If I double the side of a square, does the area double too?

No. Doubling the side doubles the perimeter but takes the area ×4. A square of side 3 cm has area 3 × 3 = 9 cm²; a square of side 6 cm has area 6 × 6 = 36 cm², which is ×4, not ×2. Four copies of the small square tile the big one, 2 along by 2 up. Under a length scale factor k, length scales by k, area by k² and volume by k³, so one ratio never fits all three.

Do two shapes with the same area always have the same perimeter?

No. A rectangle 8 cm by 5 cm has area 8 × 5 = 40 cm² and perimeter 2(8 + 5) = 26 cm. A rectangle 10 cm by 4 cm also has area 10 × 4 = 40 cm² but perimeter 2(10 + 4) = 28 cm. Same area, different perimeters. Equal area does not force equal perimeter, so the answer to a perimeter ratio of two equal-area rectangles is not 1 : 1.

Does surface area scale the same way as volume?

No. Surface area scales by the factor squared and volume by the factor cubed. A cube of side 3 cm has surface area 6 × 3² = 54 cm² and volume 3³ = 27 cm³. Double the side to 6 cm: surface area 6 × 6² = 216 cm² (×4) and volume 6³ = 216 cm³ (×8). So if two solids have a volume ratio of 1 : 8, the length factor is 2 and the surface area ratio is 1 : 4, not 1 : 8.

GCSE Maths Foundation · AQA · Geometry

Area & volume don't scale like length

When a shape is scaled, students often expect length, area and volume to grow by the same factor. Double the side, so the area must double too. But it does not. Doubling the side of a 3 cm square doubles the perimeter ( $12 \to 24$ cm) yet takes the area from $9$ cm² to $36$ cm², a factor of 4: four small squares tile the big one.

The thirty-second fix: under a length scale factor $k$ , length scales by $k$ , area by $k^2$ and volume by $k^3$ . One ratio does not fit all three. Square it for area, cube it for volume.

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How to spot it in your own work

You doubled a length and expected the area to double too, writing ×2 where it should be $\times 2^2 = \times 4$ .
You applied a length ratio straight to an area or volume (using 3 where it should be $3^2 = 9$ or $3^3 = 27$ ).
You assumed two shapes with the same area must have the same perimeter, answering a perimeter ratio as 1 : 1.
You scaled the surface area like the volume (both ×8 when a side doubles), when surface area only goes $\times 4$ .

An exam question that triggers it

Here is a canonical AQA Foundation trigger (Nov24 P3 Q21 flavour, calculator):

Rectangle A is 8 cm by 5 cm. Rectangle B is 10 cm by 4 cm.

They have the same area. Write the ratio of the perimeter of A to the perimeter of B in its simplest form.

The misconception answer is $1 : 1$ : equal areas, so equal perimeters. But the areas being equal says nothing about the perimeters.

Rectangle A has perimeter $2(8 + 5) = 26$ cm; rectangle B has perimeter $2(10 + 4) = 28$ cm. The ratio is $26 : 28 = 13 : 14$ , not $1 : 1$ .

Why students fall for this

This is the illusion of linearity: the deeply intuitive sense that if you scale one thing, everything scales the same way (De Bock, Van Dooren, Verschaffel and Janssens; Tourniaire and Pulos). Length is the measure students meet first, and its behaviour, double the length, double the length, gets carried silently onto area and volume.

The meaning fixes it. Area covers two directions, so under a length scale factor $k$ it scales by $k \times k = k^2$ . Volume fills three directions, so it scales by $k \times k \times k = k^3$ . Double the side of a cube and eight small cubes fill the big one, so the volume goes ×8; its surface, made of flat faces, only goes ×4. One length ratio cannot be the ratio for all three.

AQA Foundation exploits this directly: comparing equal-area rectangles by perimeter (Nov24 P3 Q21, the 1 : 1 trap), asking how many times bigger one circle's area is (Jun22 P1 Q25, circumference instead of area), a shaded annulus where a length ratio is applied straight to the area (Nov24 P1 Q26), checking similarity by side ratio not area (Jun23 P2 Q22), and asking whether a surface area ratio matches a volume ratio (Jun23 P3 Q19, the 1 : 8 trap).

The fix: Length ×k, area ×k², volume ×k³: one ratio does not fit all three

Name the length scale factor. If every length is multiplied by $k$ , that is the length factor, and it is the only thing that scales linearly.

Square it for area. Area scales by $k^2$ . Double the side ( $k = 2$ ) and the area goes $\times 2^2 = \times 4$ : $9 \to 36$ cm². Treble it ( $k = 3$ ) and the area goes $\times 9$ .

Cube it for volume. Volume scales by $k^3$ . Double the side and the volume goes $\times 2^3 = \times 8$ : $8 \to 64$ cm³.

Do not assume one measure forces another. Equal area does not force equal perimeter: an $8 \times 5$ and a $10 \times 4$ rectangle both have area 40 cm² but perimeters 26 cm and 28 cm.

Worked example

A cube has side 3 cm. You double the side to 6 cm. How do the surface area and the volume change? This is the trap behind Jun23 P3 Q19: surface area and volume do not scale together.

Name the length factor. Every length is doubled, so $k = 2$ .
Surface area scales by $k^2$ . A cube's surface is six faces, $6 s^2$ :
$6 \times 3^2 = 54 \text{ cm}^2 \;\to\; 6 \times 6^2 = 216 \text{ cm}^2, \quad \text{so } \times 4$
Volume scales by $k^3$ .
$3^3 = 27 \text{ cm}^3 \;\to\; 6^3 = 216 \text{ cm}^3, \quad \text{so } \times 8$
Eight small cubes fill the big one.
Read off the trap. Surface area went $\times 4$ but volume went $\times 8$ . So if two solids have a volume ratio of $1 : 8 = 1^3 : 2^3$ , the length factor is 2 and the surface area ratio is $1^2 : 2^2 = 1 : 4$ , never $1 : 8$ .

The same powers settle the equal-area question: rectangle A $8 \times 5 = 40$ cm² has perimeter 26 cm, and rectangle B $10 \times 4 = 40$ cm² has perimeter 28 cm, ratio $13 : 14$ , not $1 : 1$ .

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Common questions

If I double the side of a square, does the area double too?: No. Doubling the side doubles the perimeter but takes the area $\times 4$ . A square of side 3 cm has area $3 \times 3 = 9$ cm²; a square of side 6 cm has area $6 \times 6 = 36$ cm², which is $\times 4$ , not $\times 2$ . Four copies of the small square tile the big one, 2 along by 2 up. Under a length scale factor $k$ , length scales by $k$ , area by $k^2$ and volume by $k^3$ , so one ratio never fits all three.
Do two shapes with the same area always have the same perimeter?: No. A rectangle 8 cm by 5 cm has area $8 \times 5 = 40$ cm² and perimeter $2(8 + 5) = 26$ cm. A rectangle 10 cm by 4 cm also has area $10 \times 4 = 40$ cm² but perimeter $2(10 + 4) = 28$ cm. Same area, different perimeters. Equal area does not force equal perimeter, so the answer to a perimeter ratio of two equal-area rectangles is not $1 : 1$ .
Does surface area scale the same way as volume?: No. Surface area scales by the factor squared and volume by the factor cubed. A cube of side 3 cm has surface area $6 \times 3^2 = 54$ cm² and volume $3^3 = 27$ cm³. Double the side to 6 cm: surface area $6 \times 6^2 = 216$ cm² ( $\times 4$ ) and volume $6^3 = 216$ cm³ ( $\times 8$ ). So if two solids have a volume ratio of $1 : 8$ , the length factor is 2 and the surface area ratio is $1 : 4$ , not $1 : 8$ .

Related misconceptions

Perimeter vs area: computing the wrong measureThe neighbouring trap: naming whether the question wants the boundary length or the space inside before you compute.
Triangle area: base × height ÷ 2, not base × heightA triangle is half its bounding rectangle, so the area is base × perpendicular height ÷ 2, never the un-halved product.
Ratio: additive vs multiplicative scalingThe same scaling instinct elsewhere: keeping a relationship in proportion by multiplying, not adding.

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