Area

By Stage 9 you have met area formulae for many shapes. The goal now is to hold them all in one toolkit and pick the right one without hesitation.

Here are the formulae you should know by heart:

• Rectangle: $\text{Area} = \text{length} \times \text{width}$
• Triangle: $\text{Area} = \tfrac{1}{2} \times \text{base} \times \text{height}$
• Parallelogram: $\text{Area} = \text{base} \times \text{height}$
• Trapezium: $\text{Area} = \tfrac{1}{2} \times (a + b) \times h$
• Circle: $\text{Area} = \pi \times r^2$

The thread running through the straight-sided shapes is the perpendicular height — always measured at right angles to the base, never along a slanting side.

A habit worth keeping: write the formula down first, then substitute, then calculate. That single extra line stops you forgetting to halve a triangle, or squaring before multiplying by π in a circle. With these tools ready, even a complicated Stage 9 problem becomes a series of small, calm steps.

A sector is a 'slice' of a circle — the shape you get between two radii and the curved arc that joins them. A pizza slice and a slice of cake are everyday sectors.

The clever idea is that a sector is just a fraction of the whole circle. The fraction is decided by the angle at the centre. A full circle is $360\degree$, so a sector with a centre angle of $\theta$ is $\frac{\theta}{360}$ of the circle.

So the area of a sector is: $\text{Sector area} = \frac{\theta}{360} \times \pi r^2.$

For the sector above — radius $6\text{ cm}$, centre angle $90\degree$: $\frac{90}{360} \times \pi \times 6^2 = \tfrac{1}{4} \times \pi \times 36 \approx 28.3\text{ cm}^2.$

A quick sense-check: a $90\degree$ sector should be a quarter of the full circle area $\pi \times 36 \approx 113$, and $113 \div 4 \approx 28.3$. The numbers agree.

A semicircle ($180\degree$) is just a sector that is half the circle, and a quarter circle ($90\degree$) is a sector worth one quarter.

A compound shape is built from simpler shapes joined together. In Stage 9 these often mix straight edges with curved ones — a rectangle topped with a semicircle, or a square with a quarter-circle corner cut away.

The method never changes: split the shape, find each piece with the right formula, then add the pieces (or subtract a hole).

For a shape like this — a rectangle $14\text{ cm}$ by $9\text{ cm}$ with a semicircle of radius $4.5\text{ cm}$ on each end:

• Rectangle: $14 \times 9 = 126\text{ cm}^2$.
• Two semicircles make one full circle: $\pi \times 4.5^2 \approx 63.6\text{ cm}^2$.
• Total area $\approx 126 + 63.6 = 189.6\text{ cm}^2$.

A clear plan for any compound shape:

• Draw your splitting lines lightly and label every piece.
• Work out any missing lengths by adding or subtracting known lengths.
• For a shape with a hole, use "whole minus hole" — find the whole, then subtract the missing part.
• Keep every measurement in the same unit before you combine the pieces.

Stage 9 problems often turn things around. Instead of giving the lengths, they give you the area and ask for a missing length.

The skill is to substitute what you know into the formula, then rearrange it one operation at a time.

For a parallelogram with area $72\text{ cm}^2$ and base $9\text{ cm}$: $\text{height} = \text{area} \div \text{base} = 72 \div 9 = 8\text{ cm}.$

For a triangle with area $54\text{ cm}^2$ and base $12\text{ cm}$, undo the formula step by step: $\tfrac{1}{2} \times 12 \times h = 54 \;\Rightarrow\; 6h = 54 \;\Rightarrow\; h = 9\text{ cm}.$

You can even work backwards with a circle. If a circle has area $\approx 78.5\text{ cm}^2$, then $r^2 = 78.5 \div \pi \approx 25$, so $r = \sqrt{25} = 5\text{ cm}$.

The whole skill is staying calm and undoing the formula in reverse — exactly the inverse-operations idea you use when solving equations.

Area is one of the most practical ideas in maths — it answers real questions whenever a flat surface has to be covered, costed or compared.

Suppose a gardener needs grass seed for the lawn above — a rectangle $8\text{ m}$ by $5\text{ m}$ with a circular pond of radius $1.5\text{ m}$:

• Lawn rectangle: $8 \times 5 = 40\text{ m}^2$.
• Pond: $\pi \times 1.5^2 \approx 7.1\text{ m}^2$.
• Grass area: $40 - 7.1 = 32.9\text{ m}^2$.

If seed costs \3$per square metre, the seed bill is$32.9 \times 3 \approx $98.70$.

A reliable plan for any worded area problem:

• Sketch the situation and label every measurement given.
• Decide whether you are adding pieces or using 'whole minus hole'.
• Check every length is in the same unit before you calculate.
• Round sensibly at the end, and include the square units in your answer.

The area skills you have sharpened in Stage 9 carry straight into the maths ahead:

• Surface area — the surface area of any solid is the total of the areas of its faces.
• Volume of prisms and cylinders — volume is a cross-section area multiplied by a length.
• Circles and sectors — sector area pairs naturally with arc length and the circumference.
• In everyday life — flooring, painting, landscaping, packaging and design all rely on area.

If a later measurement topic feels hard, an area skill is often underneath it. Keep this guide close — confident, careful splitting gets easier every time you do it.