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

This study guide follows section 10.03 (Area calculations) of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026): 10.03a to 10.03e. Most of it is Foundation and Higher content. One result, the area of a triangle as ½ab sin C, is marked Higher tier only. Because the specification allows any content to appear on either type of paper, practise it by hand and with a calculator.

Use it with the revision notes and the practice questions. The OCR GCSE Mathematics hub, the printable checklist and the 10-minute diagnostics help you plan your revision.

Scope of the unit

Ref	Skill	Tier
10.03a	Know and apply area of a triangle = ½ × base × height	Foundation and Higher
10.03a	Use ½ab sin C for the area of a triangle	Higher tier only
10.03b	Know and apply area of a parallelogram = base × height (includes a rectangle)	Foundation and Higher
10.03c	Calculate the area of a trapezium	Foundation and Higher
10.03d	Know and apply area of a circle = πr²	Foundation and Higher
10.03d	Find the area of a sector from its angle and radius	Foundation and Higher
10.03e	Apply area formulae to composite 2D shapes	Foundation and Higher

Area is measured in square units: cm², m², mm². Every area answer needs its unit.

10.03a: Area of a triangle

The perpendicular height is measured at 90° to the base, not along a sloping side. Any side can be the base, but the height must be the matching perpendicular distance.

Area of a triangle = ½ × base × height

Why it works: two copies of the same triangle fit together to make a parallelogram, and the parallelogram has area base × height. One triangle is half of that.

        /|
       / |
      /  | h
     /   |
    /____|______
       base

Worked example 1. A triangle has base 14 cm and perpendicular height 9 cm.

Area = ½ × 14 × 9
     = 7 × 9
     = 63 cm²

Worked example 2 (height outside the triangle). A triangular sail has a base of 8.5 m along the bottom edge. Its top corner is 6.2 m vertically above the line of the base, but not above the base itself, because the triangle leans over. The height is still 6.2 m.

Area = ½ × 8.5 × 6.2
     = 26.35 m²

Do not use the slanting side of the sail as the height. It is longer than the perpendicular height, so it gives an area that is too big.

Higher tier only: area = ½ab sin C

If two sides and the angle between them are given, the area follows without needing the height.

Area = ½ab sin C, where a and b are two sides and C is the angle between them.

Worked example 3. A triangle has sides 9.4 cm and 6.8 cm with an angle of 52° between them.

Area = ½ × 9.4 × 6.8 × sin 52°
     = 31.96 × 0.78801...
     = 25.18...
     ≈ 25.2 cm² (3 s.f.)

Worked example 4 (finding an angle). A triangle has sides 10 cm and 8 cm and area 30 cm². Find angle C between them.

30 = ½ × 10 × 8 × sin C
30 = 40 sin C
sin C = 0.75
C = 48.6° (1 d.p.)

If the question does not say that C is acute, the other possible angle is 180° − 48.6° = 131.4°. Both give the same sine. A diagram or a statement such as “C is obtuse” tells you which one to use.

10.03b: Area of a parallelogram and a rectangle

Area of a parallelogram = base × height, again using the perpendicular height. The sloping side is not the height.

A rectangle is a parallelogram with a right angle, so its height is simply its width: area = length × width.

Worked example 5. A parallelogram has a base of 12.5 cm and a sloping side of 9 cm. The perpendicular distance between the two long sides is 7.2 cm.

Area = 12.5 × 7.2
     = 90 cm²

The 9 cm is a distractor. It is not used.

Worked example 6 (working backwards). A parallelogram has area 84 cm² and base 12 cm. Find its perpendicular height.

84 = 12 × h
h = 84 ÷ 12
h = 7 cm

10.03c: Area of a trapezium

A trapezium has one pair of parallel sides. The height is the perpendicular distance between them.

Area of a trapezium = ½ × (a + b) × h, with a and b as the lengths of the two parallel sides.

Why it works: two copies of the trapezium, one turned upside down, fit together to make a parallelogram with base (a + b) and height h. One trapezium is half of it. You can also see the formula as “average of the parallel sides × height”.

      a
    _______
   /       \
  /         \  h
 /___________\
        b

Worked example 7. The parallel sides of a trapezium are 11 cm and 17 cm, and they are 6 cm apart.

Area = ½ × (11 + 17) × 6
     = ½ × 28 × 6
     = 84 cm²

Worked example 8 (working backwards). A trapezium has parallel sides 9.5 cm and 14.5 cm and area 96 cm². Find the height.

96 = ½ × (9.5 + 14.5) × h
96 = 12 × h
h = 8 cm

10.03d: Area of a circle and of a sector

Area of a circle = πr². Square the radius first, then multiply by π. If you are given the diameter, halve it to get the radius.

Worked example 9. Find the area of a circle with radius 6.5 cm.

Area = π × 6.5²
     = π × 42.25
     = 132.73...
     ≈ 133 cm² (3 s.f.)

Worked example 10 (diameter, and an answer in terms of π). A circular table top has diameter 18 cm.

r = 18 ÷ 2 = 9
Area = π × 9²
     = 81π cm²
     ≈ 254 cm² (3 s.f.)

Leave the answer as 81π when the question asks for it “in terms of π”. It is exact.

Worked example 11 (finding the radius). A circle has area 200 cm². Find its radius.

πr² = 200
r² = 200 ÷ π = 63.66...
r = √63.66... = 7.98 cm (3 s.f.)

Area of a sector

A sector is a slice of a circle between two radii. If the angle at the centre is θ degrees, the sector is the fraction θ/360 of the whole circle.

Area of a sector = (θ/360) × πr²

Worked example 12. A sector has radius 8 cm and angle 135°.

Fraction = 135/360 = 3/8
Area = 3/8 × π × 8²
     = 3/8 × 64π
     = 24π cm²
     ≈ 75.4 cm² (3 s.f.)

Worked example 13 (finding the angle). A sector of radius 12 cm has area 40π cm².

Whole circle = π × 12² = 144π
Fraction = 40π ÷ 144π = 5/18
Angle = 5/18 × 360 = 100°

Common sectors: a semicircle is 180° (half), a quarter circle is 90° (a quarter).

10.03e: Composite shapes

A composite shape is made from simpler shapes. The method is the same every time.

1. Split the shape into rectangles, triangles, trapezia, circles or sectors.
2. Work out any missing lengths from the information given.
3. Find each area.
4. Add the areas for a shape built up from parts. Subtract for a shape with a part cut out.
5. Give the unit and round only at the end.

Worked example 14 (adding). A rectangle measures 12 cm by 7 cm. A semicircle is attached to one of the 7 cm sides, with that side as its diameter.

Rectangle = 12 × 7 = 84
Semicircle radius = 3.5
Semicircle = ½ × π × 3.5² = 19.24...
Total = 84 + 19.24... = 103.24...
      ≈ 103 cm² (3 s.f.)

Worked example 15 (subtracting). A metal plate is a rectangle 20 cm by 14 cm with a circular hole of radius 3 cm.

Rectangle = 20 × 14 = 280
Hole = π × 3² = 9π = 28.27...
Plate = 280 − 28.27... = 251.72...
      ≈ 252 cm² (3 s.f.)

Worked example 16 (sector minus triangle). A quarter circle of radius 10 cm has its two radii joined to the arc’s ends by a straight chord. Find the area between the chord and the arc.

Quarter circle = ¼ × π × 10² = 25π = 78.53...
Triangle (right angle at centre) = ½ × 10 × 10 = 50
Area between chord and arc = 25π − 50 = 28.53...
                           ≈ 28.5 cm² (3 s.f.)

Worked example 17 (a ring). A circular ring has outer radius 9 cm and inner radius 6 cm.

Area = π × 9² − π × 6²
     = 81π − 36π
     = 45π
     ≈ 141 cm² (3 s.f.)

Subtract the two areas. Do not take 3 from 9 and square the difference.

Units and costs

Area questions often continue with a cost or a quantity: paint, turf or glass at a price per square metre. Find the area first, make sure the unit matches the rate, then multiply. If you must round the number of tins or boxes, round up, because rounding down leaves part of the area uncovered.

Square units convert with the square of the length factor: 1 m² = 100 × 100 = 10 000 cm².

Common errors

• Using the sloping side of a triangle, parallelogram or trapezium as the height.
• Forgetting the ½ for a triangle or a trapezium.
• Writing πr² as (πr)² or as π × 2r. Square the radius first.
• Using the diameter in place of the radius.
• In ½ab sin C (Higher tier only), using an angle that is not between the two sides.
• Giving a sector’s area as a fraction of the circumference formula instead of the area formula.
• Rounding π to 3.14 early, or rounding part-way through a multi-step answer.
• Leaving off the unit, or writing cm instead of cm².

Next steps

Condense this unit with the revision notes, then test yourself on the practice questions. For general paper technique, see the exam preparation page.

Official syllabus

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