Enlargement and Reduction

Enlarge or reduce a figure by a scale factor $k$ : multiply every length by $k$ , keep every angle, find $k$ from a pair of lengths, draw the image on a square grid, and use the scale factor with photographs, models and the perimeters of similar figures.

Before you start

Similar figures have equal corresponding angles and corresponding sides in the same ratio, the scale factor (lesson 8.2).
Multiplying by a number less than 1 makes a number smaller: $8 \times 0.75 = 6$ and $8 \times \frac{1}{2} = 4$ .
A map with scale $1 : n$ shows every distance as $\frac{1}{n}$ of the actual distance (chapter 7).

The Scale Factor of an Enlargement

To enlarge a figure is to draw it again larger, with the same shape; to reduce it is to draw it again smaller. The figure you start with is the object, and the new figure is its image. The image is similar to the object.

Enlargement and reduction

In an enlargement or reduction with scale factor $k$ :

every length of the image $= k \times$ the corresponding length of the object, so $k = \frac{\text{image length}}{\text{object length}}\text{;}$

every angle of the image is equal to the corresponding angle of the object.

If $k > 1$ the image is an enlargement; if $0 < k < 1$ it is a reduction; if $k = 1$ it is the same size, and congruent to the object.

Scale factor	The image
$k = 3$	an enlargement: every length 3 times as long
$k = 1.5$	an enlargement: every length one and a half times as long
$k = 1$	the same size: congruent to the object
$k = \frac{1}{2}$	a reduction: every length half as long
$k = 0.2$	a reduction: every length $\frac{1}{5}$ as long

Example 1: Image lengths from the scale factor

A triangle has sides 4 cm, 6 cm and 7 cm. Find the sides of its image (a) when it is enlarged with scale factor $2.5$ , (b) when it is reduced with scale factor $0.75$ .

Solution

(a) Multiply each length by $2.5$ : $4 \times 2.5 = 10$ , $6 \times 2.5 = 15$ , $7 \times 2.5 = 17.5$ . The bottom row checks that each image length divided by its object length is $2.5$ .

Object (cm)	$4$	$6$	$7$
Image (cm)	$10$	$15$	$17.5$
Image ÷ object	$2.5$	$2.5$	$2.5$

(b) Multiply each length by $0.75$ : $4 \times 0.75 = 3$ , $6 \times 0.75 = 4.5$ , $7 \times 0.75 = 5.25$ .

Object (cm)	$4$	$6$	$7$
Image (cm)	$3$	$4.5$	$5.25$
Image ÷ object	$0.75$	$0.75$	$0.75$

(a) $10$ cm, $15$ cm and $17.5$ cm (b) $3$ cm, $4.5$ cm and $5.25$ cm

Example 2: Finding the scale factor

A rectangle 8 cm by 5 cm is changed into a similar rectangle whose longer side is 6 cm. Find the scale factor, say whether it is an enlargement or a reduction, and find the shorter side of the image.

Solution

The longer side of the image matches the longer side of the object.

$k = \frac{\text{image length}}{\text{object length}} = \frac{6}{8} = 0.75\text{.}$

$k$ is between 0 and 1, so the image is a reduction.

Shorter side of the image $= 0.75 \times 5 = 3.75$ cm. Every angle is still $90^{\circ}$ .

$k = 0.75$ , a reduction; the shorter side is $3.75$ cm.

Example 3: Angles do not change

$\triangle ABC$ has $AB = 10$ cm, $BC = 7$ cm, $CA = 8$ cm, $\angle A = 44^{\circ}$ , $\angle B = 53^{\circ}$ and $\angle C = 83^{\circ}$ . It is reduced with scale factor $0.6$ to $\triangle A'B'C'$ . Find the sides and angles of $\triangle A'B'C'$ .

Solution

Sides: $A'B' = 0.6 \times 10 = 6$ cm, $B'C' = 0.6 \times 7 = 4.2$ cm, $C'A' = 0.6 \times 8 = 4.8$ cm.

Angles do not change: $\angle A' = 44^{\circ}$ , $\angle B' = 53^{\circ}$ , $\angle C' = 83^{\circ}$ .

$A'B' = 6$ cm, $B'C' = 4.2$ cm, $C'A' = 4.8$ cm; $\angle A' = 44^{\circ}$ , $\angle B' = 53^{\circ}$ , $\angle C' = 83^{\circ}$ .

Enlarging and Reducing on a Square Grid

On a square grid, lengths are counted in squares. Go round the object one side at a time and count each step: so many squares across and so many up or down. Multiply every step by $k$ , then draw the image, starting from any grid point and taking the new steps in the same order.

A slanting side is a step across and a step up together: a side that goes 1 across and 2 up becomes, with $k = 2$ , a side that goes 2 across and 4 up. Its length is multiplied by $k$ too, but a length alone does not tell you where the side ends on the grid, so multiply its two steps.

Example 4: Enlarging with scale factor 2

Each square of the grid is 1 cm by 1 cm. Enlarge the shape with scale factor 2.

Solution

Steps round the object from its bottom left corner: 3 right, 2 up, 2 left, then 1 left and 2 down back to the start.

Multiply each by 2: 6 right, 4 up, 4 left, then 2 left and 4 down.

Check: the bottom side was 3 cm and is now 6 cm, the right side was 2 cm and is now 4 cm, and the right angles are still right angles.

The image is drawn above: every step of the object doubled.

Example 5: Reducing with scale factor one half

Each square of the grid is 1 cm by 1 cm. Reduce the triangle with scale factor $\frac{1}{2}$ .

Solution

Steps from the left end of the base: base 6 right; top 2 right and 4 up.

Multiply each by $\frac{1}{2}$ : base 3 right; top 1 right and 2 up.

The image is drawn above: base 3 cm, top 1 cm across and 2 cm up from the left end of the base.

Photographs and Models

A photograph printed larger is an enlargement; a model of a train or a building is a reduction. A map is a reduction too: a map with scale $1 : n$ is the ground reduced with scale factor $k = \frac{1}{n}$ . On a $1 : 50\,000$ map, $k = \frac{1}{50\,000}$ .

Example 6: Enlarging a photograph

A photograph 8 cm by 12 cm is enlarged so that its shorter side becomes 28 cm. (a) Find the scale factor and the longer side of the enlargement. (b) The same photograph is also reduced, for a name card, so that its longer side becomes 6 cm. Find the scale factor and the shorter side of this copy.

Solution

(a) $k = \frac{28}{8} = 3.5$ . Longer side $= 3.5 \times 12 = 42$ cm.

(b) $k = \frac{6}{12} = 0.5$ , a reduction. Shorter side $= 0.5 \times 8 = 4$ cm.

Both copies are similar to the photograph: $\frac{42}{28} = \frac{12}{8} = \frac{6}{4} = 1.5\text{,}$ the longer side over the shorter side in each.

(a) $k = 3.5$ ; $42$ cm (b) $k = 0.5$ ; $4$ cm

Example 7: A model MRT train

A model of an MRT train is made to the scale $1 : 50$ . (a) What is the scale factor from the real train to the model? (b) A real carriage is 23 m long. How long is the model carriage, in centimetres? (c) A door of the model is 2.8 cm wide. How wide is the real door, in metres?

Solution

(a) Every length of the model is $\frac{1}{50}$ of the real length, so $k = \frac{1}{50} = 0.02$ , a reduction.

(b) $23\text{ m} = 2300$ cm. Model length $= 2300 \times \frac{1}{50} = 46$ cm.

(a) $k = \frac{1}{50}$ (b) $46$ cm (c) $1.4$ m

Perimeters of Similar Figures

An enlargement or a reduction multiplies every side by $k$ , so the sides add up to $k$ times as much: a triangle with sides $a$ , $b$ , $c$ becomes one with sides $ka$ , $kb$ , $kc$ , and $ka + kb + kc = k(a + b + c)\text{.}$

Perimeters

If two figures are similar with scale factor $k$ , then perimeter of the second $= k \times$ perimeter of the first. The ratio of the perimeters is the scale factor.

This holds for lengths only (sides, perimeters, heights, diagonals). Areas grow differently: you will meet them in Secondary 3.

Example 8: A perimeter from the scale factor

A triangle has sides 5 cm, 8 cm and 9 cm. A similar triangle has its shortest side 10 cm. Find the perimeter of the second triangle.

Solution

The shortest sides correspond: $k = \frac{10}{5} = 2$ .

Perimeter of the first $= 5 + 8 + 9 = 22$ cm, so the perimeter of the second $= 2 \times 22 = 44$ cm.

Check through the sides: $10 + 16 + 18 = 44$ cm.

The perimeter is $44$ cm.

Example 9: The scale factor from the perimeters

Two similar pentagons have perimeters 60 cm and 45 cm. A side of the larger pentagon is 12 cm. Find the corresponding side of the smaller pentagon.

Solution

Scale factor from the larger to the smaller: $k = \frac{45}{60} = 0.75$ .

Corresponding side $= 0.75 \times 12 = 9$ cm.

The side is $9$ cm.

Common Mistakes

Writing the scale factor upside down. $k = \frac{\text{image}}{\text{object}}\text{:}$ an 8 cm side whose image is 6 cm gives $k = \frac{6}{8} = 0.75$ , not $\frac{8}{6}$ .
Calling every change of size an enlargement. With $k$ between 0 and 1 the image is smaller: a reduction.
Multiplying the angles by $k$ . Angles never change: only lengths are multiplied.
Adding instead of multiplying: enlarging a 3 cm by 2 cm rectangle by adding 2 cm to each side gives 5 cm by 4 cm, which is not similar.
On a grid, multiplying only the across step of a slanting side, or drawing it from a measured length. Multiply both of its steps.

Practice Questions

Try each question before you open its solution. Write the scale factor as image ÷ object, and say whether it enlarges or reduces.

Question 1

Is the image an enlargement, a reduction or congruent to the object when the scale factor is (a) $3$ , (b) $0.6$ , (c) $1$ , (d) $\frac{5}{4}$ , (e) $\frac{2}{5}$ ?

ShowHide solution

(a) $3 > 1$ : enlargement. (b) $0 < 0.6 < 1$ : reduction. (c) $k = 1$ : congruent.

(d) $\frac{5}{4} = 1.25 > 1$ : enlargement. (e) $\frac{2}{5} = 0.4 < 1$ : reduction.

(a) enlargement (b) reduction (c) congruent (d) enlargement (e) reduction

Question 2

(a) A rectangle 6 cm by 4 cm is enlarged with scale factor $1.5$ . Find the length and width of the image and the size of each of its angles. (b) A triangle with sides 10 cm, 8 cm and 6 cm is reduced with scale factor $0.3$ . Find the sides of the image.

ShowHide solution

(a) $6 \times 1.5 = 9$ cm and $4 \times 1.5 = 6$ cm. The angles do not change: each is $90^{\circ}$ .

(b) $10 \times 0.3 = 3$ cm, $8 \times 0.3 = 2.4$ cm, $6 \times 0.3 = 1.8$ cm.

(a) $9$ cm by $6$ cm, with every angle $90^{\circ}$ (b) $3$ cm, $2.4$ cm and $1.8$ cm

Question 3

Find the scale factor, and say whether the image is an enlargement or a reduction. (a) A side of 12 cm has an image of 18 cm. (b) A side of 20 cm has an image of 5 cm.

ShowHide solution

(a) $k = \frac{18}{12} = 1.5$ , more than 1: an enlargement.

(b) $k = \frac{5}{20} = 0.25$ , between 0 and 1: a reduction.

(a) $1.5$ , enlargement (b) $0.25$ , reduction

Question 4

A triangle has angles $35^{\circ}$ and $65^{\circ}$ . It is enlarged with scale factor 3. Find all three angles of the image.

ShowHide solution

Third angle of the object $= 180^{\circ} - 35^{\circ} - 65^{\circ} = 80^{\circ}$ (∠ sum of △).

An enlargement does not change angles, so the image has the same three angles.

$35^{\circ}$ , $65^{\circ}$ and $80^{\circ}$

Question 5

Each square is 1 cm by 1 cm. Enlarge the triangle with scale factor 3.

ShowHide solution

Base 2 across becomes 6 across; upright side 1 up becomes 3 up. The slanting side goes 6 across and 3 up.

The image has a base of 6 cm and an upright side of 3 cm, as drawn.

Question 6

Each square is 1 cm by 1 cm. Reduce the parallelogram with scale factor $\frac{1}{2}$ .

ShowHide solution

Base 4 across becomes 2 across; each slanting side, 2 across and 2 up, becomes 1 across and 1 up.

The image has a base of 2 cm and slanting sides 1 across and 1 up, as drawn.

Question 7

Each square is 1 cm by 1 cm. The rectangle Q is an image of the rectangle P. Find the scale factor from P to Q.

ShowHide solution

P is 2 cm by 4 cm and Q is 3 cm by 6 cm.

$k = \frac{3}{2} = 1.5$ and $\frac{6}{4} = 1.5$ . The two ratios agree, so Q is an enlargement of P.

$k = 1.5$

Question 8

A photograph 12 cm by 18 cm is enlarged into a poster whose longer side is 45 cm. Find the scale factor and the shorter side of the poster.

ShowHide solution

$k = \frac{45}{18} = 2.5$ . Shorter side $= 2.5 \times 12 = 30$ cm.

$k = 2.5$ ; the shorter side is $30$ cm.

Question 9

An architect's model of an HDB block is made to the scale $1 : 200$ . The model is 21 cm tall. (a) How tall is the real block, in metres? (b) The real block is 30 m wide. How wide is the model?

ShowHide solution

(a) Real height $= 21 \times 200 = 4200$ cm $= 42$ m.

(b) $30\text{ m} = 3000$ cm. Model width $= 3000 \times \frac{1}{200} = 15$ cm.

(a) $42$ m (b) $15$ cm

Question 10

An object has a side of 8 cm, and the corresponding side of its image is 6 cm. Ken says the scale factor is $\frac{8}{6} = 1.33\ldots$ , so the image is an enlargement. Explain his mistake.

ShowHide solution

The scale factor is image ÷ object, not object ÷ image.

$k = \frac{6}{8} = 0.75$ , which is less than 1. The image is smaller than the object, as $6 < 8$ shows.

He divided the wrong way round: $k = 0.75$ , a reduction.

Question 11

(a) Two quadrilaterals are similar with scale factor $2.5$ from the smaller to the larger. The perimeter of the smaller is 26 cm. Find the perimeter of the larger. (b) Two similar triangles have perimeters 36 cm and 54 cm. The shortest side of the smaller triangle is 8 cm. Find the shortest side of the larger.

ShowHide solution

(a) Perimeter of the larger $= 2.5 \times 26 = 65$ cm.

(b) $k = \frac{54}{36} = 1.5$ , so the shortest side of the larger is $1.5 \times 8 = 12$ cm.

(a) $65$ cm (b) $12$ cm

Question 12

A figure is enlarged with scale factor 3, and the image is then reduced with scale factor $\frac{1}{2}$ . A side of the first figure is 4 cm. (a) Find this side in the final figure. (b) What single scale factor takes the first figure straight to the final one?

ShowHide solution

(a) After the enlargement: $4 \times 3 = 12$ cm. After the reduction: $12 \times \frac{1}{2} = 6$ cm.

(b) $k = \frac{6}{4} = 1.5$ . This is $3 \times \frac{1}{2}$ : one change after another multiplies the scale factors.

(a) $6$ cm (b) $1.5$

Key Points

An enlargement or reduction with scale factor $k$ multiplies every length by $k$ , and the image is similar to the object.
$k = \frac{\text{image length}}{\text{object length}}\text{.}$
$k > 1$ : enlargement; $0 < k < 1$ : reduction; $k = 1$ : the same size (congruent).
Angles do not change in an enlargement or a reduction.
On a grid, multiply every step across and every step up or down by $k$ , including the two steps of a slanting side.
A map or model with scale $1 : n$ is a reduction with $k = \frac{1}{n}$ .
The perimeter of the image is $k$ times the perimeter of the object: the perimeters are in the ratio $1 : k$ , like every other pair of corresponding lengths.