Mensuration

50 maths questions with worked answers with answers for OCR 10 Mensuration in OCR GCSE Maths (spec J560). Practise with flashcards, a topic quiz, a random quiz across the course, 5 crosswords, or read every question and answer below. Not official OCR material.

1. Question 1 · Units · OCR 10.01a

   Convert 3.75 kg to grams.

   Answer:

   1 kg = 1000 g.
   3.75 × 1000 = 3750 g.

2. Question 2 · Units · OCR 10.01a

   Convert 2450 m to km.

   Answer:

   1 km = 1000 m.
   2450 ÷ 1000 = 2.45 km.

3. Question 3 · Units · OCR 10.01a

   Convert 4.2 litres to millilitres.

   Answer:

   1 litre = 1000 ml.
   4200 ml.

4. Question 4 · Time · OCR 10.01a

   Write 2.4 hours in hours and minutes.

   Answer:

   0.4 hours = 0.4 × 60 = 24 minutes.
   2 hours 24 minutes.

5. Question 5 · Compound measures · OCR 10.01b

   A car travels 150 km in 2.5 hours. Find its average speed.

   Answer:

   Speed = distance ÷ time.
   150 ÷ 2.5 = 60 km/h.

6. Question 6 · Compound measures · OCR 10.01b

   A force of 60 N acts on an area of 0.5 m². Find the pressure.

   Answer:

   Pressure = force ÷ area.
   60 ÷ 0.5 = 120 N/m².

7. Question 7 · Compound measures · OCR 10.01b

   Water flows at 12 litres per minute. How long does it take to fill a 300-litre tank?

   Answer:

   Time = volume ÷ rate.
   300 ÷ 12 = 25 minutes.

8. Question 8 · Scale · OCR 10.01c

   A model car is built to a scale of 1 : 24. The model is 18 cm long. How long is the real car?

   Answer:

   18 × 24.
   432 cm = 4.32 m.

9. Question 9 · Scale drawings · OCR 10.01c

   On a plan, 1 cm represents 2 m. A room is 4.5 m long. How long is it on the plan?

   Answer:

   4.5 ÷ 2.
   2.25 cm.

10. Question 10 · Bearings · OCR 10.01c

    What are the three rules for bearings?

    Answer:

    Measure from north.
    Measure clockwise, and write three figures (e.g. 065°).

11. Question 11 · Bearings · OCR 10.01c

    The bearing of B from A is 070°. Find the bearing of A from B.

    Answer:

    Back bearing = bearing ± 180°.
    070 + 180 = 250°.

12. Question 12 · Perimeter · OCR 10.02a

    A rectangle is 8 cm by 3 cm. Find its perimeter.

    Answer:

    Perimeter = 2 × (length + width).
    2 × (8 + 3) = 22 cm.

13. Question 13 · Perimeter · OCR 10.02a

    A square has side length 7 m. Find its perimeter.

    Answer:

    A square has four equal sides.
    Perimeter = 4 × 7 = 28 m.

14. Question 14 · Circles · OCR 10.02b

    Find the circumference of a circle with radius 5 cm, to 1 d.p.

    Answer:

    C = 2πr = 10π.
    31.4 cm.

15. Question 15 · Circles · OCR 10.02b

    A circle has circumference 30 cm. Find its diameter to 1 d.p.

    Answer:

    d = C ÷ π.
    30 ÷ π = 9.5 cm.

16. Question 16 · Arcs · OCR 10.02b

    Find the length of an arc of a circle with radius 9 cm and angle 80°, in terms of π.

    Answer:

    Arc = 80360 × 2π × 9.
    4π cm.

17. Question 17 · Circles · OCR 10.02b

    A circle has radius 5 cm. Find its circumference in terms of π.

    Answer:

    Circumference = 2πr.
    2π × 5 = 10π cm.

18. Question 18 · Sectors · OCR 10.02c · Higher only

    Find the perimeter of a sector with radius 10 cm and angle 72°, to 1 d.p.

    Answer:

    Arc = 72360 × 20π = 4π = 12.57 cm.
    Perimeter = 12.57 + 20 = 32.6 cm.

19. Question 19 · Area · OCR 10.03a

    Find the area of a triangle with base 12 cm and height 7 cm.

    Answer:

    Area = 12 × base × height.
    0.5 × 12 × 7 = 42 cm².

20. Question 20 · Area · OCR 10.03b

    A parallelogram has base 8 m and perpendicular height 5 m. Find its area.

    Answer:

    Parallelogram area = base × perpendicular height.
    8 × 5 = 40 m².

21. Question 21 · Area · OCR 10.03c

    Find the area of a trapezium with parallel sides 6 cm and 10 cm and height 4 cm.

    Answer:

    Area = 12(a + b)h.
    0.5 × 16 × 4 = 32 cm².

22. Question 22 · Circles · OCR 10.03d

    Find the area of a semicircle with radius 6 cm in terms of π.

    Answer:

    Full circle = 36π.
    18π cm².

23. Question 23 · Circles · OCR 10.03d

    Find the radius of a circle with area 154 cm², using π = 227.

    Answer:

    r² = 154 ÷ 227 = 49.
    r = 7 cm.

24. Question 24 · Area · OCR 10.03d

    A square has area 81 cm². Find its side length.

    Answer:

    Side length squared is 81.
    The positive side length is √81 = 9 cm.

25. Question 25 · Composite shapes · OCR 10.03e

    A shape is a 10 cm by 6 cm rectangle with a semicircle of diameter 6 cm on one end. Find its area to 1 d.p.

    Answer:

    Rectangle 60 cm² and semicircle 12 × π × 3² = 14.14 cm².
    74.1 cm².

26. Question 26 · Compound measures · OCR 10.04a

    A metal block has mass 540 g and volume 200 cm³. Find its density.

    Answer:

    Density = mass ÷ volume.
    540 ÷ 200 = 2.7 g/cm³.

27. Question 27 · Cylinders · OCR 10.04a

    Find the volume of a cylinder with radius 3 cm and height 10 cm in terms of π.

    Answer:

    V = πr²h.
    π × 9 × 10 = 90π cm³.

28. Question 28 · Volume · OCR 10.04a

    Find the volume of a cuboid 5 cm by 4 cm by 3 cm.

    Answer:

    Volume = length × width × height.
    60 cm³.

29. Question 29 · Volume · OCR 10.04a

    Find the volume of a prism with cross-sectional area 18 cm² and length 10 cm.

    Answer:

    Volume = area of cross-section × length.
    180 cm³.

30. Question 30 · Surface area · OCR 10.04a

    Find the surface area of a cube with side 4 cm.

    Answer:

    6 faces each 16 cm².
    96 cm².

31. Question 31 · Frustums · OCR 10.04b · Higher only

    How do you find the volume of a frustum of a cone?

    Answer:

    Find the volume of the whole cone.
    Subtract the volume of the small cone removed from the top.

32. Question 32 · Cones · OCR 10.04b · Higher only

    Find the curved surface area of a cone with radius 4 cm and slant height 9 cm in terms of π.

    Answer:

    Curved SA = πrl.
    π × 4 × 9 = 36π cm².

33. Question 33 · Pyramids · OCR 10.04c · Higher only

    Find the volume of a square-based pyramid with base 6 cm and height 10 cm.

    Answer:

    V = 13 × base area × height.
    13 × 36 × 10 = 120 cm³.

34. Question 34 · Pythagoras · OCR 10.05a

    The hypotenuse is 13 cm and one side is 5 cm. Find the other side.

    Answer:

    a² = 13² − 5² = 169 − 25 = 144.
    a = 12 cm.

35. Question 35 · Pythagoras · OCR 10.05a

    Find the distance between (1, 2) and (7, 10).

    Answer:

    Horizontal 6, vertical 8. d² = 36 + 64 = 100.
    d = 10.

36. Question 36 · Pythagoras · OCR 10.05a

    Find the diagonal of a 9 cm by 12 cm rectangle.

    Answer:

    Pythagoras: diagonal² = 9² + 12² = 81 + 144 = 225.
    Diagonal = √225 = 15 cm.

37. Question 37 · Pythagoras · OCR 10.05a · Higher only

    A cuboid measures 2 cm by 3 cm by 6 cm. Find its space diagonal.

    Answer:

    Apply Pythagoras in three dimensions: diagonal² = 2² + 3² + 6² = 49.
    Diagonal = √49 = 7 cm.

38. Question 38 · Pythagoras · OCR 10.05b

    Find the hypotenuse of a right-angled triangle with shorter sides 9 cm and 12 cm.

    Answer:

    c² = 9² + 12² = 81 + 144 = 225.
    c = 15 cm.

39. Question 39 · Pythagoras · OCR 10.05b

    Is a triangle with sides 7, 24 and 25 right-angled?

    Answer:

    7² + 24² = 49 + 576 = 625 = 25².
    Yes.

40. Question 40 · Trigonometry · OCR 10.05b

    In a right-angled triangle, the hypotenuse is 10 cm and the angle is 30°. Find the opposite side.

    Answer:

    sin 30° = opp10.
    opp = 10 × 0.5 = 5 cm.

41. Question 41 · Trigonometry · OCR 10.05b

    Adjacent = 8 cm, hypotenuse = 10 cm. Find the angle to 1 d.p.

    Answer:

    cos θ = 810 = 0.8.
    θ = cos⁻¹ 0.8 = 36.9°.

42. Question 42 · Exact trig values · OCR 10.05c

    What is the exact value of tan 45°?

    Answer:

    An isosceles right-angled triangle has equal opposite and adjacent sides.
    tan 45° = 1.

43. Question 43 · Exact trig values · OCR 10.05c

    State sin 0° and cos 0°.

    Answer:

    At 0°, the opposite side has zero length.
    sin 0° = 0 and cos 0° = 1.

44. Question 44 · Trigonometry · OCR 10.05c

    State the exact value of sin 30°.

    Answer:

    In a 30°, 60°, 90° triangle, the side opposite 30° is half the hypotenuse.
    sin 30° = 12.

45. Question 45 · Sine rule · OCR 10.05d · Higher only

    State the sine rule.

    Answer:

    asin A = bsin B = csin C.
    Use it when you know a side and its opposite angle.

46. Question 46 · Sine rule · OCR 10.05d · Higher only

    In triangle ABC, a = 10, A = 30° and B = 45°. Find b to 1 d.p.

    Answer:

    bsin 45° = 10sin 30°, so b = 10 × 0.70710.5.
    b = 14.1.

47. Question 47 · Sine rule · OCR 10.05d · Higher only

    In triangle ABC, A = 30°, B = 90° and side a = 5 cm. Find side b.

    Answer:

    The sine rule gives bsin 90° = 5sin 30°.
    sin 90° = 1 and sin 30° = 12.
    Therefore b = 5 ÷ 12 = 10 cm.

48. Question 48 · Cosine rule · OCR 10.05e · Higher only

    State the cosine rule for finding side a.

    Answer:

    a² = b² + c² − 2bc cos A.
    Use it with two sides and the included angle.

49. Question 49 · Cosine rule · OCR 10.05e · Higher only

    Find a when b = 5, c = 8 and A = 60°.

    Answer:

    a² = 25 + 64 − 2 × 5 × 8 × 0.5 = 89 − 40 = 49.
    a = 7.

50. Question 50 · Cosine rule · OCR 10.05e · Higher only

    A triangle has sides b = 3 cm and c = 4 cm with included angle A = 90°. Find side a.

    Answer:

    Cosine rule: a² = b² + c² − 2bc cos A.
    a² = 3² + 4² − 2 × 3 × 4 × 0 = 25.
    a = 5 cm.

More OCR GCSE Maths topics

• OCR 1 Number Operations and Integers
• OCR 2 Fractions, Decimals and Percentages
• OCR 3 Indices and Surds
• OCR 4 Approximation and Estimation
• OCR 5 Ratio, Proportion and Rates Of Change
• OCR 6 Algebra
• OCR 7 Graphs of Equations and Functions
• OCR 8 Basic Geometry
• OCR 9 Congruence and Similarity
• OCR 10 Mensuration
• OCR 11 Probability
• OCR 12 Statistics

Other Maths courses: AQA GCSE Maths, Edexcel GCSE Maths.