GCSE · Geometry and measures

Similar Shapes: Area and Volume

Double every length of a photo and you need four times as much paper. Double every length of a model boat and it needs eight times as much wood. When shapes are similar, lengths, areas and volumes all grow, but at different rates. This page shows exactly why, using tiles and cubes, and how to go backwards from an area or volume to a length.

GCSE subject content G6 and G19: apply the concepts of congruence and similarity. G19 (Higher): the relationships between lengths, areas and volumes in similar figures.

1 · Recap: similar shapes and the length scale factor

Two shapes are similar if one is an enlargement of the other: same angles, and every length multiplied by the same number, the length scale factor k. (Revise with the KS3 pages Congruence and similarity and Scale factors.) To find k, divide a length on the big shape by the matching length on the small one.

2 · Area scale factor = k² Higher

Enlarge a shape by scale factor k. Each little unit square inside it becomes a k by k square, which holds k × k = k² unit squares. So every area is multiplied by k², not k. Watch the tiles: with k = 3, each teal square becomes nine.

The big trap: using k for areas. If the lengths are 3 times bigger, the area is 3² = 9 times bigger, not 3. Lengths include perimeters, heights, radii and diagonals (all × k); areas include surface area, the area of a face and the amount of paint (all × k²).

3 · Volume scale factor = k³ Higher

In three dimensions, each unit cube becomes a k by k by k block of cubes: k³ of them. So volumes (and anything that depends on volume, such as capacity or the mass of a solid made of the same material) are multiplied by k³.

Length × k
Area × k²
Volume × k³
lengths × k areas × k² volumes × k³
ratio of lengths a : b → areas a² : b² → volumes a³ : b³

4 · Going backwards Higher

Often you're given two areas or two volumes and need a length. Undo the power: if the area scale factor is k², then k = √(area factor); if the volume factor is k³, then k = ∛(volume factor). To get from areas to volumes, always go through k: square-root first, then cube.

Type a number or a fraction, e.g. 27/8.

5 · Length, area or volume? Higher

In word problems the hard part is spotting which kind of quantity you have. Ask: is it measured in cm, cm² or cm³ (or does it depend on one of those)? Sort these quantities for two similar solids.

6 · Mixed problems Higher

Strategy: (1) find k from two matching lengths, or by square-rooting or cube-rooting a given ratio; (2) decide whether the unknown is a length, area or volume; (3) multiply (or divide, going from big to small) by k, k² or k³. Answers to 3 s.f.

Worked examples

Two triangles are similar. The small one has sides 4 cm and 6 cm; the matching sides of the large one are 10 cm and x cm. Find x.
k = 10 ÷ 4 = 2.5 (big ÷ small, using a matching pair).
x = 6 × 2.5 = 15 cm.
Two similar shapes have matching lengths 3 cm and 6 cm. The small one has area 20 cm². Find the area of the large one. Higher
k = 6 ÷ 3 = 2. Area factor = k² = 4.
Area = 20 × 4 = 80 cm². (Not 40: that uses k for an area.)
Two similar cylinders have heights 10 cm and 15 cm. The small one has volume 400 cm³. Find the volume of the large one. Higher
k = 15 ÷ 10 = 1.5. Volume factor = 1.5³ = 3.375.
Volume = 400 × 3.375 = 1350 cm³.
Two similar solids have surface areas 50 cm² and 200 cm². The small one has volume 30 cm³. Find the volume of the large one. Higher
Area factor = 200 ÷ 50 = 4.
Back to k: k = √4 = 2. Volume factor = 2³ = 8.
Volume = 30 × 8 = 240 cm³. (Using 4 for volume, 120 cm³, is the classic error.)
Two similar bottles have volumes 128 cm³ and 250 cm³. The small one is 12 cm tall and the large one has surface area 150 cm². Find the height of the large bottle and the surface area of the small one. Higher
Volume ratio 128 : 250 = 64 : 125 (÷ 2).
Length ratio: ∛64 : ∛125 = 4 : 5, so k = 5/4 = 1.25.
Height = 12 × 1.25 = 15 cm.
Area ratio = 4² : 5² = 16 : 25. Going from big to small: 150 × 16/25 = 96 cm².
Two similar statues are 20 cm and 50 cm tall. The small one has mass 1.2 kg and needs 0.3 litres of paint. Find the mass of the large one and the paint it needs. Higher
k = 50 ÷ 20 = 2.5.
Mass depends on volume (same material): 1.2 × 2.5³ = 1.2 × 15.625 = 18.75 kg.
Paint covers the surface, an area: 0.3 × 2.5² = 0.3 × 6.25 = 1.875 litres. (Paint is measured in litres, but it's an area problem.)

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