A \(7\) by \(3\) rectangle with a triangle of base \(7\) and height \(4\) on top has area \(21 + 14 = 35\) square units. This Year 8 maths page shows how.

Method 1: add the parts. Split the shape with a dashed line into simple shapes \(1\), \(2\), and so on. Work out each area on its own line, then add.

Method 2: take away the hole. Find the area of the whole outside shape, then subtract the part cut out. For \(n\) equal holes, \(A = A_1 - n \times A_2\). Use one method and check with the other.

Missing lengths first. When every corner is a right angle, the short sides across a shape add to the long one, so a missing length is a subtraction.

Use the right formula for each part: rectangle \(lw\), square \(s^2\), triangle \(\dfrac{1}{2}bh\), parallelogram \(bh\), trapezium \(\dfrac{1}{2}(a + b)h\), kite or rhombus \(\dfrac{1}{2}xy\).

Strips that cross are added, then the crossing piece is subtracted once: \(A = A_1 + A_2 - A_3\).

Then answer the question: cost \(=\) area \(\times\) price per square metre; tiles \(=\) area \(\div\) one tile’s area, in the same square unit.

Courtyard split into two rectangles A courtyard 9 m by 7 m with a 4 m by 3 m corner taken out. A dashed line splits it into rectangle 1, 9 m by 4 m, and rectangle 2, 5 m by 3 m. 9 7 4 3 5 1 2 4
Add the parts: the missing lengths are \(9 - 4 = 5\) and \(7 - 3 = 4\), so \(A = 9 \times 4 + 5 \times 3 = 51~\text{m}^2\).
Take the hole away A rectangle with a smaller rectangular hole cut out of it (white). The shaded area is the whole rectangle minus the hole: A = A1 - A2. hole A = A₁ − A₂
Take away: the shaded area is the whole rectangle minus the hole.

Two methods:

MethodArea
add the parts\(A = A_1 + A_2 + \ldots\)
take away a hole\(A = A_1 - A_2\)
crossing strips\(A = A_1 + A_2 - A_3\)
Count every part exactly once. Two paths \(30\) m by \(2\) m and \(20\) m by \(2\) m cross in a \(2\) m square: \(60 + 40 - 4 = 96~\text{m}^2\).
Diagrams are not to scale. Never measure a side; work every missing length out from the given ones.

How to find a composite area

  1. Split the shape into simple shapes, or see it as a big shape with a hole.
  2. Find any missing lengths, all in one unit.
  3. Work out each area, then add or subtract, and answer the question asked.
Example 1 — Add the parts
Find the area of the composite shape.
Rectangle with a triangle on top A rectangle 7 units wide and 3 units tall with a triangle on top whose dashed height is 4 units. 7 3 4
Solution
\(\text{rectangle}\)\(=\)\(7 \times 3 = 21\)
\(\text{triangle}\)\(=\)\(\dfrac{1}{2} \times 7 \times 4 = 14\)
\(A\)\(=\)\(21 + 14 = 35\)
35 square units

\(35\) square units

Example 2 — Take away
Find the area of the shaded region.
Rectangle with a trapezium removed An 8 by 6 rectangle with a trapezium cut out of the bottom: the trapezium has parallel sides 5 (top) and 8 (bottom, dashed) and height 4. The shaded region is what is left. 8 6 5 4
Solution
\(\text{rectangle}\)\(=\)\(8 \times 6 = 48\)
\(\text{trapezium}\)\(=\)\(\dfrac{1}{2} \times (5 + 8) \times 4 = 26\)
\(A\)\(=\)\(48 - 26 = 22\)
22 square units

\(22\) square units

Example 3 — Missing length, then add
A courtyard is \(9\) m by \(7\) m with a \(4\) m by \(3\) m corner taken out. Every corner is a right angle. Find its area.
Solution
\(A_1\)\(=\)\(9 \times 4 = 36\)
\(A_2\)\(=\)\(5 \times 3 = 15\)
\(A\)\(=\)\(36 + 15 = 51\)
51 m

\(51~\text{m}^2\). Check by taking away: \(9 \times 7 - 4 \times 3 = 51\).

Example 4 — Two cut-outs
A wall is \(4.5\) m by \(2.4\) m. It has a door \(0.9\) m by \(2.1\) m and a window \(1.6\) m by \(1.2\) m. Find the area to be painted.
Solution
\(A_1\)\(=\)\(4.5 \times 2.4 = 10.8\)
\(A_2\)\(=\)\(0.9 \times 2.1 = 1.89\)
\(A_3\)\(=\)\(1.6 \times 1.2 = 1.92\)
\(A\)\(=\)\(10.8 - 1.89 - 1.92 = 6.99\)
6.99 m

\(6.99~\text{m}^2\)

Example 5 — Trapezium and triangle
A shade sail is a trapezium (parallel sides \(5\) m and \(9\) m, height \(4\) m) joined along its \(9\) m edge to a triangle of height \(3\) m. Find its area.
Solution
\(A_1\)\(=\)\(\dfrac{1}{2} \times (5 + 9) \times 4 = 28\)
\(A_2\)\(=\)\(\dfrac{1}{2} \times 9 \times 3 = 13.5\)
\(A\)\(=\)\(28 + 13.5 = 41.5\)
41.5 m

\(41.5~\text{m}^2\)

Example 6 — Work back from a cost
A tabletop is a \(1.8\) m by \(1.2\) m rectangle with a rhombus cut out; one diagonal of the rhombus is \(0.8\) m. Varnishing the timber left cost \(\$68.60\) at \(\$35\) per square metre. Find the other diagonal \(y\).
Solution
\(68.60 \div 35\)\(=\)\(1.96\)
\(\text{hole}\)\(=\)\(2.16 - 1.96 = 0.2\)
\(\dfrac{1}{2} \times 0.8 \times y\)\(=\)\(0.2\)
\(y\)\(=\)\(0.5\)
y =0.5 m

The tabletop is \(1.8 \times 1.2 = 2.16~\text{m}^2\), so the hole is \(0.2~\text{m}^2\) and \(y = 0.5\) m.

Common pitfalls

Overlapping parts. Split along a line so the parts do not overlap, and count each part once.
Guessing a missing length. Work it out from the given lengths.
Mixing units. A \(30\) cm strip on a plan in metres is \(0.3\) m.
Adding a hole instead of subtracting it. A hole is taken away from the whole shape.

Frequently asked questions

What is a composite shape?

A shape made from two or more simple shapes, such as a rectangle and a triangle.

How do I find the area of a composite shape?

Split it into simple shapes and add their areas, or take the missing parts away from a bigger shape.

How do I find a missing length?

When the corners are right angles, the short sides across the shape add to the long side opposite, so subtract.

How do I find the number of tiles needed?

Divide the floor area by the area of one tile, both in the same unit, and round up.

How do I check my answer?

Use the other method: if you added parts, check by taking away from a bigger rectangle.