OCR GCSE Mathematics: Units, compound measures, maps, bearings, perimeter, circumference and arc length (J560)

This guide follows sections 10.01 and 10.02 of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026). It covers units and measurement (10.01a to 10.01c) and perimeter calculations (10.02a to 10.02c). Everything here is examined on both tiers: the specification lists no Higher tier only content in these sections. For quick recall use the revision notes, then test yourself with the practice questions. The OCR GCSE Mathematics hub and the printable checklist show where this unit sits. If you are unsure of your starting point, try the 10-minute diagnostics.

What this unit covers

Ref	What you must be able to do	Tier
10.01a	Use and convert standard units of length, area, volume/capacity, mass, time and money; do the same in algebraic contexts	Foundation and Higher
10.01b	Use and convert compound units (speed, rates of pay, unit pricing, density, pressure); know and apply speed = distance ÷ time and density = mass ÷ volume; use compound units in algebraic contexts	Foundation and Higher
10.01c	Use the scale of a map; work with bearings; construct and interpret scale drawings	Foundation and Higher
10.02a	Find the perimeter of rectilinear shapes	Foundation and Higher
10.02b	Know and use circumference = 2πr = πd; find the arc length of a sector given its angle and radius	Foundation and Higher
10.02c	Use perimeter formulae for composite 2D shapes	Foundation and Higher

In the specification, the algebraic contexts, density, pressure and arc length sit in the column headed “Foundation tier learners should also be able to”, so Higher tier learners need them too. The specification does not allow a calculator for Paper 2 on the Foundation tier or Paper 5 on the Higher tier, and says any content may be assessed on either calculator or non-calculator papers. Practise conversions and π work by hand as well.

10.01a Units of measurement

You need the metric links between units, plus the time and money facts the specification expects you to know.

Quantity	Facts to know
Length	1 km = 1000 m, 1 m = 100 cm, 1 cm = 10 mm
Area	1 m² = 10 000 cm², 1 cm² = 100 mm²
Volume	1 m³ = 1 000 000 cm³, 1 cm³ = 1000 mm³
Capacity	1 litre = 1000 ml = 1000 cm³, 1 m³ = 1000 litres
Mass	1 tonne = 1000 kg, 1 kg = 1000 g
Time	1 min = 60 s, 1 h = 60 min, 1 day = 24 h, 1 year = 365 days

Why area and volume factors are squared and cubed. A square 1 m by 1 m is also 100 cm by 100 cm. Its area is 100 × 100 = 10 000 cm², not 100 cm². Volume works the same way with three lengths.

Worked example 1. Convert 3.7 m² to cm².

1 m² = 100 × 100 = 10 000 cm²
3.7 × 10 000 = 37 000 cm²

Worked example 2. Give 0.35 m³ in cm³ and in litres: 0.35 × 1 000 000 = 350 000 cm³, and 0.35 × 1000 = 350 litres.

Time. Hours and minutes are not decimals. 3 h 18 min is 3 + 18/60 = 3.3 hours, and 2.35 h is 2 h 21 min (0.35 × 60 = 21). For the 24-hour clock add 12 to afternoon times, so 6.40 pm is 18:40. From 22:50 to 03:15 is 1 h 10 min to midnight plus 3 h 15 min, which is 4 h 25 min.

Money. Multiply or divide by the exchange rate. If £1 = €1.17, then £85 is 85 × 1.17 = €99.45, and €234 is 234 ÷ 1.17 = £200.

Algebraic contexts. The same conversions work when a length is a letter. Keep the unit with the letter.

Worked example 3. A ribbon is x metres long. It is cut into pieces of 20 cm. How many pieces?

x metres = 100x cm
Number of pieces = 100x ÷ 20 = 5x

10.01b Compound units

Quantity	Relationship	Units used
Speed	distance divided by time	km/h or m/s
Density	mass divided by volume	g/cm³ or kg/m³
Pressure	force divided by area	N/m² or N/cm²
Rate of pay	pay divided by hours worked	£ per hour
Unit price	cost divided by quantity	pence per gram

Rearrange each one as needed: distance = speed × time, time = distance ÷ speed, mass = density × volume, force = pressure × area.

Worked example 4 (speed). A cyclist rides 15 km in 40 minutes. Find the speed in km/h.

40 min = 40/60 h = 2/3 h
Speed = 15 ÷ (2/3) = 22.5 km/h

Converting a compound unit. Convert the top and the bottom separately.

Worked example 5. Change 54 km/h into m/s.

54 km = 54 000 m and 1 h = 3600 s
54 000 ÷ 3600 = 15 m/s

Short cut: km/h to m/s divides by 3.6.

Pay and unit pricing. At £9.60 per hour, 7 h 30 min earns 9.60 × 7.5 = £72.00. For best buys, find the cost of the same amount for each. A 750 g pack costs £2.10 and a 1.2 kg pack costs £3.24. That is 210 ÷ 750 = 0.28 p per gram and 324 ÷ 1200 = 0.27 p per gram, so the larger pack is the better buy.

Worked example 6 (density). A block has mass 2.16 kg and volume 800 cm³. Find its density in g/cm³ and in kg/m³.

2.16 kg = 2160 g
Density = 2160 ÷ 800 = 2.7 g/cm³
1 g/cm³ = 1000 kg/m³, so density = 2700 kg/m³

Worked example 7 (pressure). A force of 840 N acts on an area of 0.35 m². Find the pressure in N/m², then in N/cm².

Pressure = 840 ÷ 0.35 = 2400 N/m²
1 m² = 10 000 cm², so 2400 N/m² = 2400 ÷ 10 000 = 0.24 N/cm²

Worked example 8 (algebraic contexts). A car travels d km in t minutes. Write its speed in km/h and in m/s.

km/h: distance d km, time t/60 h, so speed = d ÷ (t/60) = 60d/t
m/s: distance 1000d m, time 60t s, so speed = 1000d/(60t) = 50d/(3t)

10.01c Maps, bearings and scale drawings

With a scale of 1 : 20 000, each 1 cm measured on the map stands for 20 000 cm on the ground, which is 200 m.

Worked example 9. On a 1 : 20 000 map, a footpath measures 7.3 cm. A cycle route is 2.8 km long in real life. Find the real length of the footpath and the map length of the cycle route.

7.3 × 20 000 = 146 000 cm = 1460 m = 1.46 km
2.8 km = 280 000 cm, and 280 000 ÷ 20 000 = 14 cm

Bearings. A bearing says which way to go from one point to another. Measure the turn clockwise from the north line, and always give three figures, such as 065° or 308°. Put your pencil on the place you travel from, draw a north line there, then measure clockwise to the line you want.

The bearing of A from B starts at B. To get a back bearing, add 180° if the bearing is below 180°, otherwise subtract 180°. If B is on a bearing of 072° from A, then A is on a bearing of 252° from B, because north lines are parallel and the angles between them add to 180°.

Worked example 10 (scale drawing). A boat sails 12 km on a bearing of 040°, then 9 km on a bearing of 130°. Find how far it is from the start and the bearing of the start from the boat.

Scale 1 cm : 3 km, so the legs are 4 cm and 3 cm.
Between bearings 040° and 130° the turn is 90°, so the legs meet at a right angle.
Distance = √(12² + 9²) = √225 = 15 km (5 cm on the drawing)
Angle at the start between the first leg and the line home = tan⁻¹(9/12) = 36.9°
Bearing of the boat from the start = 040° + 36.9° = 076.9°
Bearing of the start from the boat = 076.9° + 180° = 256.9°, so about 257°

On a scale drawing, measure the final line and the angle with a sharp pencil, ruler and protractor, and mark each north line.

10.02a Perimeter of rectilinear shapes

Perimeter is the distance round the edge. A rectilinear shape has only straight sides at right angles. Find any missing lengths, then add every side.

Worked example 11. A 14 cm by 9 cm rectangle has a 5 cm by 4 cm notch cut from a corner. Find the perimeter left.

The notch is at the top right corner. It shortens the top edge by 5 cm and the right edge by 4 cm. It also adds a step of 5 cm across and 4 cm down. The lost length and the added length match, so the outline is as long as the outer rectangle.

Perimeter = 2 × (14 + 9) = 46 cm

This shortcut works for any notch cut from a corner. It does not work for a notch cut from the middle of an edge, where you must add the two extra sides.

10.02b Circumference and arc length

For a circle with radius r and diameter d:

circumference C = πd = 2πr

Worked example 12. Find the circumference of a circle with diameter 14 cm, as a multiple of π and as a decimal to 3 significant figures.

C = π × 14 = 14π = 43.98... = 44.0 cm (3 s.f.)

Arc length. A sector is a slice of a circle. Its arc is the curved edge. The arc is a fraction of the whole circumference, and the fraction is the angle over 360°.

arc length = (angle/360) × 2πr

Worked example 13. Find the arc length of a sector of radius 9 cm and angle 80°.

Arc = 80/360 × 2 × π × 9 = 2/9 × 18π = 4π cm
4π = 12.57 cm (to 2 decimal places)

10.02c Perimeter of composite shapes

Split the outline into straight pieces and arcs. Count only the outside edge.

Worked example 14 (running track). A track has two straight sides of 60 m and two semicircular ends. Each end has diameter 30 m.

Two straights: 2 × 60 = 120 m
Two semicircles make one circle: π × 30 = 94.25 m
Perimeter = 120 + 94.25 = 214.2 m (1 d.p.)

Worked example 15 (window). A window is a rectangle 1.2 m wide and 1.5 m high with a semicircle on the top edge.

Diameter of semicircle = 1.2 m. The top of the rectangle is inside the shape, so leave it out.
Perimeter = 1.2 (bottom) + 2 × 1.5 (sides) + ½ × π × 1.2 (arc)
          = 4.2 + 1.885... = 6.08 m (3 s.f.)

Worked example 16 (sector). Find the perimeter of a sector of radius 7 cm and angle 130°.

Arc = 130/360 × 2 × π × 7 = 15.88 cm
Perimeter = arc + two radii = 15.88 + 14 = 29.9 cm (3 s.f.)

Worked example 17 (quarter circle removed). A rectangle is 10 cm by 8 cm. A quarter circle of radius 4 cm is removed from one corner.

Remaining straight edges: 10 + 8 + (10 − 4) + (8 − 4) = 28 cm
Arc = ¼ × 2π × 4 = 2π cm
Perimeter = 28 + 2π = 34.3 cm (3 s.f.)

Common errors

• Using 100 for area conversion between m² and cm², instead of 10 000.
• Treating 3 h 18 min as 3.18 hours.
• Multiplying when you should divide by an exchange rate, or the reverse.
• Giving a bearing with fewer than three figures, such as 65° instead of 065°.
• Using the diameter in 2πr, or the radius in πd.
• Giving only the arc as the perimeter of a sector, or counting the inside diameter of a semicircle.
• Rounding in the middle of a multi-step answer, or leaving out units.

Official syllabus

OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026), sections 10.01 and 10.02, published by OCR.