Comparing and Scaling in 2D and 3D Revision Notes | GCSE AQA Higher Maths

1. Why Scaling Matters — and Where Students Go Wrong

Imagine a model car that is $\frac{1}{20}$ the length of the real thing. You might assume the model uses $\frac{1}{20}$ of the paint. It actually uses $\frac{1}{400}$ of the paint — because paint covers area, not length. That gap between intuition and reality is exactly what this topic is about.

Scaling in 2D and 3D is one of the most reliably tested topics on GCSE Higher papers. Questions appear on both questions without a calculator and calculator-allowed exam questions, and they almost always catch students who apply the linear scale factor to areas or volumes without squaring or cubing it first.

By the end of this lesson you will have a reliable method for every situation: comparing lengths, areas, and volumes using ratio notation, and moving fluently between scale factors in all three dimensions.

2. The Core Idea — Linear, Area and Volume Scale Factors

Start with something concrete. Two similar rectangles: one has a length of 3 cm, the other has a length of 9 cm. The lengths are in the ratio $3 : 9$, which simplifies to $1 : 3$. That ratio is the linear scale factor — it tells you how many times bigger each length is in the larger shape.

Now think about what happens to area when you scale up by a factor of 3. The rectangle's width also triples, so the area multiplies by $3 \times 3 = 9$. Area grows in two directions simultaneously, so you square the linear scale factor.

Volume adds a third dimension. If every length triples, the volume multiplies by $3 \times 3 \times 3 = 27$. You cube the linear scale factor to get the volume scale factor. This is the relationship that surprises students most — a shape that looks only a little bigger can hold dramatically more.

Here is the key anchor to remember: lengths scale by $k$, areas scale by $k^2$, volumes scale by $k^3$. Every question on this topic reduces to identifying which of these three you need.

This is where it gets interesting: the table below shows how dramatically the ratios diverge as $k$ increases. A shape with $k = 4$ has areas 16 times bigger and volumes 64 times bigger than the original.

Notice the last row: scale factors can be less than 1 when the new shape is smaller than the original. A linear scale factor of $\frac{1}{2}$ means the new shape has areas $\frac{1}{4}$ of the original and volumes $\frac{1}{8}$ of the original. The same squaring and cubing rules apply in both directions.

3. Working with Scale Factors — Step-by-Step Methods

There are two types of problem you will meet. The first gives you two lengths and asks for an area or volume. The second gives you an area or volume ratio and asks you to find a length or the other area/volume. Both types follow the same logical chain — identify $k$, then apply the correct power.

Theory is easy, but exam questions are tricky. Let's walk through a standard question of each type so you can see the method in action.

Worked example: Finding an Area from a Linear Scale Factor

Two similar triangles, P and Q, have base lengths of 4 cm and 10 cm respectively. The area of triangle P is 12 cm². Calculate the area of triangle Q. [3 marks]

1. $k = \frac{10}{4} = 2.5$ — Divide the length in Q by the corresponding length in P. [M1] — identifying the correct ratio earns a method mark.
2. $k^2 = 2.5^2 = 2.5 \times 2.5 = 6.25$ — Square the linear scale factor to get the area scale factor. Note: $2.5 \times 2.5 = 6.25$, not 5. Always expand the multiplication.
3. $\text{Area of Q} = 12 \times 6.25 = 75 \text{ cm}^2$ — [M1] for multiplying by $k^2$. [A1] for the correct answer with units. Missing cm² costs a mark.
4. $k = 2.5$, so Q is bigger than P. $75 > 12$ ✓ — Always check the direction: if $k > 1$, the new area must be larger than the original.

Answer: Area of triangle Q = 75 cm²

Now let's tackle the reverse: you are given the volume ratio and need to find a length. This requires working backwards — taking a cube root to recover $k$ from the volume scale factor.

Worked example: Finding a Length from a Volume Ratio

Two similar cones, A and B, have volumes in the ratio $27 : 64$. The height of cone A is 9 cm. Find the height of cone B. [3 marks]

1. $\text{VSF} = \frac{64}{27}$ — The ratio $27 : 64$ tells us the VSF from A to B is $\frac{64}{27}$.
2. $k = \sqrt[3]{\frac{64}{27}} = \frac{\sqrt[3]{64}}{\sqrt[3]{27}} = \frac{4}{3}$ — [M1] — cube-rooting the VSF to get $k$. Note: $\sqrt[3]{64} = 4$ because $4 \times 4 \times 4 = 64$, and $\sqrt[3]{27} = 3$ because $3 \times 3 \times 3 = 27$.
3. $\text{Height of B} = 9 \times \frac{4}{3} = \frac{36}{3} = 12 \text{ cm}$ — [M1] for multiplying by $k$. [A1] for 12 cm. The height of B is larger because $k > 1$.
4. $\left(\frac{12}{9}\right)^3 = \left(\frac{4}{3}\right)^3 = \frac{64}{27}$ ✓ — Substituting back confirms the volume ratio matches. This check takes 10 seconds and secures the marks.

Answer: Height of cone B = 12 cm

Notice the pattern: to go from length to area, square $k$. To go from length to volume, cube $k$. To go backwards from area to length, square-root the ASF. To go backwards from volume to length, cube-root the VSF.

Worked example: Finding a Volume from an Area Ratio

Two similar prisms, X and Y, have surface areas in the ratio $9 : 49$. The volume of prism X is 54 cm³. Find the volume of prism Y. [4 marks]

1. $\text{ASF} = \frac{49}{9}$ — The ratio of surface areas gives the ASF directly.
2. $k = \sqrt{\frac{49}{9}} = \frac{\sqrt{49}}{\sqrt{9}} = \frac{7}{3}$ — [M1] — square-rooting the ASF. $\sqrt{49} = 7$ and $\sqrt{9} = 3$.
3. $k^3 = \left(\frac{7}{3}\right)^3 = \frac{7 \times 7 \times 7}{3 \times 3 \times 3} = \frac{343}{27}$ — [M1] — cubing $k$ to get the VSF. $7 \times 7 = 49$, then $49 \times 7 = 343$. $3 \times 3 = 9$, then $9 \times 3 = 27$.
4. $\text{Volume of Y} = 54 \times \frac{343}{27} = 2 \times 343 = 686 \text{ cm}^3$ — [M1] for multiplying by $k^3$. [A1] for 686 cm³. Note: $54 \div 27 = 2$, so $54 \times \frac{343}{27} = 2 \times 343 = 686$.

Answer: Volume of prism Y = 686 cm³

4. Expressing Comparisons Using Ratio Notation

GCSE Higher exam questions often ask you to compare lengths, areas, or volumes using ratio notation rather than finding a specific numerical answer. The method is the same — you still identify $k$ and apply the correct power — but you express the result as a simplified ratio.

Earlier we saw how to move from a length ratio to a volume ratio. Now we are seeing the same pattern applied to expressing comparisons. If two similar shapes have lengths in the ratio $a : b$, their areas are in the ratio $a^2 : b^2$ and their volumes are in the ratio $a^3 : b^3$. You raise every term in the ratio to the appropriate power.

For example, if lengths are in the ratio $3 : 5$, areas are in the ratio $3^2 : 5^2 = 9 : 25$ and volumes are in the ratio $3^3 : 5^3 = 27 : 125$. There is no need to find $k$ as a decimal — working with the ratio directly is cleaner and less prone to rounding errors.

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When the ratio is not made up of integers, convert to a fraction first. For instance, if lengths are in the ratio $1.5 : 4$, write this as $\frac{3}{2} : 4 = 3 : 8$ before raising to the required power. Working with integers avoids decimal errors in the squaring or cubing step.

5. Deliberate Practice — Try These Yourself

The only way to make this automatic is to practise the decision — length, area, or volume? — until it becomes instinctive. Work through these problems before checking the solutions. Cover the answer and commit to a method first.

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The graph above shows how $k^2$ (area scale factor) and $k^3$ (volume scale factor) grow as the linear scale factor $k$ increases from 0 to 5. Notice how quickly $k^3$ pulls away from $k^2$ — this is why volume grows so much faster than area when you enlarge a shape.

6. Exam Traps, Common Mistakes and Mark-Winning Technique

This topic has a small number of very predictable errors. Examiners know students make them, which is why the questions are designed to tempt you into them. Knowing the traps in advance is the most efficient revision you can do here.

The most common error is applying the linear scale factor directly to an area or volume. If $k = 3$ and the original area is 20 cm², many students write $20 \times 3 = 60$ cm² instead of $20 \ times 9 = 180$ cm². The examiner has almost certainly included 60 as a distractor option in a multiple-choice question, or it will be a value that appears naturally in the working if you make this mistake.

The second trap is working in the wrong direction. If you are given the volume ratio and asked for the area ratio, you must go via $k$: cube-root the volume ratio to get $k$, then square $k$ to get the area ratio. Trying to convert directly from volume ratio to area ratio without finding $k$ first leads to errors.

In questions without a calculator, you will need to square and cube simple fractions or integers by hand. Practise expanding these longhand: $\left(\frac{3}{4}\right)^3 = \frac{3 \times 3 \times 3}{4 \times 4 \times 4} = \frac{27}{64}$. Never try to do this in one mental step — write out each multiplication separately.