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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 10.03 (Area calculations) of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026): 10.03a to 10.03e. Questions 11 and 12 are Higher tier only; the rest are Foundation and Higher content. Diagrams are described in words, so sketch each one before you start. Round answers to 3 significant figures unless a question gives other instructions.

Read the study guide first if a topic is new, or use the revision notes. See also the OCR GCSE Mathematics hub and the printable checklist.

Questions

1. (a) A triangle has base 15 cm and perpendicular height 7.6 cm. Calculate its area. [2] (b) Another triangle has area 54 cm² and base 12 cm. Calculate its perpendicular height. [2]

2. Parallelogram ABCD has AB = 14 cm and AD = 8.5 cm. The perpendicular distance between AB and DC is 7.2 cm. (a) Calculate the area of the parallelogram. [2] (b) A rectangle has the same area and a width of 9 cm. Calculate its length. [2]

3. (a) A trapezium has parallel sides of 8.4 cm and 13.6 cm, which are 5 cm apart. Calculate its area. [2] (b) A different trapezium has parallel sides of 12 cm and 18 cm and an area of 105 cm². Calculate the distance between its parallel sides. [3]

4. (a) A circle has diameter 15.2 cm. Calculate its area. [3] (b) A second circle has area 50 cm². Calculate its radius. [3]

5. A circular pond has radius 4 m. A path of width 1.5 m goes all the way round it. Find the area of the path, first in terms of π and then to 3 significant figures. [4]

6. (a) A sector has radius 9 cm and angle 80°. Find its area in terms of π. [2] (b) Another sector has angle 210° and area 21π cm². Find its radius. [3]

7. A window is a rectangle 1.8 m wide and 1.2 m high, with a semicircle on top. The diameter of the semicircle is the 1.8 m top edge of the rectangle. Glass costs £48 per square metre. Find the cost of the glass for the whole window, to the nearest penny. [5]

8. ABCD is a square of side 14 cm. A quarter circle, centre A and radius 14 cm, passes through B and D. Find the area of the part of the square that is outside the quarter circle. [4]

9. A plot of land is a trapezium. Its parallel sides are 32 m and 20 m, and they are 14 m apart. A triangular corner of the plot, with base 8 m and perpendicular height 5 m, is paved. The rest is lawn. One box of lawn seed covers 40 m². How many boxes are needed to cover the lawn? [5]

10. O is the centre of a circle of radius 14 cm. Points P and Q are on the circle with angle POQ = 90°. Find the area of the region between the chord PQ and the arc PQ. [4]

11. (Higher tier only) In triangle XYZ, XY = 12.4 cm, XZ = 9.8 cm and angle YXZ = 38°. (a) Calculate the area of the triangle. [3] (b) Another triangle has sides of 11 cm and 8.5 cm, and an area of 41 cm². The angle between those two sides is not stated to be acute or obtuse. Find both possible sizes of that angle, to 1 decimal place. [4]

12. (Higher tier only) A circle has centre C and radius 11 cm. Points M and N are on the circle with angle MCN = 80°. Calculate the area of the region between the chord MN and the arc MN. [6]

Answers

1. (a) Area = ½ × 15 × 7.6 [1] = 57 cm² [1] (b) 54 = ½ × 12 × h = 6h [1], so h = 9 cm [1] Examiner insight: Write the formula with numbers substituted before the answer. Dividing 54 by 12 and stopping at 4.5 forgets that the 54 is already half of base × height.

2. (a) Area = base × perpendicular height = 14 × 7.2 [1] = 100.8 cm² [1] (b) Length = 100.8 ÷ 9 [1] = 11.2 cm [1] Examiner insight: The 8.5 cm side is slanted and is not needed. Using it in place of the perpendicular distance is a very common slip.

3. (a) Area = ½ × (8.4 + 13.6) × 5 [1] = 55 cm² [1] (b) 105 = ½ × (12 + 18) × h [1], so 105 = 15h [1], h = 7 cm [1] Examiner insight: Add the two parallel sides before multiplying by the height. Showing the sum 12 + 18 makes your method easy to follow.

4. (a) Radius = 15.2 ÷ 2 = 7.6 [1]. Area = π × 7.6² [1] = 181.45… = 181 cm² [1] (b) πr² = 50 [1], r² = 50 ÷ π = 15.91… [1], r = 3.99 cm [1] Examiner insight: Halving the diameter is the step most often missed in part (a). In part (b), undo the square with a square root only after dividing by π.

5. Outer radius = 4 + 1.5 = 5.5 m [1] Path = π × 5.5² − π × 4² [1] = 30.25π − 16π = 14.25π m² [1] = 44.76… = 44.8 m² [1] Examiner insight: Subtract the two circle areas. Squaring the width of the path (1.5) does not give its area.

6. (a) Area = 80/360 × π × 9² [1] = 2/9 × 81π = 18π cm² [1] (b) 210/360 × πr² = 21π [1], so 7/12 × r² = 21 [1], r² = 36, r = 6 cm [1] Examiner insight: The factor π appears on both sides in part (b), so it cancels. Keeping the fraction 210/360 as 7/12 avoids decimal rounding.

7. Rectangle = 1.8 × 1.2 = 2.16 m² [1] Semicircle radius = 0.9 m, so area = ½ × π × 0.9² [1] = 1.272… m² [1] Total area = 3.432… m² [1] Cost = 3.432… × 48 = £164.75 [1] Examiner insight: Keep the unrounded area in your calculator for the cost. Using 3.43 first gives £164.64, which is a penny-level slip when the question asks for the nearest penny.

8. Square = 14² = 196 cm² [1] Quarter circle = ¼ × π × 14² = 49π = 153.93… [1] Required area = 196 − 153.93… [1] = 42.1 cm² [1] Examiner insight: Say clearly which area is being subtracted from which. Getting 153.9 − 196 gives a negative area, which is a signal to reverse the subtraction.

9. Trapezium = ½ × (32 + 20) × 14 [1] = 364 m² [1] Triangle = ½ × 8 × 5 = 20 m² [1] Lawn = 364 − 20 = 344 m² [1] 344 ÷ 40 = 8.6, so 9 boxes [1] Examiner insight: Round up for a quantity that must cover the whole area. Writing 8.6 as the final answer or rounding to 9 without the division shown leaves the reasoning unclear.

10. Sector = ¼ × π × 14² = 49π = 153.93… cm² [1] Triangle OPQ is right-angled at O, so area = ½ × 14 × 14 = 98 cm² [1] Region = 49π − 98 [1] = 55.9 cm² [1] Examiner insight: The two radii are the base and height of the triangle because they meet at 90°. No extra length is needed.

11. (a) Area = ½ × 12.4 × 9.8 × sin 38° [1] = 60.76 × 0.6156… [1] = 37.4 cm² [1] (b) 41 = ½ × 11 × 8.5 × sin C [1], so sin C = 82 ÷ 93.5 = 0.877… [1] C = 61.3° [1] or 180° − 61.3° = 118.7° [1] Examiner insight: Use the angle that sits between the two sides you multiply. When no diagram fixes the angle, give the obtuse value as well as the acute one from your calculator.

12. Sector = 80/360 × π × 11² [1] = 84.47… cm² [1] Triangle MCN = ½ × 11 × 11 × sin 80° [1] = 59.58… cm² [1] Region = 84.47… − 59.58… [1] = 24.9 cm² [1] Examiner insight: This question joins three ideas: a sector, ½ab sin C and subtraction. Writing each part with its own heading keeps the working easy to follow, and keeping unrounded values avoids a slip in the last digit.

Where marks are usually lost

• Using a slanting side in place of the perpendicular height.
• Forgetting the ½ in a triangle or trapezium.
• Using the diameter as the radius in πr².
• Rounding π or a sub-total early, so the last digit of the answer changes.
• Not reading whether the answer should be in terms of π or rounded.
• Subtracting in the wrong order when finding a path, ring or cut-out area.
• Leaving off the unit, or writing cm instead of cm².
• Stopping at 8.6 boxes, tins or bags when a whole number is needed.
• Higher tier only: giving only the acute angle when a sine value has two possible angles.

Next steps

• Re-read the revision notes for the method boxes.
• Go back to the study guide for any question that went wrong.
• Browse the OCR GCSE Mathematics hub and tick off topics on the printable checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

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