Similarity: Area & Volume Scale Factors

Triangle ABC is similar to triangle PQR. In triangle ABC, angle B = 90° and angle A = 28°. In triangle PQR, angle Q = 90°. What is the size of angle P?

• A62°
• B14°
• C118°
• D28°✓

Answer: Similar triangles have equal corresponding angles, so matching vertices carry the same angle. Since angle Q corresponds to angle B (both 90°), angle P corresponds to angle A, so angle P = 28°. Using the complement, adding it to the right angle, or halving it gives the other results.

Two similar right-angled triangles both contain an angle where the tangent ratio is 3/4. In the smaller triangle the side corresponding to the '4' is 4 cm; in the larger triangle the corresponding side is 12 cm. Find the length of the side in the larger triangle that corresponds to the smaller triangle's '3' side.

• A9 cm✓
• B6 cm
• C12 cm
• D15 cm

Answer: In similar triangles, corresponding sides scale by the same factor, and the tangent ratio itself stays unchanged between them. Since the '4' side scales from 4 cm to 12 cm, the scale factor is 3, so the '3' side scales to 3 × 3 = 9 cm. Applying the scale factor to the wrong side, or not applying it at all, gives the other results.

Two similar right-angled triangles both contain an angle θ where sinθ = 5/13. The smaller triangle has hypotenuse 13 cm and the larger similar triangle has hypotenuse 39 cm. Find the length of the side opposite θ in the larger triangle.

• A13 cm
• B15 cm✓
• C5 cm
• D36 cm

Answer: In similar right-angled triangles, the sine ratio stays constant while the actual side lengths scale by the length scale factor. The hypotenuse scales from 13 cm to 39 cm, a factor of 3, so the opposite side scales from 5 cm to 3 × 5 = 15 cm. Leaving a length unscaled, or scaling the wrong side, gives the other results.

Two similar rectangles have a length scale factor of 3 from the smaller to the larger. The smaller rectangle has an area of 8 cm². Find the area of the larger rectangle.

• A72 cm²✓
• B48 cm²
• C216 cm²
• D24 cm²

Answer: When two similar shapes have a length scale factor of k, their areas scale by k². Here k = 3, so the area scale factor is 3² = 9, giving 8 × 9 = 72 cm². Using k³ = 27 instead of k² gives 216 cm², confusing it with the volume scale factor; applying the scale factor without squaring it gives 24 cm². The remaining distractor comes from a similar miscalculation.

Two similar polygons have areas of 18 cm² and 162 cm². Find the linear scale factor from the smaller polygon to the larger.

• A3✓
• B9
• C1.5
• D4.5

Answer: When two similar shapes have areas in a known ratio, the length scale factor is the square root of the area ratio. Here 162 ÷ 18 = 9, and √9 = 3. Forgetting to take the square root, or halving the area ratio instead, gives the other results.

Two similar triangles have corresponding sides of 6 cm and 15 cm. The smaller triangle has an area of 20 cm². Find the area of the larger triangle.

• A150 cm²
• B125 cm²✓
• C100 cm²
• D50 cm²

Answer: The length scale factor between the triangles is 15 ÷ 6 = 2.5, so the area scale factor is 2.5² = 6.25. The larger area is therefore 20 × 6.25 = 125 cm². Using the length scale factor directly without squaring it gives 50 cm² instead, a common error. The other totals come from similar mistakes.

Two similar circles have radii of 4 cm and 10 cm. Find the ratio of their areas in the form 1 : n.

• A1 : 5
• B1 : 2.5
• C1 : 2
• D1 : 6.25✓

Answer: The length ratio of the radii is 4 : 10, which simplifies to 2 : 5. Since area scales with the square of length, the area ratio is 2² : 5² = 4 : 25, which is equivalent to 1 : 6.25. Using the length ratio directly without squaring it gives 1 : 2.5, and a further mis-simplification gives the other results.

Two similar solids have a length scale factor of 2 from the smaller to the larger. The smaller solid has a volume of 15 cm³. Find the volume of the larger solid.

• A60 cm³
• B120 cm³✓
• C30 cm³
• D45 cm³

Answer: When two similar solids have a length scale factor of k, their volumes scale by k³. Here k = 2, so the volume scale factor is 2³ = 8, giving 15 × 8 = 120 cm³. Applying the scale factor without cubing it gives 30 cm³ instead, a common error. The other totals come from similar mistakes.

Two similar cylinders have volumes of 27 cm³ and 216 cm³. Find the linear scale factor from the smaller cylinder to the larger.

• A2✓
• B0.5
• C1.5
• D2.83

Answer: When two similar solids have volumes in a known ratio, the length scale factor is the cube root of the volume ratio. Here 216 ÷ 27 = 8, and the cube root of 8 is 2. Taking the square root instead of the cube root gives 2.83, and inverting the ratio before rooting gives 0.5; the remaining distractor comes from a similar slip.

Two similar cones have base radii of 3 cm and 9 cm. The smaller cone has a volume of 20 cm³. Find the volume of the larger cone.

• A180 cm³
• B60 cm³
• C540 cm³✓
• D4320 cm³

Answer: The length scale factor of the radii is 9 ÷ 3 = 3, so the volume scale factor is 3³ = 27. The larger volume is therefore 20 × 27 = 540 cm³. Confusing the volume scale factor with the area scale factor gives 180 cm³, and using the scale factor directly without cubing it gives 60 cm³; doubling the scale factor by mistake before cubing gives the remaining total.

Two similar containers have heights of 5 cm and 20 cm. Find the ratio of their volumes in the form 1 : n.

• A1 : 0.02
• B1 : 256
• C1 : 64✓
• D1 : 400

Answer: The length ratio of the heights is 5 : 20, which simplifies to 1 : 4. Since volume scales with the cube of length, the volume ratio is 1³ : 4³ = 1 : 64. Raising the ratio to the fourth power instead of the third gives 1 : 256; expressing the ratio as if out of 100 instead of 1 gives 1 : 400, and inverting the ratio gives the remaining distractor.

Two similar solids have surface areas in the ratio 4 : 9. Find the ratio of their volumes.

• A16 : 81
• B2 : 3
• C8 : 27✓
• D4 : 9

Answer: An area ratio of 4 : 9 corresponds to a length ratio of √4 : √9 = 2 : 3, since area scales with the square of length. Volume scales with the cube of length, so the volume ratio is 2³ : 3³ = 8 : 27. Leaving the ratio as the length ratio without cubing it gives 2 : 3; squaring the area ratio again instead of converting to a length ratio gives 16 : 81, and leaving the area ratio unconverted gives 4 : 9.

Two similar statues have volumes in the ratio 8 : 125. Find the ratio of their surface areas.

• A20 : 50
• B64 : 625
• C8 : 125
• D4 : 25✓

Answer: A volume ratio of 8 : 125 corresponds to a length ratio of ∛8 : ∛125 = 2 : 5, since volume scales with the cube of length. Area scales with the square of length, so the area ratio is 2² : 5² = 4 : 25. Leaving the length ratio unsquared and unsimplified gives 20 : 50, and squaring the volume ratio directly without first taking the cube root gives 64 : 625; the remaining distractor uses the volume ratio unchanged.

A scale model of a building is similar to the real building with a length scale factor of 1 : 50. The model has a surface area of 600 cm². Find the surface area of the real building in square metres.

• A15000 m²
• B30000 m²
• C1500000 m²
• D150 m²✓

Answer: The length scale factor of 1 : 50 gives an area scale factor of 50² = 2500. The real surface area is therefore 600 × 2500 = 1,500,000 cm², which converts to 150 m² since 1 m² = 10,000 cm². Forgetting to convert cm² to m² at all gives 1500000 m²; dividing by 100 instead of 10,000 gives 15000 m², and applying the length scale factor without squaring it (600 × 50, left unconverted) gives 30000 m².

A scale model of a car is built at a length scale factor of 1 : 20 compared with the real car, using the same material throughout. The model has a mass of 150 g. Find the mass of the real car in kilograms.

• A120 kg
• B3 kg
• C1200 kg✓
• D1200000 kg

Answer: Since the model and the real car are made from the same material, they have the same density, so mass scales in the same way as volume: by the cube of the length scale factor. The volume scale factor is 20³ = 8000, so the real mass is 150 g × 8000 = 1,200,000 g = 1200 kg. Forgetting to convert grams to kilograms at the end gives 1200000 kg, and applying the length scale factor directly without cubing it gives 3 kg; the remaining distractor comes from a similar slip.

A model boat is similar to the real boat with a length scale factor of 1 : 25, using the same material throughout. The real boat has a mass of 5000 kg. Find the mass of the model, correct to 3 significant figures.

• A8.00 kg
• B200 kg
• C0.320 kg✓
• D78125000 kg

Answer: Since the model and the real boat share the same material and therefore the same density, mass scales by the cube of the length scale factor. The volume scale factor is 25³ = 15,625, so the model's mass is 5000 kg ÷ 15,625 = 0.320 kg. Multiplying by the scale factor instead of dividing gives 78125000 kg, and using k² instead of k³ gives 8.00 kg; the remaining distractor divides by the scale factor directly without cubing it.