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

This guide is written for the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1 (for exams May/June 2018 onwards). It covers statements G17 and G18 in section 3.3.2, Mensuration and calculation. Both statements appear only in the Extension content column, so every skill here is Extension only. The specification also says all content can be assessed on the Extension papers.

Work through the guide in order, then use the revision notes and the practice questions. Course pages: the OxfordAQA IGCSE Mathematics hub and the printable checklist. To see where you stand first, try the free 10-minute diagnostics.

What this unit covers

Ref	What you must be able to do	Tier
G17	Use the relationships between lengths, areas and volumes in similar figures	Extension only
G18	Calculate arc lengths of sectors of circles	Extension only
G18	Calculate angles of sectors of circles	Extension only
G18	Calculate areas of sectors of circles	Extension only

Area and circumference of a full circle, and volumes of solids, belong to other statements (G15 and G16). Here you simply use the circle formulas you already know.

G17: similar figures

Two figures are similar when one is an enlargement of the other. Every pair of matching angles is equal, and every pair of matching lengths is in the same ratio. That ratio is the scale factor, written k here.

The key results:

Quantity	Multiplied by
Any length (side, height, radius, perimeter)	k
Any area (surface area, cross-section, face)	k²
Any volume (capacity)	k³

If the ratio of lengths is a : b, the ratio of areas is a² : b² and the ratio of volumes is a³ : b³.

Two things to hold on to. First, the rule applies to matching measurements only: match the base to the base and the height to the height. Second, the rule needs the figures to be truly similar. A tall thin tin and a short wide tin are not similar, even if both are cylinders.

Finding a scale factor

Always divide new by old, with both quantities of the same type.

• From lengths: k = new length ÷ old length.
• From areas: k² = new area ÷ old area, so k = √(area ratio).
• From volumes: k³ = new volume ÷ old volume, so k = ∛(volume ratio).

Worked example 1: lengths

Triangle ABC is similar to triangle PQR. AB = 14 cm and BC = 22 cm. PQ = 35 cm, and PQ matches AB. Find QR.

k = PQ ÷ AB = 35 ÷ 14 = 2.5
QR = BC × k = 22 × 2.5 = 55 cm

Worked example 2: from lengths to area

Two similar plots of land have matching sides in the ratio 3 : 7. The smaller plot has area 54 m². Find the area of the larger plot.

Length ratio 3 : 7, so k = 7/3
Area scale factor = k² = 49/9
Larger area = 54 × 49/9 = 6 × 49 = 294 m²

Worked example 3: capacity and surface area

Two similar bottles have heights 12 cm and 18 cm. The smaller holds 240 ml. Its label has area 96 cm². Find the capacity of the larger bottle and the area of its label.

k = 18 ÷ 12 = 3/2
Volume scale factor = (3/2)³ = 27/8
Capacity = 240 × 27/8 = 30 × 27 = 810 ml
Area scale factor = (3/2)² = 9/4
Label = 96 × 9/4 = 216 cm²

Worked example 4: from area back to volume

Two similar solids have surface areas 75 cm² and 108 cm². The larger has volume 1296 cm³. Find the volume of the smaller.

Area ratio, large : small = 108 : 75 = 36 : 25
Length ratio = √36 : √25 = 6 : 5
Volume ratio = 6³ : 5³ = 216 : 125
Smaller volume = 1296 × 125/216 = 6 × 125 = 750 cm³

The path is always the same: convert whatever you are given into a length ratio, then raise it to the power you need.

Worked example 5: from volume to height

Two solid statues are cast from the same metal and are similar. The small one has mass 2 kg and height 15 cm. The large one has mass 16 kg. Find the height of the large statue.

Same metal means mass is proportional to volume.

Volume ratio = 16 : 2 = 8 : 1
Length ratio = ∛8 : ∛1 = 2 : 1
Height of large statue = 15 × 2 = 30 cm

Worked example 6: a scale model

A model ship is built to a scale of 1 : 80. The sail area on the model is 0.45 m². Find the real sail area.

Length scale factor = 80
Area scale factor = 80² = 6400
Real sail area = 0.45 × 6400 = 2880 m²

Common errors in G17

• Applying k to an area. If k = 3, areas are multiplied by 9, not 3.
• Using the area ratio as if it were the length ratio. Take the square root first.
• Dividing the wrong way round. Decide which figure is “new” before you divide.
• Mixing units. Convert both lengths to the same unit before forming k.
• Assuming two shapes are similar because they look alike. The question has to say so, or give matching angles and proportional sides.

G18: arcs and sectors

A sector is the region between two radii and the arc that joins their ends. Its angle θ is measured at the centre of the circle. The sector is a fraction of the whole circle:

fraction = θ/360

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

Remember the perimeter: the two straight edges are radii, so add 2r to the arc.

You may be asked for answers in terms of π. In that case, cancel the fraction first and leave π as a symbol.

Worked example 7: arc length and sector area

A sector has radius 9 cm and angle 40°. Find its arc length and its area, in terms of π.

Fraction = 40/360 = 1/9
Arc length = 1/9 × 2π × 9 = 2π cm
Area = 1/9 × π × 81 = 9π cm²

As decimals these are 6.28 cm and 28.3 cm² to 3 significant figures.

Worked example 8: perimeter

A sector has radius 12 cm and angle 105°. Find its perimeter to 3 significant figures.

Fraction = 105/360 = 7/24
Arc = 7/24 × 2π × 12 = 7π = 21.99 cm
Perimeter = 21.99 + 2 × 12 = 45.99
Perimeter = 46.0 cm (3 s.f.)

Finding the angle

Put the known information into the formula and solve for θ. A quicker view: the angle is 360° multiplied by (arc ÷ circumference), or by (sector area ÷ circle area).

Worked example 9: angle from arc length

A sector of radius 10 cm has arc length 15 cm. Find the angle.

Circumference = 2π × 10 = 20π
θ = 15/(20π) × 360 = 85.94...
θ = 85.9° (1 d.p.)

Worked example 10: angle from area

A sector of radius 8 cm has area 40 cm². Find the angle.

Circle area = π × 64 = 64π
θ = 40/(64π) × 360 = 71.62...
θ = 71.6° (1 d.p.)

Finding the radius

Substitute θ and the given arc or area, then solve. For an area, you reach r² first, so take a square root.

Worked example 11: radius from arc length

A 75° sector has arc length 11 cm. Find the radius.

11 = 75/360 × 2π × r
r = 11 × 360 / (75 × 2π) = 8.403...
r = 8.40 cm (3 s.f.)

Worked example 12: radius from area

A 150° sector has area 60π cm². Find the radius.

60π = 150/360 × π × r²
60 = 5/12 × r²
r² = 144
r = 12 cm

Worked example 13: part of a ring

Two concentric arcs have radii 14 cm and 9 cm. Two radii at 70° cut off the region between them. Find its area.

Treat it as the big sector minus the small sector. Both have angle 70°, so factorise:

Area = 70/360 × π × (14² − 9²)
     = 7/36 × π × (196 − 81)
     = 7/36 × 115π
     = (805/36)π = 70.2 cm² (3 s.f.)

Worked example 14: using proportion

A sector of angle 120° has area 75 cm². A second sector has the same radius and angle 45°. Find its area.

Same circle, so area is proportional to angle:

Area = 75 × 45/120 = 28.125 cm²

Linking G17 and G18

Two sectors with the same angle are similar. Take sector A with radius 6 cm and angle 50°, and sector B with radius 15 cm and the same angle. Then k = 15/6 = 2.5.

	Sector A	Sector B	Ratio B : A
Arc length	5π/3 cm	25π/6 cm	2.5
Area	5π cm²	31.25π cm²	6.25

The arc is a length, so it scales by k. The area scales by k². Checking this link is a useful way to catch slips.

Common errors in G18

• Forgetting to add 2r when the question asks for the perimeter.
• Using the diameter in a formula that needs the radius.
• Using the full-circle formula and forgetting the fraction θ/360.
• Rounding π to 3.14 too early. Keep full values until the last line.
• Reporting an angle larger than 360°, or one that does not fit the picture. A small arc should give a small angle.

Exam technique for this unit

• Write the fraction θ/360 on its own line before you substitute.
• Underline the word that tells you what the question wants: arc, area or perimeter.
• On a similar-figures problem, write “k = “ and what it is before using it.
• Give exact answers in terms of π when asked; otherwise round only at the end.
• Put units on every answer: cm, cm² and cm³ are different things.

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.