OxfordAQA International GCSE Mathematics: Similar figures and arcs, sectors (9260) – Revision Notes

These notes condense statements G17 and G18 of section 3.3.2, Mensuration and calculation, in the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1 (for exams May/June 2018 onwards). Both statements are Extension only. For full explanations and longer worked examples, read the study guide.

When you have revised, test yourself with the practice questions. Course pages: the OxfordAQA IGCSE Mathematics hub and the printable checklist. For a quick level check, use the free 10-minute diagnostics.

Key facts at a glance

Ref	Skill	Tier
G17	Lengths, areas and volumes in similar figures	Extension only
G18	Arc lengths, angles and areas of sectors	Extension only

G17: similar figures

Similar means one figure is an enlargement of the other: matching angles equal, matching lengths in one ratio. Call the length scale factor k.

Measure	Scale factor
Length	k
Area	k²
Volume	k³

Length ratio	Area ratio	Volume ratio
a : b	a² : b²	a³ : b³
2 : 3	4 : 9	8 : 27
3 : 5	9 : 25	27 : 125
1 : 10	1 : 100	1 : 1000

Method in steps

1. Write down what you are given: lengths, areas or volumes.
2. Turn it into a length ratio. From areas take a square root. From volumes take a cube root.
3. Raise k to the power you need: 1 for length, 2 for area, 3 for volume.
4. Multiply or divide, then check the size of the answer.

Small reminders

• Same material means mass is proportional to volume.
• A map or model scale of 1 : n is a length scale. Areas scale by n² and volumes by n³.
• Perimeter is a length, so it scales by k.
• Convert lengths to one unit before forming k, and convert area or volume answers carefully afterwards: 1 m² is 10 000 cm² and 1 m³ is 1 000 000 cm³.
• Decide first whether the unknown figure is larger or smaller. If it is larger, k is greater than 1 and the answer must grow.

Reminder example. Two similar jugs have heights 9 cm and 12 cm. The small one holds 270 ml.

k = 12/9 = 4/3
Volume factor = 64/27
Larger jug = 270 × 64/27 = 640 ml

G18: arcs and sectors

A sector is a fraction θ/360 of a full circle, where θ is the angle at the centre in degrees.

Quantity	Formula
Arc length	(θ/360) × 2πr
Area of sector	(θ/360) × πr²
Perimeter of sector	arc + 2r
Angle from arc	θ = arc ÷ (2πr) × 360
Angle from area	θ = area ÷ (πr²) × 360

Method in steps

1. Write the fraction θ/360 and cancel it if it simplifies.
2. Decide: arc (length), sector (area) or perimeter (arc plus two radii).
3. Substitute into the matching formula.
4. For an unknown angle or radius, rearrange before you calculate.
5. Round only at the end. Leave π as a symbol when asked for an exact answer.

Reminder example. Sector radius 15 cm, angle 48°.

Fraction = 48/360 = 2/15
Arc = 2/15 × 30π = 4π cm
Area = 2/15 × 225π = 30π cm²

Reminder example (radius). A sector has angle 90° and area 49π cm².

49π = 1/4 × π × r²
r² = 196
r = 14 cm

More worked reminders

From area to volume. Two similar vases have label areas 45 cm² and 80 cm². The smaller holds 405 ml.

Area ratio 45 : 80 = 9 : 16
Length ratio 3 : 4, so k = 4/3
Volume factor = 64/27
Larger vase = 405 × 64/27 = 960 ml

A scale model. A model of a statue is 24 cm tall. The real statue is 3.6 m tall.

Real height = 360 cm, so k = 360 ÷ 24 = 15
Area factor = 225, volume factor = 3375

A model surface of 1500 cm² therefore matches 1500 × 225 = 337 500 cm² = 33.75 m² on the real statue.

Part of a ring. Radii 14 cm and 9 cm, angle 70°.

Area = 70/360 × π × (14² − 9²) = 7/36 × 115π = 70.2 cm²

Using proportion. Same circle, so area and arc length are proportional to the angle. A 120° sector with area 75 cm² gives a 45° sector with area 75 × 45/120 = 28.125 cm².

Angle from an arc. Radius 12 cm, arc 20 cm.

θ = 20/(24π) × 360 = 95.5° (1 d.p.)

Order of attack for mixed problems

1. Sketch or label the figure, and mark which measurements match.
2. Decide whether the problem is G17 (similar figures) or G18 (arc or sector). Some are both, for example two sectors with the same angle.
3. For G17, write k as a fraction and simplify it. A fraction such as 3/2 is easier to cube than 1.5.
4. For G18, write the fraction θ/360 and simplify it.
5. Do the calculation, round only at the end, and write the unit.
6. Check with a sense test: a bigger figure has a bigger area, volume and arc, and a smaller angle gives a smaller arc.

Must-know distinctions

• Arc length versus sector area. Arc length is a length (cm). Sector area is an area (cm²).
• Arc versus perimeter. The perimeter of a sector includes both radii.
• Radius versus diameter. Both circle formulas need the radius.
• Length ratio versus area ratio. If areas are in the ratio 16 : 25, lengths are in the ratio 4 : 5.
• Proportion in one circle. For a fixed radius, arc length and sector area are both proportional to the angle.
• Two sectors with equal angles are similar. Their arcs scale by k and their areas by k².

Quick self-test

1. The matching lengths of two similar shapes are in the ratio 2 : 5. State the ratio of their areas.
2. Two similar solids have lengths in the ratio 3 : 4. The smaller has volume 135 cm³. Find the volume of the larger.
3. Two similar flags have areas 20 cm² and 45 cm². The shorter flag has length 8 cm. Find the matching length on the larger flag.
4. Two similar tanks have volumes 81 litres and 192 litres. Find the ratio of their surface areas, in its simplest form.
5. A model is built at scale 1 : 50. A wall on the real building has area 600 m². Find the area of the matching wall on the model, in cm².
6. A sector has radius 6 cm and angle 60°. Give the arc length in terms of π.
7. A sector has radius 10 cm and angle 36°. Give its area in terms of π.
8. A sector has radius 5 cm and angle 144°. Find its perimeter to 3 significant figures.
9. A sector of radius 9 cm has arc length 12 cm. Find the angle to 1 decimal place.
10. An 80° sector has area 40π cm². Find the radius.
11. Two similar cans have heights 10 cm and 15 cm. The label on the smaller can has area 80 cm². Find the label area on the larger can.
12. A sector has radius 8 cm and angle 45°. Give its arc length and its area in terms of π.

Answers

1. 4 : 25.
2. k = 4/3, so volume factor is 64/27. 135 × 64/27 = 320 cm³.
3. Area ratio 20 : 45 = 4 : 9, so k = 3/2. Length = 8 × 3/2 = 12 cm.
4. Volume ratio 81 : 192 = 27 : 64, so length ratio 3 : 4. Area ratio 9 : 16.
5. Length factor 50, so area factor 2500. 600 ÷ 2500 = 0.24 m² = 2400 cm².
6. 60/360 × 12π = 2π cm.
7. 36/360 × 100π = 10π cm².
8. Arc = 144/360 × 10π = 4π = 12.566. Perimeter = 12.566 + 10 = 22.6 cm.
9. θ = 12/(18π) × 360 = 76.4°.
10. 80/360 × πr² = 40π, so r² = 180, r = 13.4 cm (3 s.f.).
11. k = 3/2, so area factor 9/4. 80 × 9/4 = 180 cm².
12. Fraction 1/8. Arc = 1/8 × 16π = 2π cm. Area = 1/8 × 64π = 8π cm².

Where marks are usually lost

• Scaling an area by k instead of k².
• Taking the volume ratio and forgetting to cube root when finding a length.
• Dividing the wrong way round, so a larger figure gets a smaller answer.
• Leaving the radii out of the perimeter of a sector.
• Substituting the diameter where the formula needs the radius.
• Rounding π to 3.14 early, which can shift the third significant figure.
• Giving a decimal when the question asked for an answer in terms of π.
• Leaving out units or mixing cm², m² and cm³.
• Giving no written method for a multi-step answer. A clear line of working lets an examiner see which ideas you used even if the final value slips.

Official syllabus

OxfordAQA International GCSE Mathematics (9260), specification Version 5.1 (for exams May/June 2018 onwards), published by OxfordAQA. Statements G17 and G18, section 3.3.2 Mensuration and calculation.