OCR GCSE Mathematics: Area: triangles, parallelograms, trapezia, circles, sectors and composite shapes (J560) – Revision Notes

These revision notes condense section 10.03 (Area calculations) of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026). For full explanations and more worked examples, read the study guide. Everything here is Foundation and Higher content except the one result marked Higher tier only. Check your progress against the OCR GCSE Mathematics hub and the printable checklist, and find weak spots with the 10-minute diagnostics.

Formulas to know

Ref	Shape	Area	Tier
10.03a	Triangle	½ × base × height	Foundation and Higher
10.03a	Triangle from two sides and the angle between them	½ab sin C	Higher tier only
10.03b	Parallelogram	base × height	Foundation and Higher
10.03b	Rectangle	length × width	Foundation and Higher
10.03c	Trapezium	½ × (a + b) × h	Foundation and Higher
10.03d	Circle	πr²	Foundation and Higher
10.03d	Sector, angle θ degrees	(θ/360) × πr²	Foundation and Higher
10.03e	Composite shape	add or subtract the parts	Foundation and Higher

Be ready to leave an answer as a multiple of π as well as to round it.

Key words

• Perpendicular height: the distance measured at 90° to the base. It is never the sloping side.
• Parallel sides (trapezium): the two sides used as a and b. The height is the distance between them.
• Sector: a slice of a circle between two radii. The fraction of the circle is θ/360.
• Composite shape: a shape made from simpler shapes joined together or cut out of each other.
• In terms of π: leave π in the answer, for example 24π. This is exact.

Method boxes

Any straight-sided area

1. Identify the shape, or split it into simple shapes.
2. Mark the perpendicular height. Ignore sloping lengths unless you need them to find a missing length.
3. Write the formula, substitute, then calculate.
4. Write the unit squared.

Circle

1. Find the radius. If you have the diameter, halve it.
2. Square the radius.
3. Multiply by π, or write the answer as a multiple of π.
4. Round only at the end.

Sector

1. Write the fraction θ/360 and simplify it if you can.
2. Multiply by πr².
3. To find the angle, divide the sector area by the whole-circle area, then multiply by 360.
4. To find the radius, rearrange (θ/360) × πr² = area for r².

Composite shape

1. Split it into rectangles, triangles, trapezia, circles or sectors. Draw the split lines.
2. Work out any missing lengths by adding or subtracting known lengths.
3. Add the parts for a built-up shape. Subtract for a hole or a cut-out.
4. Round at the end and give the unit.

Small worked reminders

Triangle. Base 20 cm, perpendicular height 7.5 cm.

½ × 20 × 7.5 = 75 cm²

Sector. Radius 5 cm, angle 288°.

288/360 = 0.8 = 4/5
Area = 4/5 × π × 25 = 20π ≈ 62.8 cm²

Composite. A rectangle 8 m by 5 m with a triangle on top of one 8 m side. The triangle has base 8 m and height 3 m.

Rectangle = 8 × 5 = 40
Triangle = ½ × 8 × 3 = 12
Total = 52 m²

Parallelogram. Base 6.4 cm, perpendicular height 5 cm.

6.4 × 5 = 32 cm²

Trapezium, working backwards. Area 66 cm², parallel sides 8 cm and 14 cm. Find the height.

66 = ½ × (8 + 14) × h
66 = 11h
h = 6 cm

Circle, working backwards. Area 49π cm². Find the radius.

πr² = 49π
r² = 49
r = 7 cm

Ring. Outer radius 7 cm, inner radius 3 cm.

π × 7² − π × 3² = 49π − 9π = 40π ≈ 126 cm²

Cost. A trapezium-shaped panel has parallel sides 2.4 m and 3.6 m, 2.5 m apart. Paint costs £3.60 per m².

Area = ½ × (2.4 + 3.6) × 2.5 = ½ × 6 × 2.5 = 7.5 m²
Cost = 7.5 × 3.60 = £27

Higher tier only: ½ab sin C. Sides 7 cm and 12 cm with 30° between them.

½ × 7 × 12 × sin 30° = 42 × 0.5 = 21 cm²

Check that your answer is sensible

• A triangle is always smaller than the rectangle with the same base and height.
• A trapezium lies between the area of the rectangle using the shorter parallel side and the one using the longer side.
• A circle of radius r fits inside a square of side 2r, so its area is less than 4r². It is also more than 3r², because π is more than 3.
• A sector is never bigger than its whole circle. A 90° sector is a quarter of it.
• A shape with a hole has less area than the shape before the hole was cut.

If an answer fails one of these checks, look for a wrong height, a missing ½ or a diameter used as a radius.

Must-know distinctions

• Radius against diameter. The formula uses the radius. Diameter ÷ 2 = radius.
• Height against slant. Triangles, parallelograms and trapezia all use the perpendicular height.
• Triangle against trapezium. Both have a ½. In a trapezium the ½ multiplies the sum of the two parallel sides.
• Area against perimeter. Area uses square units. Perimeter and circumference use plain units. Section 10.03 is about area only.
• Ring against smaller circle. The area of a ring is πR² − πr². It is not π(R − r)².
• Exact against rounded. 24π is exact. 75.4 is a rounded value. Give the form the question asks for.

Quick self-test

Try each one before you look at the answers. Questions 9 and 10 are Higher tier only.

1. A triangle has base 18 cm and perpendicular height 5.5 cm. Find its area.
2. A parallelogram has base 9.2 cm and perpendicular height 4 cm. Find its area.
3. A trapezium has parallel sides 7 cm and 12 cm, 4 cm apart. Find its area.
4. Find the area of a circle of radius 10 cm, in terms of π.
5. A circle has diameter 7 cm. Find its area to 3 significant figures.
6. Find the area of a sector with radius 12 cm and angle 45°, to 3 significant figures.
7. A sector of radius 10 cm has area 15π cm². Find its angle.
8. An L-shape is made from a rectangle 10 cm by 6 cm and a rectangle 4 cm by 3 cm that does not overlap it. Find the total area.
9. (Higher tier only) A triangle has sides 7 cm and 12 cm with an angle of 30° between them. Find its area.
10. (Higher tier only) A triangle has area 12 cm² and sides 6 cm and 8 cm. The angle C between them is acute. Find C.
11. Three-quarters of a circle of radius 4 cm is shaded. Find the shaded area in terms of π.

Answers

1. ½ × 18 × 5.5 = 49.5 cm²
2. 9.2 × 4 = 36.8 cm²
3. ½ × (7 + 12) × 4 = ½ × 19 × 4 = 38 cm²
4. π × 10² = 100π cm²
5. Radius 3.5 cm. π × 3.5² = 38.48… = 38.5 cm²
6. 45/360 = 1/8. 1/8 × π × 144 = 18π = 56.54… = 56.5 cm²
7. Whole circle = 100π. Fraction = 15π ÷ 100π = 0.15. 0.15 × 360 = 54°
8. 60 + 12 = 72 cm²
9. ½ × 7 × 12 × sin 30° = 42 × 0.5 = 21 cm²
10. 12 = ½ × 6 × 8 × sin C, so 12 = 24 sin C and sin C = 0.5. C = 30°
11. 3/4 × π × 4² = 3/4 × 16π = 12π cm²

Where marks are usually lost

• Writing the sloping side as the height in a triangle, parallelogram or trapezium.
• Leaving out the ½ in a triangle or trapezium, or applying it to only one of the parallel sides.
• Using the diameter in πr² without halving it.
• Giving a sector’s area using the whole circle, or using θ/360 on the wrong quantity.
• Rounding π or an interim value early, so the final digits drift.
• Not adding or subtracting the parts of a composite shape in the way the diagram shows, for example adding a hole instead of removing it.
• Missing a length: a composite shape often needs one side found by subtraction before you can start.
• Leaving off the unit or writing cm for cm².
• Rounding when the question asks for an exact multiple of π, or leaving π in when a decimal is wanted.
• In ½ab sin C (Higher tier only), using an angle that is not between the two sides, or forgetting that an obtuse angle has the same sine as an acute one.

Writing out each formula with the numbers substituted lets whoever marks your work see what you intended, even if the final value slips.

Next steps

Test yourself on the practice questions, then return to the study guide for any step that felt uncertain.

Official syllabus

OCR GCSE (9-1) Mathematics J560, Version 2.1 (August 2026), first assessment 2017, published by OCR. These notes cover section 10.03, Area calculations.