Maths · 5 Geometry

5.3 Similarity

Before you start

Angle facts: angles in a triangle add up to $180 °$ , and vertically opposite angles are equal.
Alternate and corresponding angles on parallel lines are equal.
Writing a ratio as a fraction, and multiplying or dividing by a fraction such as $\frac{5}{3}$ .
Squares, cubes, square roots and cube roots, with and without a calculator.
Area of a rectangle and a triangle, volume of a cuboid, and the units cm² and cm³.

Syllabus objectives covered (3)

Calculate lengths of similar shapes.
Use the relationships between lengths and areas of similar shapes and lengths, surface areas and volumes of similar solids.
Solve problems and give simple explanations involving similarity.

Part 1 of 7

What similar means

Two shapes are similar when one is an enlargement of the other. They have exactly the same shape, but they can be different sizes. Questions say mathematically similar to make clear that they mean this exact idea, not just 'looks alike'.
Similar means two things are true at once. Every angle in one shape equals the matching angle in the other. And every length in one shape is multiplied by the same number to give the matching length in the other.
The matching parts are called corresponding parts: the side in the same position, the angle at the matching vertex. The number that every length is multiplied by is the scale factor, written $k$ .
For a shape with four or more sides you must check both conditions. A $4$ cm by $4$ cm square and a $4$ cm by $8$ cm rectangle have all their angles equal to $90 °$ , but one pair of sides is multiplied by $1$ and the other pair by $2$ , so they are not similar. A square and a rhombus with all sides $4$ cm have every side in the ratio $1 : 1$ , but the rhombus's angles are not $90 °$ , so they are not similar either.
graph diagram — not drawn yet
Three shapes side by side with every side labelled: a 4 by 4 square, a 4 by 8 rectangle, and a rhombus with all sides 4 and angles 60° and 120°, each with its angles marked.
Equal angles alone, or equal side ratios alone, do not make a four-sided shape similar to another.
To explain that two shapes are not similar, calculate the scale factor from two different pairs of corresponding sides and show that the factors are different. A $4$ cm by $6$ cm rectangle and a $6$ cm by $8$ cm rectangle give $6 \div 4 = 1.5$ but $8 \div 6 = \frac{4}{3}$ . The factors differ, so the rectangles are not similar.
That example hides a trap. It adds $2$ cm to every side, and adding the same amount to every side does not keep the shape. Similar shapes come from multiplying every length by the same number, never from adding.
Some shapes are always similar to each other, because their shape is fixed and only their size can change: any two squares, any two circles, any two equilateral triangles. Two rectangles, or two isosceles triangles, can be similar or not. You have to check.
If the scale factor is $1$ , the two shapes are the same size as well as the same shape. They are congruent.

Part 2 of 7

Using the scale factor to find a missing length

The scale factor from shape A to shape B is a length on B divided by the corresponding length on A.
$k = \frac{length on B}{corresponding length on A}$
To find $k$ you need one pair of corresponding sides where you know both lengths. Find that pair first. It is the only place $k$ can come from.
Then multiply by $k$ to go from A to B, and divide by $k$ to go from B back to A. For example, a triangle has sides $4$ cm, $6$ cm and $7$ cm, and a similar triangle has shortest side $10$ cm. The shortest sides correspond, so $k = 10 \div 4 = 2.5$ . The other sides of the new triangle are $6 \times 2.5 = 15$ cm and $7 \times 2.5 = 17.5$ cm.
Check the direction every time. A side on the bigger shape must come out bigger than its partner on the smaller shape. If your answer for the bigger shape is smaller, you divided when you needed to multiply.
If $k$ is not exact, such as $7 \div 3 = 2.333...$ , keep it as the fraction $\frac{7}{3}$ or keep the full value in your calculator. If you round $k$ to $2.33$ and then multiply, the final answer is slightly wrong.
You can write the same idea as an equation of two ratios. Each side of the equation is 'new length over old length' for one pair of corresponding sides. Then solve for the unknown.
$\frac{x}{7} = \frac{10}{4} \Rightarrow x = 7 \times \frac{10}{4} = 17.5$
Angles do not change. Corresponding angles in similar shapes are equal, so never multiply an angle by $k$ .

Part 3 of 7

Triangles inside triangles, and triangles that cross

Two similar triangles can share part of one diagram, and then you have to pull them apart. The first pattern: in triangle $A B C$ , $D$ is on $A B$ , $E$ is on $A C$ , and $D E$ is parallel to $B C$ . The small triangle $A D E$ sits in the corner of the big triangle $A B C$ , and the two triangles are similar.
graph diagram — not drawn yet
Triangle ABC with A at the top, B at bottom left and C at bottom right. D on AB and E on AC, with DE horizontal and parallel to BC, shown by matching arrow marks on DE and BC. Angles ADE and ABC marked with the same single arc, angles AED and ACB with the same double arc. AD and DB labelled as separate lengths, with a bracket beside them showing the whole side AB.
Triangle ADE is the corner of triangle ABC. Its side AD matches the whole of AB, not DB.
The trap is the side of the big triangle. That side is the whole of $A B$ , which is $A D + D B$ . The length $D B$ is not a side of either triangle. If $A D = 6$ cm and $D B = 4$ cm, then $A B = 10$ cm and the scale factor from small to big is $\frac{10}{6} = \frac{5}{3}$ , not $\frac{4}{6}$ .
Redraw the two triangles side by side, the same way round, and write every known length on them. It takes a few seconds and it stops you pairing the wrong sides.
The second pattern: two straight lines cross at a point $X$ , and their ends are joined by two parallel lines. This makes two triangles that meet at $X$ , and one of them is upside down compared with the other.
graph diagram — not drawn yet
Two straight lines PS and QR crossing at X, with P at top left, Q at top right, R at bottom left and S at bottom right. PQ and RS are horizontal, with matching arrow marks to show they are parallel, and PQ is shorter than RS. The equal alternate angles at P and S are marked with the same arc, and the vertically opposite angles at X are marked as equal.
The bottom triangle is the top one enlarged and turned upside down: P matches S and Q matches R.
Here you cannot match sides by their position on the page. Match them by equal angles. The side opposite an angle in one triangle corresponds to the side opposite the equal angle in the other triangle.
Write the similarity with the vertices in matching order, and the order does the matching for you. 'Triangle $P Q X$ is similar to triangle $S R X$ ' means $P$ matches $S$ , $Q$ matches $R$ and $X$ matches $X$ . So $P Q$ corresponds to $S R$ , $Q X$ to $R X$ , and $P X$ to $S X$ .

Part 4 of 7

Showing that two triangles are similar

Triangles have a shortcut. If two angles of one triangle equal two angles of another, the third angles must be equal too, because the angles in each triangle add up to $180 °$ . Triangles with equal angles are similar, so their sides are then in the same ratio without any more checking.
So to show that two triangles are similar, find two pairs of equal angles and give a geometric reason for each pair.
These are the reasons you need. A common angle: one angle that belongs to both triangles. Vertically opposite angles are equal. Alternate angles are equal, because of parallel lines. Corresponding angles are equal, because of parallel lines.
Every statement needs its reason, and a reason about parallel lines names the lines. 'Angle $A D E$ = angle $A B C$ (corresponding angles, $D E$ parallel to $B C$ )' is complete. 'Angle $A D E$ = angle $A B C$ ' with no reason shows nothing, because it does not say why the angles are equal.
Finish with a conclusion that names both triangles in matching order: 'so triangle $A D E$ is similar to triangle $A B C$ '. The order tells the reader which sides correspond.
A second test uses sides instead of angles. If all three pairs of corresponding sides are in the same ratio, the triangles are similar. Sides $3$ , $4$ , $5$ cm and $6$ , $8$ , $10$ cm give $6 \div 3 = 8 \div 4 = 10 \div 5 = 2$ , so the two triangles are similar.
These shortcuts work for triangles only. For a quadrilateral, equal angles are not enough: the square and the rectangle earlier had equal angles and were not similar.

Part 5 of 7

Areas: multiply by the scale factor squared

Suppose every length is multiplied by $k$ . A rectangle $a$ by $b$ becomes $k a$ by $k b$ . Its area changes from $a b$ to $k a \times k b$ , so the area is multiplied by $k^{2}$ .
$k a \times k b = k^{2} \times a b$
This works for every shape, not only rectangles. An area always comes from one length multiplied by another length, and each of the two lengths carries a factor of $k$ .
Picture it. A $1$ cm square enlarged by scale factor $2$ becomes a $2$ cm square, and the $2$ cm square holds four of the original squares, not two.
graph diagram — not drawn yet
Left: a 1 cm square next to a 2 cm square divided by grid lines into four 1 cm squares. Right: a 1 cm cube next to a 2 cm cube divided into eight 1 cm cubes, both cubes drawn in perspective. Labels: 'lengths × 2', 'area × 4', 'volume × 8'.
Doubling the lengths makes the area 4 times as big and the volume 8 times as big.
So for similar shapes the area scale factor is $k^{2}$ . Lengths $\times 3$ means areas $\times 9$ . Lengths $\times \frac{1}{2}$ means areas $\times \frac{1}{4}$ .
As a ratio of two similar shapes A and B:
$\frac{Area of A}{Area of B} = {(\frac{Length of A}{Length of B})}^{2}$
Every area of a similar solid follows the same rule: the total surface area, the area of one face, a cross-section, a label. But perimeters and circumferences are lengths, so they use $k$ , not $k^{2}$ .
These scale-factor rules are not on the List of formulas printed in Papers 2 and 4, so learn them.

Part 6 of 7

Volumes: multiply by the scale factor cubed

A volume comes from three lengths multiplied together. A cuboid $a$ by $b$ by $c$ becomes $k a$ by $k b$ by $k c$ , so its volume is multiplied by $k \times k \times k = k^{3}$ .
$k a \times k b \times k c = k^{3} \times a b c$
In the picture of cubes, a $1$ cm cube enlarged by scale factor $2$ becomes a $2$ cm cube, which holds $2 \times 2 \times 2 = 8$ of the original cubes.
So the volume scale factor is $k^{3}$ . This is the syllabus relationship:
$\frac{Volume of A}{Volume of B} = \frac{{(Length of A)}^{3}}{{(Length of B)}^{3}}$
Capacity is a volume, so the amount a container holds also scales by $k^{3}$ . For two solid objects made of the same material, the mass is proportional to the volume, so the mass also scales by $k^{3}$ .
Keep the three factors apart. This one line holds the whole idea:
$lengths \times k areas \times k^{2} volumes \times k^{3}$
In numbers: two similar cylinders, the larger one twice as tall. Its radius is also twice as big, its surface area is $4$ times as big, and its volume is $8$ times as big.

Part 7 of 7

Working backwards from an area or a volume

Sometimes the question gives two areas or two volumes and asks for a length. Undo the power. From areas, take the square root of the area ratio to get $k$ . From volumes, take the cube root of the volume ratio.
$k = \sqrt{\frac{Area of B}{Area of A}} k = \sqrt[3]{\frac{Volume of B}{Volume of A}}$
Example: two similar triangles have areas $20$ cm² and $45$ cm². The area factor is $45 \div 20 = 2.25$ , so $k = \sqrt{2.25} = 1.5$ . A side of $8$ cm on the smaller triangle matches a side of $8 \times 1.5 = 12$ cm on the larger one.
To get from an area to a volume, or the other way, always go through $k$ . An area factor of $\frac{9}{4}$ gives $k = \frac{3}{2}$ , so the volume factor is ${(\frac{3}{2})}^{3} = \frac{27}{8}$ . Multiplying a volume by $\frac{9}{4}$ would be using an area factor on a volume.
$area factor \frac{9}{4} \to k = \frac{3}{2} \to volume factor \frac{27}{8}$
When the root is not exact, such as $\sqrt[3]{3} = 1.442...$ , keep the full calculator value for the next step. Round only the final answer, to 3 significant figures.
Decide which way you are going before you calculate. If you are finding something on the smaller shape, the answer must be smaller, so divide by the factor.
The same idea gives you simple explanations in words. A similar container that holds $8$ times as much is only $2$ times as tall, because capacity scales by $k^{3}$ and $\sqrt[3]{8} = 2$ .

Worked examples

Example 1. Quadrilaterals $P Q R S$ and $W X Y Z$ are mathematically similar, with $P$ corresponding to $W$ , $Q$ to $X$ , $R$ to $Y$ and $S$ to $Z$ . $P Q = 4$ cm, $Q R = 7$ cm, angle $P Q R = 110 °$ , $W X = 10$ cm and $Y Z = 15$ cm. Find (a) $X Y$ , (b) $R S$ , (c) angle $W X Y$ .

Show the working

Find a pair of corresponding sides where you know both lengths. $P Q$ corresponds to $W X$ (the first two letters of each name), and both are given: $4$ cm and $10$ cm.
The scale factor from $P Q R S$ to $W X Y Z$ :
$k = \frac{10}{4} = 2.5$
(a) $X Y$ corresponds to $Q R$ . You are going from the small shape to the big one, so multiply.
$X Y = 7 \times 2.5 = 17.5$
(b) $R S$ corresponds to $Y Z$ . This time you go from the big shape back to the small one, so divide.
$R S = 15 \div 2.5 = 6$
Check: $17.5$ is bigger than $7$ , and $6$ is smaller than $15$ , so each answer is on the correct side of its partner.
(c) Angle $W X Y$ corresponds to angle $P Q R$ . Angles do not change in similar shapes.

Answer: (a) $X Y = 17.5$ cm (b) $R S = 6$ cm (c) angle $W X Y = 110 °$

Example 2. In triangle $A B C$ , $D$ is on $A B$ and $E$ is on $A C$ , with $D E$ parallel to $B C$ . $A D = 6$ cm, $D B = 4$ cm, $D E = 9$ cm and $A E = 7.2$ cm. (a) Show that triangle $A D E$ is similar to triangle $A B C$ . (b) Find $B C$ . (c) Find $E C$ .

Show the working

(a) Angle $A$ is in both triangles: angle $D A E$ = angle $B A C$ (common angle).
Angle $A D E$ = angle $A B C$ (corresponding angles, $D E$ parallel to $B C$ ).
Two pairs of angles are equal, so the third pair is equal too, because the angles in each triangle add up to $180 °$ . So triangle $A D E$ is similar to triangle $A B C$ .
(b) The side of the big triangle is the whole of $A B$ : $A D + D B = 6 + 4 = 10$ cm. $A D$ in the small triangle corresponds to $A B$ in the big one.
$k = \frac{10}{6} = \frac{5}{3}$
$B C$ corresponds to $D E$ . You go from small to big, so multiply.
$B C = 9 \times \frac{5}{3} = 15$
(c) $E C$ is not a side of either triangle. Find the whole side $A C$ first, then subtract $A E$ .
$A C = 7.2 \times \frac{5}{3} = 12, E C = 12 - 7.2 = 4.8$

Answer: (a) Angle $A$ is common and angle $A D E$ = angle $A B C$ (corresponding angles), so the triangles are similar. (b) $B C = 15$ cm (c) $E C = 4.8$ cm

Example 3. Two solid candles are mathematically similar. The smaller candle is $12$ cm tall and the larger candle is $18$ cm tall. The surface area of the smaller candle is $320$ cm². The volume of the larger candle is $1350$ cm³. Find (a) the surface area of the larger candle, (b) the volume of the smaller candle.

Show the working

The heights are corresponding lengths. The scale factor from small to large:
$k = \frac{18}{12} = 1.5$
(a) Surface area is an area, so use $k^{2}$ . Small to large, so multiply.
$320 \times {1.5}^{2} = 320 \times 2.25 = 720$
(b) Volume uses $k^{3}$ . You are going from large to small, so divide.
$1350 \div {1.5}^{3} = 1350 \div 3.375 = 400$
Check the sizes: the larger candle has the larger area, and the smaller candle has the smaller volume.

Answer: (a) $720$ cm² (b) $400$ cm³

Example 4. Two solid ornaments are mathematically similar. Their volumes are $135$ cm³ and $625$ cm³. The larger ornament is $20$ cm tall. The surface area of the smaller ornament is $180$ cm². Find (a) the height of the smaller ornament, (b) the surface area of the larger ornament.

Show the working

You are given volumes, so start with the volume factor from small to large, and simplify it.
$\frac{625}{135} = \frac{125}{27}$
Take the cube root to get the length factor.
$k = \sqrt[3]{\frac{125}{27}} = \frac{5}{3}$
(a) For the height of the smaller ornament you go from large to small, so divide by $k$ .
$20 \div \frac{5}{3} = 20 \times \frac{3}{5} = 12$
(b) Surface area uses $k^{2}$ , and you go from small to large, so multiply.
$180 \times {(\frac{5}{3})}^{2} = 180 \times \frac{25}{9} = 500$
Notice that you never used $\frac{125}{27}$ on the area. The volume factor only gets you to $k$ ; the area needs $k^{2}$ .

Answer: (a) $12$ cm (b) $500$ cm²

Where people go wrong

Mistake: Doubling the lengths of a shape and then doubling its area or volume too.

Instead: Multiply areas by $k^{2}$ and volumes by $k^{3}$ . Doubling the lengths makes areas $4$ times and volumes $8$ times as big.

An area is a length times a length, so it picks up the factor $k$ twice. A volume picks it up three times.

Mistake: In a triangle with a parallel line inside it, using $D B$ as the side of the big triangle.

Instead: Use the whole side, $A B = A D + D B$ .

$D B$ is not a side of either triangle. The side of the big triangle runs from $A$ all the way to $B$ .

Mistake: Pairing sides by where they sit on the page when one triangle is upside down.

Instead: Pair sides by equal angles, or read the pairs from the letter order in 'triangle $P Q X$ is similar to triangle $S R X$ '.

In a crossing-lines diagram one triangle is turned round, so the top side of one triangle matches the bottom side of the other.

Mistake: Thinking that adding the same length to every side gives a similar shape.

Instead: Similar shapes come from multiplying every length by the same scale factor.

Adding $2$ cm to each side of a $4$ by $6$ rectangle gives $6$ by $8$ . The factors are $1.5$ and $\frac{4}{3}$ , which differ, so the shape has changed.

Mistake: Going straight from an area factor to a volume factor by using the same number.

Instead: Take the square root of the area factor to get $k$ , then cube $k$ .

An area factor of $4$ means $k = 2$ , so the volume factor is $2^{3} = 8$ , not $4$ .

Check yourself

Explain why a $3$ cm by $5$ cm rectangle and a $6$ cm by $8$ cm rectangle are not mathematically similar.
Answer
$6 \div 3 = 2$ but $8 \div 5 = 1.6$ . The two scale factors are different, so the lengths are not all multiplied by the same number.
A triangle has sides $5$ cm, $7$ cm and $9$ cm. A similar triangle has shortest side $8$ cm. Find its longest side.
Answer
$k = 8 \div 5 = 1.6$ , so the longest side is $9 \times 1.6 = 14.4$ cm.
Lines $P S$ and $Q R$ cross at $X$ , and $P Q$ is parallel to $R S$ . (a) Give two pairs of equal angles, with reasons, that show triangles $P Q X$ and $S R X$ are similar. (b) $P Q = 6$ cm, $R S = 15$ cm and $Q X = 4$ cm. Find $R X$ .
Answer
(a) Angle $P X Q$ = angle $S X R$ (vertically opposite angles); angle $Q P X$ = angle $R S X$ (alternate angles, $P Q$ parallel to $R S$ ). (b) $k = 15 \div 6 = 2.5$ , and $Q X$ corresponds to $R X$ , so $R X = 4 \times 2.5 = 10$ cm.
The lengths of two similar shapes are in the ratio $2 : 5$ . The area of the smaller shape is $12$ cm². Find the area of the larger shape.
Answer
$k = \frac{5}{2} = 2.5$ , so the area factor is ${2.5}^{2} = 6.25$ , and the area is $12 \times 6.25 = 75$ cm².
Two similar solids have scale factor $3$ . The smaller solid has volume $20$ cm³. Find the volume of the larger solid.
Answer
Volume factor $3^{3} = 27$ , so the volume is $20 \times 27 = 540$ cm³.
Two similar triangles have areas $18$ cm² and $50$ cm². The base of the smaller triangle is $6$ cm. Find the base of the larger triangle.
Answer
Area factor $\frac{50}{18} = \frac{25}{9}$ , so $k = \sqrt{\frac{25}{9}} = \frac{5}{3}$ , and the base is $6 \times \frac{5}{3} = 10$ cm.
A model car is made to a scale of $1 : 20$ . The windscreen of the real car has area $1.2$ m². Find the area of the model's windscreen in cm².
Answer
The length factor is $\frac{1}{20}$ , so the area factor is $\frac{1}{400}$ . $1.2$ m² $= 12000$ cm², and $12000 \div 400 = 30$ cm².
Two similar solids have surface areas $100$ cm² and $225$ cm². The smaller solid has volume $40$ cm³. Find the volume of the larger solid.
Answer
Area factor $225 \div 100 = 2.25$ , so $k = \sqrt{2.25} = 1.5$ . Volume factor ${1.5}^{3} = 3.375$ , so the volume is $40 \times 3.375 = 135$ cm³.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. Do not use a calculator in this question. In triangle $A B C$ , $D$ is a point on $A B$ and $E$ is a point on $A C$ . $D E$ is parallel to $B C$ . $A D = 8$ cm, $D B = 2$ cm, $D E = 6$ cm and $E C = 2.5$ cm.

(a) Show that triangle $A D E$ is similar to triangle $A B C$ . Give a reason for each statement you make. 2 marks No calculator

Mark scheme

B1 one correct pair of equal angles with its reason, from: angle A (angle DAE = angle BAC) common to both triangles; angle ADE = angle ABC, corresponding angles [DE parallel to BC]; angle AED = angle ACB, corresponding angles [DE parallel to BC] oe
B1 a second, different correct pair of equal angles from that list, with its reason oe

(b) Calculate the length of $B C$ . 2 marks No calculator

Mark scheme

M1 6 × 10/8 oe, using AB = 10
A1 7.5 [cm]

(c) Calculate the length of $A E$ . 2 marks No calculator

Mark scheme

M1 AE/(AE + 2.5) = 8/10 oe, or 2.5 × 8/2 oe
A1 10 [cm]

(d) The area of triangle $A D E$ is $24$ cm². Calculate the area of the trapezium $D B C E$ . 3 marks No calculator

Mark scheme

M1 area factor (10/8)² or 25/16 or 1.5625 soi
M1 dep 24 × 25/16 − 24 oe [= 37.5 − 24]
A1 13.5 [cm²]

Question 2. Two bottles are mathematically similar. The smaller bottle holds $500$ cm³ and the larger bottle holds $1500$ cm³. The smaller bottle is $18$ cm tall.

(a) Calculate the height of the larger bottle. 3 marks

Mark scheme

M1 volume factor 1500 ÷ 500 [= 3] soi
M1 18 × ∛3 oe
A1 26.0 [cm] awrt

(b) The label on the smaller bottle has area $80$ cm². The label on the larger bottle is mathematically similar to it. Calculate the area of the label on the larger bottle. 2 marks

Mark scheme

M1 80 × (∛3)² oe, FT their length factor from (a)
A1 166 [cm²] awrt

(c) The larger bottle holds $3$ times as much as the smaller bottle. Explain why it is not $3$ times as tall. 1 mark

Mark scheme

B1 capacity (volume) is multiplied by the cube of the length factor, so the height is multiplied by ∛3 [≈ 1.44], not 3 oe

Question 3. Rectangle $A$ measures $12$ cm by $8$ cm. Rectangle $B$ measures $18$ cm by $10$ cm.

(a) Explain why rectangle $A$ and rectangle $B$ are not mathematically similar. 2 marks No calculator

Mark scheme

B1 18 ÷ 12 = 1.5 and 10 ÷ 8 = 1.25 oe, or 12 ÷ 8 = 1.5 and 18 ÷ 10 = 1.8 oe
B1 dep: the two values are different, so the sides are not all multiplied by the same scale factor oe

(b) Rectangle $C$ is mathematically similar to rectangle $A$ . The longer side of rectangle $C$ is $15$ cm. Find the length of the shorter side of rectangle $C$ . 2 marks No calculator

Mark scheme

M1 15 ÷ 12 [= 1.25] used as scale factor, or 8 × 15/12 oe
A1 10 [cm]

(c) Write down the ratio area of rectangle $A$ : area of rectangle $C$ in its simplest form. 1 mark No calculator

Mark scheme

B1 16 : 25 cao

How this is examined

Working back from two volumes (or two areas) to a length, then on to an area: A Calculate part about two similar containers or solids, with both volumes or capacities given and a height or a surface area to find The ratio of the two given volumes, then its cube root as the length factor (for 500 cm³ and 1500 cm³, $\sqrt[3]{3}$ ), then that factor applied to the length. For an area, the length factor squared. Using the volume ratio directly on a length is not a valid method.
Finding a missing length when one triangle sits inside another or two triangles cross: A Calculate part on a diagram where a line inside a triangle is parallel to one side, or where two lines cross between two parallel lines The scale factor from one pair of corresponding sides, using the whole side of the big triangle ( $A B = A D + D B$ ), is the method, e.g. $6 \times \frac{10}{8}$ . The length is the answer, and on the bigger triangle it must be bigger than its partner.
Turning a length scale factor into an area or volume scale factor: A Calculate part that gives two corresponding lengths and one area or one volume $k^{2}$ used for any area, including surface area, and $k^{3}$ for any volume, capacity or mass of the same material, shown in the working, e.g. $320 \times {1.5}^{2}$ . Then the value with the right unit, cm² or cm³.
Showing that two triangles are similar: A Show (that) part on a diagram with parallel lines or crossing lines Two pairs of equal angles, each with a named reason: 'angle $A$ is common to both triangles' and 'angle $A D E$ = angle $A B C$ , corresponding angles, $D E$ parallel to $B C$ '. Equal angles stated with no reasons do not show anything.
Explaining why two shapes are or are not similar: An Explain part about two rectangles, or two other shapes, with their side lengths given Two scale factors from corresponding sides calculated ( $18 \div 12 = 1.5$ and $10 \div 8 = 1.25$ ), then the statement that they are different, so the lengths are not all multiplied by the same number.

Words to know

Mathematically similar: Exactly the same shape: corresponding angles are equal and every length is multiplied by the same scale factor.
Corresponding: In the same position in two similar shapes, such as the angle at matching vertices or the side between them.
Scale factor: The number $k$ that every length is multiplied by to go from one similar shape to the other: $k = \frac{new length}{old length}$ .
Area scale factor: $k^{2}$ : the number every area, including surface area, is multiplied by when the lengths are multiplied by $k$ .
Volume scale factor: $k^{3}$ : the number every volume or capacity is multiplied by when the lengths are multiplied by $k$ .
Congruent: The same shape and the same size; similar with scale factor $1$ .
Common angle: One angle that belongs to two triangles at once, because they share it.

All Maths lessons Next: 5.4 Symmetry