Area

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.

• Keep every area formula in one mental toolkit.
• Straight-sided shapes always use a perpendicular height.
• Circle area = π × r² — square the radius first.
• Write the formula, substitute, then calculate.

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 sector is the slice between two radii and an arc.
• Sector area = (θ ÷ 360) × π r².
• The centre angle decides the fraction of the circle.
• A semicircle is a 180° sector; a quarter circle is 90°.

• 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.

• Split a compound shape into rectangles, triangles, trapezia and sectors.
• Two semicircles of equal radius make one full circle.
• Use 'whole minus hole' for shapes with a piece cut out.
• Find missing lengths by adding or subtracting known lengths.

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.

• Substitute the known area and length into the formula.
• Rearrange step by step to isolate the unknown.
• Parallelogram: missing height = area ÷ base.
• Circle: r² = area ÷ π, then take the square root.

• 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.

• Area answers 'how much surface must be covered or costed?'
• Use 'whole minus hole' for a region with a part removed.
• Multiply area by a cost-per-unit to find a total price.
• Always label units and round sensibly at the end.