What is a length of 6 enlarged by scale factor ⅓?

It is 2. Enlarging multiplies every length by the scale factor, so 6 × ⅓ = 2. A scale factor that is a fraction less than 1 makes the shape smaller, even though the word enlargement sounds like it should grow. Answering 8 comes from multiplying by 1⅓ by mistake, and answering 18 comes from multiplying by 3 instead of dividing — both ignore that ⅓ is less than 1.

Can an enlargement make a shape smaller?

Yes. In GCSE maths an enlargement with a scale factor between 0 and 1 produces a smaller, similar shape. The word enlargement only means the lengths are all multiplied by the same scale factor; it does not promise the shape grows. Scale factor 2 doubles every length, scale factor ½ halves them, and scale factor ⅓ divides them by 3. So describing a shape that has shrunk to half size, you would still call it an enlargement, with scale factor ½.

How do you find the scale factor that maps one shape onto another?

Divide a length on the image by the matching length on the original. If a side of 4 on shape A maps to a side of 2 on shape B, the scale factor is 2 ÷ 4 = ½. Because that factor is less than 1, shape B is smaller, and the transformation is an enlargement with scale factor ½ about its centre. To describe it fully you also state the centre of enlargement, found by drawing lines through pairs of corresponding points until they meet.

GCSE Maths Foundation · AQA · Transformations & vectors

Enlargement and scale factor: 6 enlarged by ⅓ is 2, not 8 or 18

Asked to enlarge a length of $6$ by scale factor $\tfrac13$ , students assume an enlargement must grow and answer 8 or 18. But enlarging just multiplies every length by the scale factor, and $\tfrac13$ is less than 1, so the shape shrinks: $6 \times \tfrac13 = 2$ .

The thirty-second fix: an enlargement multiplies every length by the scale factor, and a fraction between 0 and 1 makes the shape smaller. So a base of 6 becomes $6 \times \tfrac13 = 2$ , not the 8 from multiplying by $1\tfrac13$ or the 18 from multiplying by 3.

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How to spot it in your own work

You assumed an enlargement must make the shape bigger, so a fractional scale factor felt wrong.
You multiplied by 3 instead of by $\tfrac13$ , turning a base of 6 into $18$ instead of $2$ .
You used $1\tfrac13$ by mistake, getting $6 \times \tfrac43 = 8$ instead of multiplying by the true $\tfrac13$ .
When describing an enlargement that shrank a shape, you called it a translation or said it “just got smaller” instead of an enlargement with scale factor less than 1.

An exam question that triggers it

Here is a canonical AQA Foundation trigger (enlarge by a fractional factor):

Shape B has a base of length

$6\ \text{cm}.$

Shape B is enlarged by scale factor $\tfrac13$ . Work out the length of the base of the enlarged shape.

The misconception is to read “enlarged” as “made bigger” and reach for a number above 6. But the scale factor is $\tfrac13$ , which is less than 1, so the image is smaller.

Multiply the length by the scale factor: $6 \times \tfrac13 = 2$ . The new base is $2\ \text{cm}$ .

Why students fall for this

The everyday meaning of “enlarge” is “make bigger”, so the word fights the maths. In GCSE, an enlargement only means every length is multiplied by the same scale factor — and that factor can be less than 1. When the scale factor is $\tfrac13$ , students cannot believe the shape should shrink, so they multiply by 3 to get $18$ , or slip to $1\tfrac13$ and get $8$ , anything to make the answer grow.

Multiplying by the actual scale factor settles it. A scale factor of $\tfrac13$ divides every length by 3: $6 \times \tfrac13 = 2$ . The image is a smaller, similar copy. Scale factor 2 would double the lengths, scale factor 1 would leave them unchanged, and any fraction between 0 and 1 brings them down. The shape below shows B alongside its $\tfrac13$ image, which is genuinely smaller.

The same idea runs the other way when you describe a transformation. If shape A maps to a smaller shape B, you find the scale factor by dividing an image length by the matching original — and a factor of $\tfrac12$ still counts as an enlargement, here about the centre $(1,\ -7)$ . Calling it “just smaller” or a translation misses the named transformation.

The fix: Multiply every length by the scale factor — a fraction below 1 shrinks the shape

Multiply, don't guess the direction. An enlargement scales every length by the scale factor. For a base of 6 and scale factor $\tfrac13$ : $6 \times \tfrac13 = 2$ .

Read the size of the scale factor. A factor above 1 makes the shape bigger, a factor of 1 keeps it the same, and a factor between 0 and 1 makes it smaller. $\tfrac13$ is below 1, so the image shrinks.

Watch the two classic slips. Multiplying by 3 gives $18$ (you inverted the fraction); using $1\tfrac13$ gives $8$ (you mis-read the factor). Multiply by exactly $\tfrac13$ .

To describe an enlargement, give scale factor and centre. If A maps to a half-size B, divide image by original to get scale factor $\tfrac12$ , and state the centre, e.g. $(1,\ -7)$ . It is still an enlargement, not a translation.

Worked example

Enlarge a base of $6\ \text{cm}$ by scale factor $\tfrac13$ . The trap is to make it bigger; the fix is to multiply by the scale factor exactly.

Read the scale factor. It is $\tfrac13$ , which is less than 1, so the image must be smaller than the original.
Multiply the length by the scale factor.
$6 \times \tfrac13 = \tfrac{6}{3} = 2$
The new base is $2\ \text{cm}$ .
Reject the two traps. Multiplying by 3 gives $6 \times 3 = 18$ (the factor inverted); using $1\tfrac13$ gives $6 \times \tfrac43 = 8$ (the wrong factor). Neither uses the stated $\tfrac13$ .
Check the description direction. If a shape A maps to a smaller shape B, the transformation is an enlargement with scale factor below 1, e.g. $\tfrac12$ about centre $(1,\ -7)$ — found by dividing an image length by the matching original.

So a base of 6 enlarged by $\tfrac13$ is $2\ \text{cm}$ — not the 8 from $1\tfrac13$ or the 18 from multiplying by 3.

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Common questions

What is a length of 6 enlarged by scale factor $\tfrac13$ ?: It is $2$ . Enlarging multiplies every length by the scale factor, so $6 \times \tfrac13 = 2$ . A scale factor that is a fraction less than 1 makes the shape smaller, even though the word enlargement sounds like it should grow. Answering 8 comes from multiplying by $1\tfrac13$ by mistake, and answering 18 comes from multiplying by 3 instead of by $\tfrac13$ — both ignore that $\tfrac13$ is less than 1.
Can an enlargement make a shape smaller?: Yes. In GCSE maths an enlargement with a scale factor between 0 and 1 produces a smaller, similar shape. The word enlargement only means the lengths are all multiplied by the same scale factor; it does not promise the shape grows. Scale factor 2 doubles every length, scale factor $\tfrac12$ halves them, and scale factor $\tfrac13$ divides them by 3. So describing a shape that has shrunk to half size, you would still call it an enlargement, with scale factor $\tfrac12$ .
How do you find the scale factor that maps one shape onto another?: Divide a length on the image by the matching length on the original. If a side of 4 on shape A maps to a side of 2 on shape B, the scale factor is $2 \div 4 = \tfrac12$ . Because that factor is less than 1, shape B is smaller, and the transformation is an enlargement with scale factor $\tfrac12$ about its centre. To describe it fully you also state the centre of enlargement, such as $(1,\ -7)$ , found by drawing lines through pairs of corresponding points until they meet.

Related misconceptions

Reflecting in the given lineThe neighbouring transformations skill: the named line is the fold, so (4, −1) reflected in x = 6 is (8, −1), not a reflection in an axis.
Column vectors and reversingThe neighbouring vectors skill: reversing a vector negates each component in place, so (−3, 7) becomes (3, −7), not (7, −3).

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