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This page sets out the method for similar shapes: area & volume in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Similar shapes: area & volume topic page.
When two similar shapes or solids are given and you need a missing area or volume, always find (the length factor) before anything else.
When you are given two areas and asked about volumes (or the other way round), you must pass through the length factor.
Length scale factor 3. The area scale factor is 9.
Length scale factor 2. The volume scale factor is 8.
Area scale factor 9. The length scale factor is 3, the square root.
Volume scale factor 8. The length scale factor is 2, the cube root.
A scale factor of 3 does not mean add 3 to the area.
Area uses the square. Volume uses the cube. Then multiply the area or volume by that scale factor.
Write the three multipliers in a column before you touch the question. Lengths use k. Areas, including surface area and the area of a face, use k^2. Volumes use k^3. For k = 3, those multipliers are 3, 9 and 27. A face of 4 cm^2 becomes 4 × 9 = 36 cm^2. A volume of 10 cm^3 becomes 10 × 27 = 270 cm^3. The angles are not multiplied at all.
The powers match the units. Length is one dimension, so the power is 1. Area is measured in cm^2, so the power is 2. Volume is measured in cm^3, so the power is 3. If you multiply a volume by k^2 you have used an area multiplier on a volume. The number may look plausible and still be wrong. Say the word area or volume out loud before you choose the power.
Method: 1. Multiply lengths by k, areas by k^2, and volumes by k^3. 2. For k=3, the multipliers are 3, 9 and 27. 3. A face of 4cm^2 becomes 4× 9=36cm^2, and a volume of 10cm^3 becomes 270cm^3.
Divide corresponding lengths. Two similar bottles have heights 8 cm and 12 cm. From the smaller to the larger, k = 12/8 = 1.5. The volume scale factor is 1.5^3 = 3.375. A smaller bottle of 200 ml therefore holds 200 × 3.375 = 675 ml when enlarged to the larger height. The area scale factor would be 1.5^2 = 2.25, which you would use for a label, not for the liquid.
Keep one direction for the whole question. If k = 1.5 takes you from small to large, then a large measurement is the small one multiplied by the right power of 1.5. To go back, divide by that power, or use 1/1.5 as the scale factor from large to small. Mixing the two directions is the usual reason a large bottle comes out smaller than the small one. After the multiplication, the larger solid must have the larger volume.
Method: 1. Divide matching lengths to get k before you touch an area or a volume. 2. Heights 12cm and 8cm give k=12/8=1.5. 3. A 200ml bottle then holds 200× 1.5^3=675ml.
Sometimes no length is given. If two areas are 36 cm^2 and 100 cm^2, the area scale factor is 100/36. The length scale factor is the square root, 10/6 = 5/3. You square-root because you are undoing the square that took lengths to areas. Only then may you cube k if a volume is needed. Square-rooting a volume ratio, or cube-rooting an area ratio, uses the wrong power.
If the volumes are 27 cm^3 and 125 cm^3, the volume scale factor is 125/27. Then k^3 = 125/27, so k = 5/3. The area scale factor is (5/3)^2 = 25/9. A small surface area of 54 cm^2 becomes 54 × 25/9 = 150 cm^2 on the large solid. Find k first, square it second, multiply third. Do not multiply 54 by 125/27. That would treat an area as a volume.
Method: 1. Square-root an area ratio to recover k. 2. Areas 100cm^2 and 36cm^2 give k=10/6=5/3. 3. Cube-root a volume ratio: volumes 125cm^3 and 27cm^3 also give k=5/3.
Stay in one unit system. Heights in cm give areas in cm^2 and volumes in cm^3. One millilitre is one cubic centimetre, so a capacity in ml is scaled by k^3 in the same way as a volume. Do not convert ml to litres in the middle of the scale-factor line unless the question asks for litres. A scale factor itself has no units, because it is a length divided by a length.
Fractions are often cleaner than decimals. 15/6 = 5/2 = 2.5, and (5/2)^3 = 125/8 = 15.625. On a calculator you may use 2.5^3 in one go. Round only if the question asks, and round the final amount, not the scale factor, if the scale factor is exact. A check: k > 1 must make areas and volumes larger. If your large volume is smaller than the small one, invert k and start again.
Method: 1. Keep centimetres with cm^2 and cm^3, and scale millilitres by k^3. 2. A scale factor has no units, because it is a length divided by a length. 3. Cube an exact fraction in one go, such as (5/2)^3=125/8.
Exam questions often give one ratio and ask for another. If the surface areas of two similar solids are in the ratio 9 : 25, you cannot jump straight to the volumes. Take the square root of each part to get the length ratio 3 : 5, then cube each part to get the volume ratio 27 : 125.
The three steps are a chain: area ratio, square root to length ratio, then cube to volume ratio. Going the other way, a volume ratio of 8 : 27 has a cube root giving lengths 2 : 3, and squaring gives areas 4 : 9. Always pass through the length ratio, because that is the one ratio that matches the scale factor k.
Method: 1. Write the ratio you are given with the correct shapes in order. 2. Convert to a length ratio: square root for an area ratio, cube root for a volume ratio. 3. Raise the length ratio to the power you need: squared for area, cubed for volume. 4. Simplify the final ratio if it asks for the simplest form.
Tip: The length ratio is the key. Every area and volume question passes through it.
In a pouring problem, one container is filled and emptied into a second similar container. The number of times you must fill the first is the volume ratio. For example, if the surface areas are in the ratio 1 : 9, the lengths are in the ratio 1 : 3, the volumes are in the ratio 1 : 27, and the larger container holds 27 of the smaller.
The same reasoning applies to capacity (a bottle holding 250 ml), to the mass of two blocks of the same material, and to the cost of paint (which depends on surface area, not volume). Decide first whether the quantity is a length, an area or a volume, and then use k, k^2 or k^3.
Method: 1. Decide whether the quantity is a length, an area or a volume. 2. Find the length scale factor k from the information given. 3. Multiply by k, k^2 or k^3 as appropriate. 4. Write the answer in context, such as the number of fills or the mass in kilograms.
Sometimes you must find k before anything else. If two similar solids have volumes of 96 cm^3 and 324 cm^3, the volume scale factor is 324/96 = 3.375. The length scale factor is the cube root: k = sqrt[3]3.375 = 1.5. Use the cube root key on your calculator, or raise to the power 1/3.
Once you have k, find any missing length by multiplying or dividing. If the missing length is in the smaller solid, divide by k. A cross-section area given together with a base lets you find a height first. Keep the unrounded value of k on the calculator and round only the final answer.
Method: 1. Divide the larger value by the smaller to get the area or volume factor. 2. Take the square root for an area factor or the cube root for a volume factor to get k. 3. Multiply by k going to the larger solid, or divide by k going to the smaller solid. 4. Round at the end and give the units.
Real contexts build on the same rules. The mass of two similar objects made of the same material is in the ratio of their volumes. The amount of paint, the area of a label or the cost of covering a surface follows the ratio of areas. The height of a model follows the length ratio.
Read each part of the question and write which factor applies before you calculate. A common slip is to use the length scale factor for everything. For a statue made 6 times as tall as a model, the mass is 6^3 = 216 times as much, but the surface area is only 6^2 = 36 times as large.
Method: 1. Find the length scale factor k from two matching lengths. 2. Label each quantity: length, area or volume. 3. Apply k, k^2 or k^3 to the matching quantity. 4. State the answer with units: kilograms, square centimetres or pounds.
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