Composite Areas and Problems on Areas
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.
Two methods:
| Method | Area |
|---|---|
| add the parts | \(A = A_1 + A_2 + \ldots\) |
| take away a hole | \(A = A_1 - A_2\) |
| crossing strips | \(A = A_1 + A_2 - A_3\) |
How to find a composite area
- Split the shape into simple shapes, or see it as a big shape with a hole.
- Find any missing lengths, all in one unit.
- Work out each area, then add or subtract, and answer the question asked.
| \(\text{rectangle}\) | \(=\) | \(7 \times 3 = 21\) |
| \(\text{triangle}\) | \(=\) | \(\dfrac{1}{2} \times 7 \times 4 = 14\) |
| \(A\) | \(=\) | \(21 + 14 = 35\) |
\(35\) square units
| \(\text{rectangle}\) | \(=\) | \(8 \times 6 = 48\) |
| \(\text{trapezium}\) | \(=\) | \(\dfrac{1}{2} \times (5 + 8) \times 4 = 26\) |
| \(A\) | \(=\) | \(48 - 26 = 22\) |
\(22\) square units
| \(A_1\) | \(=\) | \(9 \times 4 = 36\) |
| \(A_2\) | \(=\) | \(5 \times 3 = 15\) |
| \(A\) | \(=\) | \(36 + 15 = 51\) |
\(51~\text{m}^2\). Check by taking away: \(9 \times 7 - 4 \times 3 = 51\).
| \(A_1\) | \(=\) | \(4.5 \times 2.4 = 10.8\) |
| \(A_2\) | \(=\) | \(0.9 \times 2.1 = 1.89\) |
| \(A_3\) | \(=\) | \(1.6 \times 1.2 = 1.92\) |
| \(A\) | \(=\) | \(10.8 - 1.89 - 1.92 = 6.99\) |
\(6.99~\text{m}^2\)
| \(A_1\) | \(=\) | \(\dfrac{1}{2} \times (5 + 9) \times 4 = 28\) |
| \(A_2\) | \(=\) | \(\dfrac{1}{2} \times 9 \times 3 = 13.5\) |
| \(A\) | \(=\) | \(28 + 13.5 = 41.5\) |
\(41.5~\text{m}^2\)
| \(68.60 \div 35\) | \(=\) | \(1.96\) |
| \(\text{hole}\) | \(=\) | \(2.16 - 1.96 = 0.2\) |
| \(\dfrac{1}{2} \times 0.8 \times y\) | \(=\) | \(0.2\) |
| \(y\) | \(=\) | \(0.5\) |
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
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.