Chapter 21 · Higher tier

GCSE Maths Transformations, Congruence and Similarity (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

Reflect a shape in the lines x = a, y = b, y = x and y = −x, and describe a reflection fully
Rotate a shape about any centre through 90°, 180° or 270° and describe a rotation fully
Translate a shape using a column vector and describe a translation
Enlarge a shape using a positive or fractional scale factor and a centre, and describe an enlargement
Describe any single transformation fully, and say what changes and what stays the same
Higher: use negative scale factors, combine transformations and find invariant points
Use the conditions SSS, SAS, ASA and RHS to show that triangles are congruent
Use scale factors to find missing lengths in similar shapes, and use k, k² and k³ for lengths, areas and volumes

You should already know

Coordinates in all four quadrants (Chapter 12)
Angle facts and parallel lines (Chapter 17)
Ratio and scale factors (Chapter 15)
Area and volume of simple shapes (Chapters 18 and 19)

Where this leads

Constructions, loci and vectors (Chapter 22)
Trigonometry and similar triangles (Chapter 20 links back to this chapter)
Graph transformations (Chapter 25)
Circle theorems and proof (Chapter 26)

Edexcel GCSE (9–1) Mathematics 1MA1, Higher tier. Content references: G5, G7, G8, G19, R12.

Printable worksheets. Every lesson in this chapter has a free worksheet and an answer sheet to download and print.

21.1 Reflection: Worksheet (PDF) · Answers (PDF) · lesson page
21.2 Rotation: Worksheet (PDF) · Answers (PDF) · lesson page
21.3 Translation: Worksheet (PDF) · Answers (PDF) · lesson page
21.4 Enlargement: Worksheet (PDF) · Answers (PDF) · lesson page
21.5 Describing and identifying single transformations: Worksheet (PDF) · Answers (PDF) · lesson page
21.6 Combinations of transformations and invariant points: Worksheet (PDF) · Answers (PDF) · lesson page
21.7 Congruent shapes and triangles: Worksheet (PDF) · Answers (PDF) · lesson page
21.8 Similar shapes and scale factors: Worksheet (PDF) · Answers (PDF) · lesson page
21.9 Area and volume scale factors: Worksheet (PDF) · Answers (PDF) · lesson page
Check-up and exam-style questions: Worksheet (PDF) · Answers (PDF) · lesson page

Exercise P Do you remember?

Non-calculator

FWrite down the coordinates of the points A and B.

Done

FTwo angles in a triangle are 48° and 67°. Work out the third angle.

Done

FWrite the ratio 12 : 30 in its simplest form.

Done

FA rectangle is 6 cm long and 4.5 cm wide. Work out its area.

Done

FWrite down the equation of the vertical line that passes through the point (3, 5).

Done

21.1

Reflection

Edexcel 1MA1: G7
Grade 1-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: ReflectionComing soon

A transformation changes a shape in some way. The shape we start with is called the object and the new shape is called the image. In this chapter you will meet four transformations: reflection, rotation, translation and enlargement. The first three only move the shape, so the image is exactly the same size and shape as the object. These shapes are called congruent. Enlargement changes the size, so the image is similar to the object but not congruent.

A reflection gives a mirror image. Imagine standing a mirror on the mirror line and looking at the shape in it. Each point of the image is the same distance from the mirror line as the matching point of the object, but on the opposite side. The line joining a point to its image crosses the mirror line at a right angle. That is why the mirror line is exactly half-way between a point and its image.

Shape B is the reflection of shape A in the line x = 4. Each dashed line is crossed in its middle by the mirror line.

On a coordinate grid the mirror line is usually one of four kinds of line:

A vertical line x = a number, such as x = 4. The y-axis is the line x = 0.
A horizontal line y = a number, such as y = −1. The x-axis is the line y = 0.
The diagonal y = x, which goes through (1, 1), (2, 2), (3, 3) and so on.
The other diagonal y = −x, which goes through (1, −1), (2, −2), (−1, 1) and so on.

For a vertical or horizontal mirror line, count the squares from each vertex to the line, then count the same number of squares on the other side. For the diagonals there is a shortcut, shown in the picture: a reflection in y = x swaps the coordinates, (a, b) → (b, a), and a reflection in y = −x swaps them and changes both signs, (a, b) → (−b, −a). The axes themselves are easy as well: reflecting in the x-axis changes the sign of y, and reflecting in the y-axis changes the sign of x.

The point P(3, 1) and its four reflections.

Describing a reflection. To describe a reflection fully you must say that it is a reflection and give the equation of the mirror line. To find the line, look at a point and its image: the line is exactly half-way between them. If you have tracing paper, you can also trace the object, flip the tracing paper over and slide it until it lands on the image, then draw along the fold.

Key ideas

A reflection flips a shape over a mirror line. The image is the same size and shape (congruent) but a mirror image.
Each point and its image are the same distance from the mirror line, on opposite sides.
Vertical line x = a: the y-coordinate stays the same. Horizontal line y = b: the x-coordinate stays the same.
In y = x: (a, b) → (b, a). In y = −x: (a, b) → (−b, −a).
To describe a reflection, give the equation of the mirror line.
A point on the mirror line is invariant: it does not move.

Worked example 21.1.1

Reflect shape A in the line x = 5. Write down the coordinates of the vertices of the image B. A has vertices at (1, 1), (4, 1), (1, 3).

Key words. A reflection flips a shape over a mirror line. The object and its image are the same distance from the mirror line, on opposite sides.

Method. To reflect a shape, reflect every vertex over the mirror line x = 5, then join the image points in the same order.

Draw the mirror line. The line x = 5 is a vertical line: every point on it has x = 5. Everything in the picture will flip over this line.
Draw a vertical line through x = 5.
Reflect each vertex of A in turn. For a reflection in a vertical line x = 5, the y-coordinate stays the same. Count how far each point is from the line (sideways), then go the same distance on the other side.
(1, 1): 4 left of the line, so the image is 4 right of the line. x = 5 + 4 = 9. Image (9, 1)
(4, 1): 1 left of the line, so the image is 1 right of the line. x = 5 + 1 = 6. Image (6, 1)
(1, 3): 4 left of the line, so the image is 4 right of the line. x = 5 + 4 = 9. Image (9, 3)
Plot the image points and join them up in the same order to make shape B. The image has the same size and shape as the object but is the other way round, like a mirror image.
B has vertices at (9, 1), (6, 1), (9, 3).

Check. The mirror line should be exactly half-way between a point and its image. The midpoint of (1, 1) and (9, 1) is (1 + 9) ÷ 2, (1 + 1) ÷ 2 = (5, 1), which lies on the line x = 5. ✓

Watch out. Measure the distance from the mirror line, not from the y-axis. The line is x = 5, so distances start from there.

Worked example 21.1.2

Reflect shape A in the line y = −1. Write down the coordinates of the vertices of the image B. A has vertices at (1, 2), (3, 2), (3, 3), (2, 3), (2, 5), (1, 5).

Key words. A reflection flips a shape over a mirror line. The object and its image are the same distance from the mirror line, on opposite sides.

Method. To reflect a shape, reflect every vertex over the mirror line y = −1, then join the image points in the same order.

Draw the mirror line. The line y = −1 is a horizontal line: every point on it has y = −1. Everything in the picture will flip over this line.
Draw a horizontal line through y = −1.
Reflect each vertex of A in turn. For a reflection in a horizontal line y = −1, the x-coordinate stays the same. Count how far each point is from the line (up or down), then go the same distance on the other side.
(1, 2): 3 above the line, so the image is 3 below the line. y = −1 − 3 = −4. Image (1, −4)
(3, 2): 3 above the line, so the image is 3 below the line. y = −1 − 3 = −4. Image (3, −4)
(3, 3): 4 above the line, so the image is 4 below the line. y = −1 − 4 = −5. Image (3, −5)
(2, 3): 4 above the line, so the image is 4 below the line. y = −1 − 4 = −5. Image (2, −5)
(2, 5): 6 above the line, so the image is 6 below the line. y = −1 − 6 = −7. Image (2, −7)
(1, 5): 6 above the line, so the image is 6 below the line. y = −1 − 6 = −7. Image (1, −7)
Plot the image points and join them up in the same order to make shape B. The image has the same size and shape as the object but is the other way round, like a mirror image.
B has vertices at (1, −4), (3, −4), (3, −5), (2, −5), (2, −7), (1, −7).

Check. The mirror line should be exactly half-way between a point and its image. The midpoint of (1, 2) and (1, −4) is (1 + 1) ÷ 2, (2 + −4) ÷ 2 = (1, −1), which lies on the line y = −1. ✓

Watch out. Measure the distance from the mirror line, not from the x-axis. The line is y = −1, so distances start from there.

Worked example 21.1.3

Reflect shape A in the line y = x. Write down the coordinates of the vertices of the image B. A has vertices at (3, 1), (5, 1), (6, 3), (3, 2).

Key words. A reflection flips a shape over a mirror line. The object and its image are the same distance from the mirror line, on opposite sides.

Method. To reflect a shape, reflect every vertex over the mirror line y = x, then join the image points in the same order.

Draw the mirror line. The line y = x is a diagonal line. Every point on it has the same x and y coordinates (for example (0, 0), (1, 1), (2, 2), (3, 3)).
Draw a straight line through (0, 0), (1, 1), (2, 2) and (3, 3).
Reflect each vertex of A in turn. A mirror along y = x swaps the two coordinates: the point (a, b) becomes (b, a). This works because the diagonal is the line where the x and y coordinates are equal.
(3, 1) → (1, 3)
(5, 1) → (1, 5)
(6, 3) → (3, 6)
(3, 2) → (2, 3)
Plot the image points and join them up in the same order to make shape B. The image has the same size and shape as the object but is the other way round, like a mirror image.
B has vertices at (1, 3), (1, 5), (3, 6), (2, 3).

Check. The mirror line should be exactly half-way between a point and its image. The midpoint of (3, 1) and (1, 3) is (3 + 1) ÷ 2, (1 + 3) ÷ 2 = (2, 2), which lies on the line y = x. ✓

Watch out. Reflecting in y = x swaps the coordinates; it does not change any signs. (Changing signs is for y = −x.)

Worked example 21.1.4

Reflect shape A in the line y = −x. Write down the coordinates of the vertices of the image B. A has vertices at (−4, 1), (−1, 2), (−3, 4).

Key words. A reflection flips a shape over a mirror line. The object and its image are the same distance from the mirror line, on opposite sides.

Method. To reflect a shape, reflect every vertex over the mirror line y = −x, then join the image points in the same order.

Draw the mirror line. The line y = −x is a diagonal line sloping down to the right. Every point on it has y equal to the opposite of x (for example (−2, 2), (−1, 1), (0, 0), (1, −1), (2, −2)).
Draw a straight line through (−2, 2), (0, 0) and (2, −2).
Reflect each vertex of A in turn. A mirror along y = −x swaps the coordinates and changes both signs: the point (a, b) becomes (−b, −a). For example (2, 1) goes to (−1, −2).
(−4, 1) → (−1, 4)
(−1, 2) → (−2, 1)
(−3, 4) → (−4, 3)
Plot the image points and join them up in the same order to make shape B. The image has the same size and shape as the object but is the other way round, like a mirror image.
B has vertices at (−1, 4), (−2, 1), (−4, 3).

Check. The mirror line should be exactly half-way between a point and its image. The midpoint of (−4, 1) and (−1, 4) is (−4 + −1) ÷ 2, (1 + 4) ÷ 2 = (−2.5, 2.5), which lies on the line y = −x. ✓

Watch out. Reflecting in y = −x swaps the coordinates and changes both signs. Swapping alone gives a reflection in y = x.

Worked example 21.1.5

Describe fully the single transformation that maps shape A onto shape B.

Key words. A reflection flips a shape over a mirror line. The object and its image are the same distance from the mirror line, on opposite sides.

Method. Decide which type of transformation maps A onto B, then find the extra details each type needs (a line, a centre, an angle, a vector or a scale factor).

Decide which type of transformation it is. A reflection flips the shape over. All other transformations of this type keep the shape facing the same way.
Compare a side: it is 3 long in A and 3 long in B.
The size is unchanged, but B is the mirror image of A: if you go round the corners in the same order, one shape is clockwise and the other is anticlockwise. So the transformation is a reflection.
Pick vertices and their images, and find the midpoints. A mirror line is exactly half-way between an object point and its image, so find the midpoint of a point and its image.
(1, 1) and (11, 1): midpoint = ((1 + 11) ÷ 2, (1 + 1) ÷ 2) = (6, 1)
(4, 1) and (8, 1): midpoint = ((4 + 8) ÷ 2, (1 + 1) ÷ 2) = (6, 1)
Write the equation of the mirror line. The midpoints all have x = 6. The points are in a vertical line, so the mirror line is the vertical line x = 6.
Mirror line: x = 6
Write the full description. We have found every piece of information needed for this transformation.
Reflection in the line x = 6

Check. Reflect (4, 1) in x = 6 and we get (8, 1), which matches (8, 1) in B. ✓

Watch out. A reflection needs the equation of the mirror line. Writing just “reflection” or “reflect in the x-axis” for a line like x = 3 is wrong.

Worked example 21.1.6

Describe fully the single transformation that maps shape A onto shape B.

Key words. A reflection flips a shape over a mirror line. The object and its image are the same distance from the mirror line, on opposite sides.

Method. Decide which type of transformation maps A onto B, then find the extra details each type needs (a line, a centre, an angle, a vector or a scale factor).

Decide which type of transformation it is. A reflection flips the shape over. All other transformations of this type keep the shape facing the same way.
Compare the areas: 4 square units for A and 4 for B.
The size is unchanged, but B is the mirror image of A: if you go round the corners in the same order, one shape is clockwise and the other is anticlockwise. So the transformation is a reflection.
Compare the coordinates of a vertex and its image. For the diagonal mirror lines y = x and y = −x the coordinates change in a special way.
The coordinates have been swapped: (2, 1) became (1, 2).
Use the midpoint as a check. The mirror line must pass through the midpoint of a point and its image.
Midpoint of (2, 1) and (1, 2) = (1.5, 1.5), which lies on y = x.
Write the equation of the mirror line. A reflection that swaps x and y is a reflection in the line y = x.
Mirror line: y = x
Write the full description. We have found every piece of information needed for this transformation.
Reflection in the line y = x

Check. Reflect (5, 2) in y = x and we get (2, 5), which matches (2, 5) in B. ✓

Watch out. A reflection needs the equation of the mirror line. Writing just “reflection” or “reflect in the x-axis” for a line like x = 3 is wrong.

Watch out

Always measure from the mirror line, not from the axes (unless the axis is the mirror line). For a line like x = 4, a point at x = 1 is 3 squares from the line, so its image is at x = 4 + 3 = 7. Also, in y = −x do not forget to change both signs as well as swapping.

21.2

Rotation

Edexcel 1MA1: G7
Grade 2-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: RotationComing soon

A rotation turns a shape about a fixed point called the centre of rotation. To describe a rotation fully you must give three things: the angle of the turn, the direction (clockwise or anticlockwise) and the centre. Clockwise is the way the hands of a clock move; anticlockwise is the opposite way. A quarter turn is 90°, a half turn is 180° and a three-quarter turn is 270°. A turn of 270° anticlockwise ends up in the same place as a turn of 90° clockwise. A half turn is the same whichever way you go, so for 180° you do not need to give a direction.

Shape B is shape A rotated 90° clockwise about the origin. The dashed lines show one vertex and its image: they are the same length and make a right angle at the centre C.

How to rotate a shape. The easiest way with tracing paper is to trace the shape, press a pencil point on the centre, and turn the paper. Without tracing paper, use the move from the centre method, which works for every vertex:

Write the move from the centre C to the vertex (across, then up or down). For example, if C is (1, 2) and the vertex is (4, 3), the move is 3 right and 1 up, which we write (3, 1).
Turn the move using the rule below.
Start at C again and follow the turned move. That gives the image of the vertex.

Turning the point P(3, 1) about the origin. A quarter turn clockwise gives (b, −a), a half turn gives (−a, −b), a quarter turn anticlockwise gives (−b, a).

The rules for a move (a, b) from the centre are:

90° clockwise: (a, b) → (b, −a)
180°: (a, b) → (−a, −b)
90° anticlockwise: (a, b) → (−b, a)

Why do they work? Take the move (3, 1): 3 right and 1 up. A quarter turn clockwise makes “right” become “down” and “up” become “right”, so the move becomes 1 right and 3 down, which is (1, −3). If the centre is the origin the move is the coordinates of the point, so you can use the rules directly.

Finding the centre. The centre of rotation is the only point that does not move. Look for it in this order: (1) is a vertex of the shape in the same place on both shapes? Then that vertex is the centre. (2) Is the turn 180°? Then the centre is the midpoint of a vertex and its image. (3) Try the origin. (4) In harder cases, the centre is where the perpendicular bisectors of two lines joining a point to its image cross (Higher; see the last worked example).

Key ideas

A rotation turns a shape about a fixed point, the centre. Give the angle, the direction and the centre.
The image is congruent to the object (same size and shape).
Each point and its image are the same distance from the centre.
90° clockwise: (a, b) → (b, −a); 90° anticlockwise: (a, b) → (−b, a); 180°: (a, b) → (−a, −b). The move is measured from the centre.
270° anticlockwise = 90° clockwise. A 180° turn needs no direction.
The centre is the only invariant point of a rotation.

Worked example 21.2.1

Rotate shape A through 90° clockwise about the point C (0, 0). Write down the coordinates of the vertices of the image B. A has vertices at (1, 1), (4, 1), (1, 3).

Key words. A rotation turns a shape about a fixed point called the centre of rotation. We state the angle and the direction (clockwise or anticlockwise).

Method. To rotate a shape about C, write each vertex as a move from C, turn that move through the angle, and add it back on to C.

Mark the centre of rotation C on the grid. Every point of the shape turns around C, so we always measure from C. The centre itself never moves.
C = (0, 0)
Write each vertex as a move from C. A turn is easiest to do by thinking about the move (across, up) needed to get from C to the point.
(1, 1): (1, 1) means 1 right, 1 up
(4, 1): (4, 1) means 4 right, 1 up
(1, 3): (1, 3) means 1 right, 3 up
Turn each move by 90° clockwise. A quarter turn clockwise changes a move using the rule (a, b) → (b, −a). Distances from C stay the same; only the direction changes.
(1, 1) → (1, −1)
(4, 1) → (1, −4)
(1, 3) → (3, −1)
Add the turned move to C to find each image vertex. The turned move is measured from C, so we start at C again.
C + (1, −1) = (1, −1)
C + (1, −4) = (1, −4)
C + (3, −1) = (3, −1)
Plot the image points and join them up to make shape B. The image is the same size and shape as the object but turned.
B has vertices at (1, −1), (1, −4), (3, −1).

Check. After a quarter turn the line from C to (4, 1) and the line from C to its image (1, −4) are at right angles, so the "dot product" of the two moves is 0: (4 × 1) + (1 × −4) = 0. Both are also the same distance from C. ✓

Watch out. Turn about the centre C, not about the origin (unless C is the origin). And check the direction: clockwise is the way the hands of a clock move.

Worked example 21.2.2

Rotate shape A through 90° anticlockwise about the point C (1, 1). Write down the coordinates of the vertices of the image B. A has vertices at (1, 1), (4, 1), (3, 3), (1, 3).

Key words. A rotation turns a shape about a fixed point called the centre of rotation. We state the angle and the direction (clockwise or anticlockwise).

Method. To rotate a shape about C, write each vertex as a move from C, turn that move through the angle, and add it back on to C.

Mark the centre of rotation C on the grid. Every point of the shape turns around C, so we always measure from C. The centre itself never moves.
C = (1, 1)
Write each vertex as a move from C. A turn is easiest to do by thinking about the move (across, up) needed to get from C to the point.
(1, 1): (0, 0) means no move
(4, 1): (3, 0) means 3 right
(3, 3): (2, 2) means 2 right, 2 up
(1, 3): (0, 2) means 2 up
Turn each move by 90° anticlockwise. A quarter turn anticlockwise changes a move using the rule (a, b) → (−b, a). Distances from C stay the same; only the direction changes.
(0, 0) → (0, 0)
(3, 0) → (0, 3)
(2, 2) → (−2, 2)
(0, 2) → (−2, 0)
Add the turned move to C to find each image vertex. The turned move is measured from C, so we start at C again.
C + (0, 0) = (1, 1)
C + (0, 3) = (1, 4)
C + (−2, 2) = (−1, 3)
C + (−2, 0) = (−1, 1)
Plot the image points and join them up to make shape B. The image is the same size and shape as the object but turned.
B has vertices at (1, 1), (1, 4), (−1, 3), (−1, 1).

Check. After a quarter turn the line from C to (4, 1) and the line from C to its image (1, 4) are at right angles, so the "dot product" of the two moves is 0: (3 × 0) + (0 × 3) = 0. Both are also the same distance from C. ✓

Watch out. Turn about the centre C, not about the origin (unless C is the origin). And check the direction: clockwise is the way the hands of a clock move.

Worked example 21.2.3

Rotate shape A through 90° clockwise about the point C (3, −1). Write down the coordinates of the vertices of the image B. A has vertices at (1, 1), (4, 2), (2, 4).

Key words. A rotation turns a shape about a fixed point called the centre of rotation. We state the angle and the direction (clockwise or anticlockwise).

Method. To rotate a shape about C, write each vertex as a move from C, turn that move through the angle, and add it back on to C.

Mark the centre of rotation C on the grid. Every point of the shape turns around C, so we always measure from C. The centre itself never moves.
C = (3, −1)
Write each vertex as a move from C. A turn is easiest to do by thinking about the move (across, up) needed to get from C to the point.
(1, 1): (−2, 2) means 2 left, 2 up
(4, 2): (1, 3) means 1 right, 3 up
(2, 4): (−1, 5) means 1 left, 5 up
Turn each move by 90° clockwise. A quarter turn clockwise changes a move using the rule (a, b) → (b, −a). Distances from C stay the same; only the direction changes.
(−2, 2) → (2, 2)
(1, 3) → (3, −1)
(−1, 5) → (5, 1)
Add the turned move to C to find each image vertex. The turned move is measured from C, so we start at C again.
C + (2, 2) = (5, 1)
C + (3, −1) = (6, −2)
C + (5, 1) = (8, 0)
Plot the image points and join them up to make shape B. The image is the same size and shape as the object but turned.
B has vertices at (5, 1), (6, −2), (8, 0).

Check. After a quarter turn the line from C to (2, 4) and the line from C to its image (8, 0) are at right angles, so the "dot product" of the two moves is 0: (−1 × 5) + (5 × 1) = 0. Both are also the same distance from C. ✓

Watch out. Turn about the centre C, not about the origin (unless C is the origin). And check the direction: clockwise is the way the hands of a clock move.

Worked example 21.2.4

Describe fully the single transformation that maps shape A onto shape B.

Key words. A rotation turns a shape about a fixed point called the centre of rotation. We state the angle and the direction (clockwise or anticlockwise).

Method. Decide which type of transformation maps A onto B, then find the extra details each type needs (a line, a centre, an angle, a vector or a scale factor).

Decide which type of transformation it is. A rotation turns the shape. Compare with a translation, where all sides stay parallel to the originals.
Compare a side: it is 3 long in A and 3 long in B.
The size is unchanged, B is not flipped, but its sides are not parallel to the originals: the shape has turned. So the transformation is a rotation.
Find the centre of rotation. The centre is the same distance from a point and from its image. The origin is the first place to try. It works if the squared distances from the origin to the point and its image match.
Distance² of (1, 1) from (0, 0) = 1² + 1² = 2
Distance² of (−1, 1) from (0, 0) = −1² + 1² = 2
They match, and the same works for the other vertices, so the centre is (0, 0).
Find the angle and direction. Write the move from C to a vertex, and the move from C to its image. A quarter turn changes (a, b) in a special way: clockwise gives (b, −a) and anticlockwise gives (−b, a).
From C to (1, 1): (1, 1) (1 right, 1 up)
From C to (−1, 1): (−1, 1) (1 left, 1 up)
This matches (a, b) → (−b, a), so the turn is 90° anticlockwise.
Write the full description. We have found every piece of information needed for this transformation.
Rotation, 90° anticlockwise, centre (0, 0)

Check. Rotate (4, 1) about (0, 0) by 90° anticlockwise: we get (−1, 4), which matches the vertex (−1, 4) of B. ✓

Watch out. Give three things for a rotation: the angle, the direction (clockwise or anticlockwise) and the centre. A 180° turn does not need a direction.

Worked example 21.2.5

Describe fully the single transformation that maps shape A onto shape B.

Key words. A rotation turns a shape about a fixed point called the centre of rotation. We state the angle and the direction (clockwise or anticlockwise).

Method. Decide which type of transformation maps A onto B, then find the extra details each type needs (a line, a centre, an angle, a vector or a scale factor).

Decide which type of transformation it is. A rotation turns the shape. Compare with a translation, where all sides stay parallel to the originals.
Compare a side: it is 3 long in A and 3 long in B.
The size is unchanged, B is not flipped, but its sides are not parallel to the originals: the shape has turned. So the transformation is a rotation.
Find the centre of rotation. In a half turn, the centre is exactly half-way between a point and its image. Find the midpoint for two different vertices: they should be the same point.
(1, 1) and (7, 5): midpoint (4, 3)
(3, 1) and (5, 5): midpoint (4, 3)
Both midpoints are the same, so the centre is (4, 3).
Find the angle and direction. From C the moves to the vertex and to its image point in exactly opposite directions, so it is a half turn. For a half turn the direction does not matter.
From C to (1, 1): (−3, −2); from C to (7, 5): (3, 2). Both signs have changed, so the angle is 180°.
Write the full description. We have found every piece of information needed for this transformation.
Rotation, 180°, centre (4, 3)

Check. Rotate (3, 1) about (4, 3) by 180°: we get (5, 5), which matches the vertex (5, 5) of B. ✓

Watch out. Give three things for a rotation: the angle, the direction (clockwise or anticlockwise) and the centre. A 180° turn does not need a direction.

Worked example 21.2.6

Describe fully the single transformation that maps shape A onto shape B.

Key words. A rotation turns a shape about a fixed point called the centre of rotation. We state the angle and the direction (clockwise or anticlockwise).

Method. Decide which type of transformation maps A onto B, then find the extra details each type needs (a line, a centre, an angle, a vector or a scale factor).

Decide which type of transformation it is. A rotation turns the shape. Compare with a translation, where all sides stay parallel to the originals.
Compare a side: it is 3 long in A and 3 long in B.
The size is unchanged, B is not flipped, but its sides are not parallel to the originals: the shape has turned. So the transformation is a rotation.
Find the centre of rotation using perpendicular bisectors. The centre is the same distance from a point and from its image. So it lies on the perpendicular bisector of the line joining them (the line at right angles through the midpoint). The centre is where two such bisectors cross.
For (1, 1) → (2, 6): the points (x, y) that are equally far from both satisfy x + 5y = 19.
For (4, 1) → (2, 3): the points equally far from both satisfy x − y = 1.
Solving the two equations together gives x = 4, y = 3.
Check in both: 1(4) + 5(3) = 19 ✓; 1(4) + −1(3) = 1 ✓.
The centre is (4, 3).
Find the angle and direction. Write the move from C to a vertex, and the move from C to its image. A quarter turn changes (a, b) in a special way: clockwise gives (b, −a) and anticlockwise gives (−b, a).
From C to (1, 1): (−3, −2) (3 left, 2 down)
From C to (2, 6): (−2, 3) (2 left, 3 up)
This matches (a, b) → (b, −a), so the turn is 90° clockwise.
Write the full description. We have found every piece of information needed for this transformation.
Rotation, 90° clockwise, centre (4, 3)

Check. Rotate (4, 1) about (4, 3) by 90° clockwise: we get (2, 3), which matches the vertex (2, 3) of B. ✓

Watch out. Give three things for a rotation: the angle, the direction (clockwise or anticlockwise) and the centre. A 180° turn does not need a direction.

Watch out

Turn about the centre, which is not always the origin. If the centre is C(3, 2) and a vertex is (5, 3), the move to turn is (2, 1), not (5, 3). Also check clockwise or anticlockwise carefully: a quarter turn the wrong way puts the image on the opposite side of the centre.

21.3

Translation

Edexcel 1MA1: G7
Grade 2-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: TranslationComing soon

A translation slides a shape from one place to another. The shape is not turned, not flipped and not resized: every point moves the same distance in the same direction. We describe a translation with a column vector, which is two numbers written one above the other in brackets.

Shape B is shape A translated by 4
−2: 4 to the right and 2 down. Every vertex moves the same way.

In a column vector a
b:

the top number is the move across (the x-direction): positive means right, negative means left;
the bottom number is the move up or down (the y-direction): positive means up, negative means down.

So −3
2 means 3 left and 2 up. To translate a point, add the top number to its x-coordinate and the bottom number to its y-coordinate. For example, (5, 1) translated by −3
2 is (5 − 3, 1 + 2) = (2, 3).

Describing a translation. Choose any vertex and its image, then work out image − object for the x-coordinates (top number) and for the y-coordinates (bottom number). To undo a translation, use the opposite vector: the inverse of 4
−2 is −4
2.

Key ideas

A translation slides a shape. The image is congruent to the object and faces the same way.
A column vector a
b means a across (right if positive) and b up/down (up if positive).
Image point = object point + vector. Vector = image − object.
The vector that undoes a translation is the opposite vector (change both signs).
A translation has no invariant points.

Worked example 21.3.1

Translate shape A by the vector 4
3. Write down the coordinates of the vertices of the image B. A has vertices at (1, 1), (4, 1), (1, 3).

Key words. A translation slides a shape without turning or flipping it. A column vector gives the move: the top number is across (x), the bottom number is up or down (y).

Method. To translate a shape by a column vector, add the top number to every x-coordinate and the bottom number to every y-coordinate.

Read the column vector. The top number is the move across (x direction): positive means right, negative means left. The bottom number is the move up or down (y direction): positive means up, negative means down.
4
3 means 4 right, 3 up.
Add the vector to every vertex. Every point of the shape slides the same way, so we add the same numbers to every x and every y.
(1, 1) → (1 + 4, 1 + 3) = (5, 4)
(4, 1) → (4 + 4, 1 + 3) = (8, 4)
(1, 3) → (1 + 4, 3 + 3) = (5, 6)
Plot the image points and join them to make shape B. A translation does not turn, flip or resize the shape. The image looks exactly like the object, just in a new place.
B has vertices at (5, 4), (8, 4), (5, 6).

Check. Undo the translation: take 4
3 away from (5, 4) and we get back to (5 − 4, 4 − 3) = (1, 1). ✓

Watch out. The top number is across (x) and the bottom number is up/down (y). Do not swap them, and watch the signs: a negative top number means left.

Worked example 21.3.2

Translate shape A by the vector −5
−4. Write down the coordinates of the vertices of the image B. A has vertices at (5, 4), (7, 4), (7, 5), (6, 5), (6, 7), (5, 7).

Key words. A translation slides a shape without turning or flipping it. A column vector gives the move: the top number is across (x), the bottom number is up or down (y).

Method. To translate a shape by a column vector, add the top number to every x-coordinate and the bottom number to every y-coordinate.

Read the column vector. The top number is the move across (x direction): positive means right, negative means left. The bottom number is the move up or down (y direction): positive means up, negative means down.
−5
−4 means 5 left, 4 down.
Add the vector to every vertex. Every point of the shape slides the same way, so we add the same numbers to every x and every y.
(5, 4) → (5 − 5, 4 − 4) = (0, 0)
(7, 4) → (7 − 5, 4 − 4) = (2, 0)
(7, 5) → (7 − 5, 5 − 4) = (2, 1)
(6, 5) → (6 − 5, 5 − 4) = (1, 1)
(6, 7) → (6 − 5, 7 − 4) = (1, 3)
(5, 7) → (5 − 5, 7 − 4) = (0, 3)
Plot the image points and join them to make shape B. A translation does not turn, flip or resize the shape. The image looks exactly like the object, just in a new place.
B has vertices at (0, 0), (2, 0), (2, 1), (1, 1), (1, 3), (0, 3).

Check. Undo the translation: take −5
−4 away from (0, 0) and we get back to (0 + 5, 0 + 4) = (5, 4). ✓

Watch out. The top number is across (x) and the bottom number is up/down (y). Do not swap them, and watch the signs: a negative top number means left.

Worked example 21.3.3

Describe fully the single transformation that maps shape A onto shape B.

Key words. A translation slides a shape without turning or flipping it. A column vector gives the move: the top number is across (x), the bottom number is up or down (y).

Method. Decide which type of transformation maps A onto B, then find the extra details each type needs (a line, a centre, an angle, a vector or a scale factor).

Decide which type of transformation it is. A translation only slides the shape. If the shape had turned, its sides would no longer be parallel to the originals.
Compare a side: it is 3 long in A and 3 long in B.
The size is unchanged, B faces the same way as A, and every side is parallel to its original. So the transformation is a translation.
Pick a vertex and its image. Every point moves by the same amount, so one vertex is enough to find the move.
(1, 1) → (4, −3)
Find the move across and the move up or down. Subtract the x-coordinates, then the y-coordinates (image − object). Positive means right/up, negative means left/down.
Across: 4 − 1 = 3 (3 right)
Up/down: −3 − 1 = −4 (4 down)
Write the move as a column vector. The across move goes on top and the up/down move goes underneath.
Translation by 3
−4
Write the full description. We have found every piece of information needed for this transformation.
Translation by 3
−4

Check. Test another vertex: (4, 1) + 3
−4 = (7, −3), which is the matching vertex of B. ✓

Watch out. Write the vector as across on top, up/down underneath, and take image − object, not the other way round.

Worked example 21.3.4

Point P(2, 5) is translated to P′(−1, 1). Write down the column vector of the translation.

Key words. A translation slides a shape without turning or flipping it. A column vector gives the move: the top number is across (x), the bottom number is up or down (y).

Method. The vector is the image coordinates minus the object coordinates.

Find the move across. Image x minus object x. A negative answer means left.
−1 − 2 = −3 (3 left)
Find the move up or down. Image y minus object y. A negative answer means down.
1 − 5 = −4 (4 down)
Write the column vector. Across on the top, up/down underneath.
−3
−4

Check. Add the vector to P: (2 − 3, 5 − 4) = (−1, 1) = P′. ✓

Watch out. Take away the object from the image, not the other way round.

Worked example 21.3.5

A translation maps shape A to shape B using 3
2. B is then translated by −1
4 to give shape C. Write down the single vector that maps A to C.

Key words. A translation slides a shape without turning or flipping it. A column vector gives the move: the top number is across (x), the bottom number is up or down (y).

Method. Two translations one after the other are the same as one translation whose vector is the sum of the two vectors.

Add the top numbers. The total move across is the sum of the two moves across.
3 + (−1) = 2
Add the bottom numbers. The total move up is the sum of the two moves up.
2 + 4 = 6
Write the single vector.
2
6

Check. Start at (0, 0): move 3
2 to (3, 2), then −1
4 to (2, 6), which is the same as one move of 2
6. ✓

Watch out. You can add vectors in either order: translations are one of the few transformations where the order does not matter.

Watch out

Do not mix up the two numbers. The top number is across (x); the bottom number is up/down (y). A column vector with a negative top number moves the shape to the left, and a negative bottom number moves it down.

21.4

Enlargement

Edexcel 1MA1: G7
Grade 3-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: EnlargementComing soon

An enlargement changes the size of a shape but keeps its shape. Every length is multiplied by the same number, called the scale factor, and the angles do not change. The image is similar to the object (not congruent, unless the scale factor is 1). To describe an enlargement fully you give the scale factor and the centre of enlargement.

The centre is the fixed point that distances are measured from. Take any vertex, draw a straight line from the centre through the vertex, and extend it so that the distance from the centre is multiplied by the scale factor. For a scale factor of 2, the image vertex is twice as far from the centre as the object vertex, on the same line. On a grid it is easiest to count: if a vertex is 3 right and 1 up from the centre, then for a scale factor of 2 the image is 6 right and 2 up from the centre.

Shape B is shape A enlarged with scale factor 2 and centre C(0, 0). The dashed lines pass through each vertex and its image.

Fractional scale factors. A scale factor between 0 and 1, such as 12 or 13, makes the shape smaller. This is still called an enlargement. For a scale factor of 12, every distance from the centre is halved. If a vertex is 6 right and 4 up from the centre, its image is 3 right and 2 up from the centre. The image is closer to the centre than the object.

Shape B is shape A enlarged with scale factor 12 and centre (0, 0). B is half the size of A and is half as far from the centre.

Describing an enlargement. To find the scale factor, divide a length on the image by the matching length on the object. To find the centre, draw a straight line through each vertex of the object and its image; the lines cross at the centre. Remember that a fractional scale factor still makes an enlargement.

Negative scale factors (Higher). With a negative scale factor the image is on the opposite side of the centre, and turned upside down. The size of the number gives the stretch, and the minus sign sends each point through the centre to the other side. For a scale factor of −2 a vertex that is 3 right and 1 up from the centre goes to the point 6 left and 2 down from the centre.

Shape B is shape A enlarged with scale factor −2 and centre C(2, 2). B is on the opposite side of C and upside down.

Key ideas

An enlargement changes the size of a shape, not its shape. Angles stay the same.
To enlarge, multiply the distance from the centre of enlargement to each vertex by the scale factor.
Scale factor = length on the image ÷ matching length on the object.
A scale factor between 0 and 1 makes the shape smaller (still an enlargement).
Higher: a negative scale factor puts the image on the opposite side of the centre.
The centre of enlargement is the only invariant point.

Worked example 21.4.1

Enlarge shape A with scale factor 2 and centre C (0, 0). Write down the coordinates of the vertices of the image B. A has vertices at (1, 1), (4, 1), (1, 3).

Key words. An enlargement changes the size of a shape but not its shape. The scale factor says how many times bigger each length becomes. The centre of enlargement is the fixed point that distances are measured from.

Method. To enlarge a shape with scale factor 2 about C, multiply the distance from C to every vertex by 2, keeping the same line through C.

Mark the centre of enlargement C. Every distance in an enlargement is measured from C. C itself does not move.
C = (0, 0)
Find the move from C to each vertex. We need to know how far each point is from C, across and up (or down).
(1, 1): (1, 1) = 1 right, 1 up
(4, 1): (4, 1) = 4 right, 1 up
(1, 3): (1, 3) = 1 right, 3 up
Multiply each move by the scale factor 2. The image is bigger, so each distance from C is multiplied by 2.
(1, 1) × 2 = (2, 2)
(4, 1) × 2 = (8, 2)
(1, 3) × 2 = (2, 6)
Add each new move to C to find the image vertex. The new moves are still measured from C.
C + (2, 2) = (2, 2)
C + (8, 2) = (8, 2)
C + (2, 6) = (2, 6)
Plot the image points and join them to make shape B. The dashed lines from C pass through each vertex and its image: this is a quick way to check the picture.
B has vertices at (2, 2), (8, 2), (2, 6).

Check. Compare one side: it is 3 in A and 6 in B, and 3 × 2 = 6 = 6. ✓

Watch out. Measure from the centre of enlargement, not from the origin and not from the edge of the shape. Multiply the distance from C, not the coordinates themselves (unless C is the origin).

Worked example 21.4.2

Enlarge shape A with scale factor 2 and centre C (1, 1). Write down the coordinates of the vertices of the image B. A has vertices at (1, 1), (4, 1), (3, 3), (1, 3).

Method. To enlarge a shape with scale factor 2 about C, multiply the distance from C to every vertex by 2, keeping the same line through C.

Mark the centre of enlargement C. Every distance in an enlargement is measured from C. C itself does not move.
C = (1, 1)
Find the move from C to each vertex. We need to know how far each point is from C, across and up (or down).
(1, 1): (0, 0) = no move
(4, 1): (3, 0) = 3 right
(3, 3): (2, 2) = 2 right, 2 up
(1, 3): (0, 2) = 2 up
Multiply each move by the scale factor 2. The image is bigger, so each distance from C is multiplied by 2.
(0, 0) × 2 = (0, 0)
(3, 0) × 2 = (6, 0)
(2, 2) × 2 = (4, 4)
(0, 2) × 2 = (0, 4)
Add each new move to C to find the image vertex. The new moves are still measured from C.
C + (0, 0) = (1, 1)
C + (6, 0) = (7, 1)
C + (4, 4) = (5, 5)
C + (0, 4) = (1, 5)
Plot the image points and join them to make shape B. The dashed lines from C pass through each vertex and its image: this is a quick way to check the picture.
B has vertices at (1, 1), (7, 1), (5, 5), (1, 5).

Check. Compare one side: it is 3 in A and 6 in B, and 3 × 2 = 6 = 6. ✓

Worked example 21.4.3

Enlarge shape A with scale factor 12 and centre C (0, 0). Write down the coordinates of the vertices of the image B. A has vertices at (2, 2), (8, 2), (2, 6).

Method. To enlarge a shape with scale factor 12 about C, multiply the distance from C to every vertex by 12, keeping the same line through C.

Mark the centre of enlargement C. Every distance in an enlargement is measured from C. C itself does not move.
C = (0, 0)
Find the move from C to each vertex. We need to know how far each point is from C, across and up (or down).
(2, 2): (2, 2) = 2 right, 2 up
(8, 2): (8, 2) = 8 right, 2 up
(2, 6): (2, 6) = 2 right, 6 up
Multiply each move by the scale factor 12. The image is smaller, so each distance from C is multiplied by 12.
(2, 2) × 12 = (1, 1)
(8, 2) × 12 = (4, 1)
(2, 6) × 12 = (1, 3)
Add each new move to C to find the image vertex. The new moves are still measured from C.
C + (1, 1) = (1, 1)
C + (4, 1) = (4, 1)
C + (1, 3) = (1, 3)
Plot the image points and join them to make shape B. The dashed lines from C pass through each vertex and its image: this is a quick way to check the picture.
B has vertices at (1, 1), (4, 1), (1, 3).

Check. Compare one side: it is 6 in A and 3 in B, and 6 × 12 = 3 = 3. ✓

Watch out. Measure from the centre of enlargement, not from the origin and not from the edge of the shape. A scale factor between 0 and 1 makes the shape smaller; it is still called an enlargement.

Worked example 21.4.4

Describe fully the single transformation that maps shape A onto shape B.

Method. Decide which type of transformation maps A onto B, then find the extra details each type needs (a line, a centre, an angle, a vector or a scale factor).

Decide which type of transformation it is. Reflections, rotations and translations never change the size of a shape. If the size changes, it must be an enlargement.
Compare a side: it is 3 long in A and 6 long in B.
The sides have changed length but the shape is the same, so the transformation is an enlargement.
Find the size of the scale factor. Scale factor = length on the image ÷ matching length on the object.
A side of A is 3 and the matching side of B is 6.
6 ÷ 3 = 2
Decide whether it is positive or negative. B is the right way up and on the same side of the centre, so the scale factor is positive. It is larger than A.
Scale factor = 2
Find the centre of enlargement. Draw a straight line through a vertex of the object and the matching vertex of the image. Do the same for a second vertex. The centre is where the lines cross. (The lines are found here from their equations.)
Line through (1, 1) and (2, 2): y = x
Line through (4, 1) and (8, 2): y = 0.25x
They cross at (0, 0), so C = (0, 0).
Write the full description. We have found every piece of information needed for this transformation.
Enlargement, scale factor 2, centre (0, 0)

Check. Enlarge (1, 1) from (0, 0) with scale factor 2: (0 + 2 × 1, 0 + 2 × 1) = (2, 2), which is the matching vertex of B. ✓

Watch out. An enlargement needs both the scale factor and the centre. Do not forget that a fractional scale factor still means “enlargement”.

Worked example 21.4.5

Describe fully the single transformation that maps shape A onto shape B.

Method. Decide which type of transformation maps A onto B, then find the extra details each type needs (a line, a centre, an angle, a vector or a scale factor).

Decide which type of transformation it is. Reflections, rotations and translations never change the size of a shape. If the size changes, it must be an enlargement.
Compare a side: it is 6 long in A and 3 long in B.
The sides have changed length but the shape is the same, so the transformation is an enlargement.
Find the size of the scale factor. Scale factor = length on the image ÷ matching length on the object.
A side of A is 6 and the matching side of B is 3.
3 ÷ 6 = 12
Decide whether it is positive or negative. B is the right way up and on the same side of the centre, so the scale factor is positive. It is smaller than A, so the scale factor is a fraction between 0 and 1.
Scale factor = 12
Find the centre of enlargement. A vertex that has not moved can only be the centre of enlargement.
The vertex (7, 1) is in the same place on both shapes, so the centre is (7, 1).
Write the full description. We have found every piece of information needed for this transformation.
Enlargement, scale factor 12, centre (7, 1)

Check. Enlarge (1, 1) from (7, 1) with scale factor 12: (7 + 12 × −6, 1 + 12 × 0) = (4, 1), which is the matching vertex of B. ✓

Watch out. An enlargement needs both the scale factor and the centre. Do not forget that a fractional scale factor still means “enlargement”.

Worked example 21.4.6

Enlarge shape A with scale factor −2 and centre C (2, 2). Write down the coordinates of the vertices of the image B. A has vertices at (3, 3), (5, 3), (3, 4).

Method. To enlarge a shape with scale factor −2 about C, multiply the distance from C to every vertex by −2, keeping the same line through C.

Mark the centre of enlargement C. Every distance in an enlargement is measured from C. C itself does not move.
C = (2, 2)
Find the move from C to each vertex. We need to know how far each point is from C, across and up (or down).
(3, 3): (1, 1) = 1 right, 1 up
(5, 3): (3, 1) = 3 right, 1 up
(3, 4): (1, 2) = 1 right, 2 up
Multiply each move by the scale factor −2. The scale factor is negative, so the image is on the opposite side of C (and turned upside down). Multiply each move by −2; the signs of both numbers flip as well as the size changing.
(1, 1) × −2 = (−2, −2)
(3, 1) × −2 = (−6, −2)
(1, 2) × −2 = (−2, −4)
Add each new move to C to find the image vertex. The new moves are still measured from C.
C + (−2, −2) = (0, 0)
C + (−6, −2) = (−4, 0)
C + (−2, −4) = (0, −2)
Plot the image points and join them to make shape B. The dashed lines from C pass through each vertex and its image: this is a quick way to check the picture.
B has vertices at (0, 0), (−4, 0), (0, −2).

Check. Compare one side: it is 2 in A and 4 in B, and 2 × 2 = 4 = 4. ✓

Watch out. Measure from the centre of enlargement, not from the origin and not from the edge of the shape. A negative scale factor puts the image on the other side of the centre.

Worked example 21.4.7

Describe fully the single transformation that maps shape A onto shape B.

Method. Decide which type of transformation maps A onto B, then find the extra details each type needs (a line, a centre, an angle, a vector or a scale factor).

Decide which type of transformation it is. Reflections, rotations and translations never change the size of a shape. If the size changes, it must be an enlargement.
Compare a side: it is 3 long in A and 6 long in B.
The sides have changed length but the shape is the same, so the transformation is an enlargement.
Find the size of the scale factor. Scale factor = length on the image ÷ matching length on the object.
A side of A is 3 and the matching side of B is 6.
6 ÷ 3 = 2
Decide whether it is positive or negative. B is on the opposite side of the centre from A and is upside down, so the scale factor is negative.
Scale factor = −2
Find the centre of enlargement. Draw a straight line through a vertex of the object and the matching vertex of the image. Do the same for a second vertex. The centre is where the lines cross. (The lines are found here from their equations.)
Line through (1, 1) and (1, 4): x = 1
Line through (4, 1) and (−5, 4): y = −13x + 73
They cross at (1, 2), so C = (1, 2).
Write the full description. We have found every piece of information needed for this transformation.
Enlargement, scale factor −2, centre (1, 2)

Check. Enlarge (1, 1) from (1, 2) with scale factor −2: (1 + −2 × 0, 2 + −2 × −1) = (1, 4), which is the matching vertex of B. ✓

Watch out. An enlargement needs both the scale factor and the centre. Do not forget that a fractional scale factor still means “enlargement”.

Watch out

Measure from the centre of enlargement, not from the origin (unless that is the centre) and not from the shape’s own corner. Do not add the scale factor to the coordinates: you must multiply the distances from the centre. And a scale factor like 12 makes the shape smaller, but it is still an enlargement.

21.5

Describing and identifying single transformations

Edexcel 1MA1: G7
Grade 3-7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Describing and identifying single transformationsComing soon

In an examination the question often says “Describe fully the single transformation that maps shape A onto shape B.” You must (1) name the transformation and (2) give the extra detail that each type needs. Without the detail you do not get the marks. The table shows what you need.

Transformation	You must give	Size	Facing
Reflection	the equation of the mirror line	same (congruent)	flipped
Rotation	angle, direction (not for 180°), centre	same (congruent)	turned
Translation	the column vector	same (congruent)	same way, sides parallel
Enlargement	scale factor and centre	changed (similar)	same way (unless the scale factor is negative)

A is the object. B is a reflection of A (flipped). C is a rotation of A (turned). D is a translation of A (slid, same way up and same direction).

How to decide. Ask these questions in order. (1) Has the size changed? If yes, it is an enlargement. (2) Is the shape flipped over (a mirror image)? If yes, it is a reflection. (3) Are all the sides parallel to the sides of the original, with the shape facing the same way? If yes, it is a translation. (4) Otherwise it has turned, so it is a rotation. Then find the detail: the line, the vector, the centre and angle, or the scale factor and centre.

Invariant points. A point that does not move is called an invariant point. Every point on the mirror line of a reflection is invariant; the centre of a rotation or an enlargement is invariant; a translation has none.

Key ideas

“Describe fully” means: name the transformation and give its detail.
Reflection: equation of the mirror line. Rotation: angle, direction, centre. Translation: column vector. Enlargement: scale factor, centre.
Reflections, rotations and translations give congruent images. Enlargements give similar images.
Reflection turns the shape into a mirror image; rotation turns it; translation slides it; enlargement resizes it.
Invariant point: a point that does not move.

Worked example 21.5.1

Describe fully the single transformation that maps shape A onto shape B.

Key words. A reflection flips a shape over a mirror line. The object and its image are the same distance from the mirror line, on opposite sides.

Method. Decide which type of transformation maps A onto B, then find the extra details each type needs (a line, a centre, an angle, a vector or a scale factor).

Decide which type of transformation it is. A reflection flips the shape over. All other transformations of this type keep the shape facing the same way.
Compare a side: it is 3 long in A and 3 long in B.
The size is unchanged, but B is the mirror image of A: if you go round the corners in the same order, one shape is clockwise and the other is anticlockwise. So the transformation is a reflection.
Pick vertices and their images, and find the midpoints. A mirror line is exactly half-way between an object point and its image, so find the midpoint of a point and its image.
(1, 3) and (1, −1): midpoint = ((1 + 1) ÷ 2, (3 + −1) ÷ 2) = (1, 1)
(3, 3) and (3, −1): midpoint = ((3 + 3) ÷ 2, (3 + −1) ÷ 2) = (3, 1)
Write the equation of the mirror line. The midpoints all have y = 1. The points are in a horizontal line, so the mirror line is the horizontal line y = 1.
Mirror line: y = 1
Write the full description. We have found every piece of information needed for this transformation.
Reflection in the line y = 1

Check. Reflect (3, 3) in y = 1 and we get (3, −1), which matches (3, −1) in B. ✓

Watch out. A reflection needs the equation of the mirror line. Writing just “reflection” or “reflect in the x-axis” for a line like x = 3 is wrong.

Worked example 21.5.2

Describe fully the single transformation that maps shape A onto shape B.

Key words. A rotation turns a shape about a fixed point called the centre of rotation. We state the angle and the direction (clockwise or anticlockwise).

Method. Decide which type of transformation maps A onto B, then find the extra details each type needs (a line, a centre, an angle, a vector or a scale factor).

Decide which type of transformation it is. A rotation turns the shape. Compare with a translation, where all sides stay parallel to the originals.
Compare a side: it is 2 long in A and 2 long in B.
The size is unchanged, B is not flipped, but its sides are not parallel to the originals: the shape has turned. So the transformation is a rotation.
Find the centre of rotation. The centre is the same distance from a point and from its image. The origin is the first place to try. It works if the squared distances from the origin to the point and its image match.
Distance² of (2, 2) from (0, 0) = 2² + 2² = 8
Distance² of (−2, 2) from (0, 0) = −2² + 2² = 8
They match, and the same works for the other vertices, so the centre is (0, 0).
Find the angle and direction. Write the move from C to a vertex, and the move from C to its image. A quarter turn changes (a, b) in a special way: clockwise gives (b, −a) and anticlockwise gives (−b, a).
From C to (2, 2): (2, 2) (2 right, 2 up)
From C to (−2, 2): (−2, 2) (2 left, 2 up)
This matches (a, b) → (−b, a), so the turn is 90° anticlockwise.
Write the full description. We have found every piece of information needed for this transformation.
Rotation, 90° anticlockwise, centre (0, 0)

Check. Rotate (4, 2) about (0, 0) by 90° anticlockwise: we get (−2, 4), which matches the vertex (−2, 4) of B. ✓

Watch out. Give three things for a rotation: the angle, the direction (clockwise or anticlockwise) and the centre. A 180° turn does not need a direction.

Worked example 21.5.3

Describe fully the single transformation that maps shape A onto shape B.

Method. Decide which type of transformation maps A onto B, then find the extra details each type needs (a line, a centre, an angle, a vector or a scale factor).

Decide which type of transformation it is. Reflections, rotations and translations never change the size of a shape. If the size changes, it must be an enlargement.
Compare a side: it is 1 long in A and 3 long in B.
The sides have changed length but the shape is the same, so the transformation is an enlargement.
Find the size of the scale factor. Scale factor = length on the image ÷ matching length on the object.
A side of A is 1 and the matching side of B is 3.
3 ÷ 1 = 3
Decide whether it is positive or negative. B is the right way up and on the same side of the centre, so the scale factor is positive. It is larger than A.
Scale factor = 3
Find the centre of enlargement. Draw a straight line through a vertex of the object and the matching vertex of the image. Do the same for a second vertex. The centre is where the lines cross. (The lines are found here from their equations.)
Line through (2, 1) and (4, 1): y = 1
Line through (2, 2) and (4, 4): y = x
They cross at (1, 1), so C = (1, 1).
Write the full description. We have found every piece of information needed for this transformation.
Enlargement, scale factor 3, centre (1, 1)

Check. Enlarge (2, 1) from (1, 1) with scale factor 3: (1 + 3 × 1, 1 + 3 × 0) = (4, 1), which is the matching vertex of B. ✓

Watch out. An enlargement needs both the scale factor and the centre. Do not forget that a fractional scale factor still means “enlargement”.

Worked example 21.5.4

Write down the invariant point(s) of the rotation of 90° clockwise about the point (1, 2).

Key words. A combination of transformations means doing one transformation and then another. A point that does not move is an invariant point.

Method. Think about what the transformation does to each point, then find which points stay where they are.

Say what an invariant point is. An invariant point is a point that does not move when the transformation is done.
Invariant = unchanged
Think about what a rotation does. Every point is moved (turned or stretched) about the centre, except the centre itself: it is at distance 0 from the centre, so it cannot move.
The centre is (1, 2).
Test the centre. Apply the transformation to the centre.
(1, 2) → (1, 2)
Write the answer. Only the centre is invariant (the scale factor is not 1, and the turn is not 0° or 360°).
Invariant point: (1, 2) only.

Check. Test another point, (2, 3): it goes to (2, 1), which is different, so it is not invariant. ✓

Watch out. Do not confuse “invariant point” with “image”. An invariant point is its own image.

Worked example 21.5.5

Describe fully the single transformation that maps shape A onto shape B.

Key words. A reflection flips a shape over a mirror line. The object and its image are the same distance from the mirror line, on opposite sides.

Method. Decide which type of transformation maps A onto B, then find the extra details each type needs (a line, a centre, an angle, a vector or a scale factor).

Decide which type of transformation it is. A reflection flips the shape over. All other transformations of this type keep the shape facing the same way.
Compare the areas: 4 square units for A and 4 for B.
The size is unchanged, but B is the mirror image of A: if you go round the corners in the same order, one shape is clockwise and the other is anticlockwise. So the transformation is a reflection.
Compare the coordinates of a vertex and its image. For the diagonal mirror lines y = x and y = −x the coordinates change in a special way.
The coordinates have been swapped and both signs changed: (2, 1) became (−1, −2).
Use the midpoint as a check. The mirror line must pass through the midpoint of a point and its image.
Midpoint of (2, 1) and (−1, −2) = (0.5, −0.5), which lies on y = −x.
Write the equation of the mirror line. A reflection that swaps x and y and changes both signs is a reflection in the line y = −x.
Mirror line: y = −x
Write the full description. We have found every piece of information needed for this transformation.
Reflection in the line y = −x

Check. Reflect (5, 2) in y = −x and we get (−2, −5), which matches (−2, −5) in B. ✓

Watch out. A reflection needs the equation of the mirror line. Writing just “reflection” or “reflect in the x-axis” for a line like x = 3 is wrong.

Watch out

Never write just “reflection” or “enlargement”. Each type needs its detail: the mirror line, the centre and scale factor, and so on. Also do not write two transformations: the question asks for a single one. If the shape has got bigger, it is an enlargement, even if it has also moved.

21.6

Combinations of transformations and invariant points

Edexcel 1MA1: G8
Grade 6-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Combinations of transformations and invariant pointsComing soon

Sometimes a shape is moved by two or more transformations, one after the other. For example, “reflect A in the x-axis to give B, then rotate B through 90° about the origin to give C”. We work out each image in turn: the image of the first transformation is the object for the second. The order matters: reflecting and then rotating usually gives a different result from rotating and then reflecting.

A combination of transformations can often be replaced by a single transformation that does the whole job. Three facts are useful:

Two reflections in parallel lines make a translation. The translation is perpendicular to the mirror lines and is twice the distance between them.
Two reflections in lines that cross make a rotation about the point where the lines cross. The angle of the rotation is twice the angle between the mirror lines.
Two rotations about the same centre make a rotation about that centre whose angle is the sum of the two angles. Two translations make a translation whose vector is the sum of the two vectors.

Reflect A in x = 1 to get B, then B in x = 4 to get C. The mirror lines are 3 apart, and C is 2 × 3 = 6 to the right of A: a translation by 6
0.

Reflect A in the x-axis to get B, then B in the y-axis to get C. The two lines cross at the origin at 90°, so C is a rotation of A through 2 × 90° = 180° about the origin.

Invariant points. A point that is unchanged by a transformation is called an invariant point. For a combination, test the single equivalent transformation. For example, a half turn about O has only one invariant point, O. A reflection has a whole line of invariant points (the mirror line). A translation has none. When you must find the invariant point of a combination, you can also look for the point whose image is itself.

Key ideas

Do the transformations in the order given: the image of the first is the object of the second.
Two reflections in parallel lines = a translation (twice the distance between the lines).
Two reflections in lines that cross = a rotation about the crossing point (twice the angle between the lines).
In general the order matters; for two translations it does not.
An invariant point is its own image: the mirror line (reflection), the centre (rotation, enlargement); none for a translation.

Worked example 21.6.1

A is mapped to B by a reflection in the line x = 1. B is mapped to C by a reflection in the line x = 4. Describe fully the single transformation that maps A onto C. A has vertices at (−3, 1), (−1, 1), (−3, 4).

Key words. A combination of transformations means doing one transformation and then another. A point that does not move is an invariant point.

Method. Carry out the transformations one at a time, in the order given, using the image of one as the object for the next. Then compare the final image with the original to find the single transformation that does the whole job.

Do the first transformation: reflection in the line x = 1. Reflecting in a vertical line x = 1: the y-coordinate stays the same and the new x is 2 × 1 − x (the same distance on the other side of the line). Do this to every vertex of shape A.
Shape B:
(−3, 1) → (2 × 1 − (−3), 1) = (5, 1)
(−1, 1) → (2 × 1 − (−1), 1) = (3, 1)
(−3, 4) → (2 × 1 − (−3), 4) = (5, 4)
Then do the second transformation: reflection in the line x = 4. Reflecting in a vertical line x = 4: the y-coordinate stays the same and the new x is 2 × 4 − x (the same distance on the other side of the line). Do this to every vertex of shape B.
Shape C:
(5, 1) → (2 × 4 − 5, 1) = (3, 1)
(3, 1) → (2 × 4 − 3, 1) = (5, 1)
(5, 4) → (2 × 4 − 5, 4) = (3, 4)
Plot the shapes. Joining the image points in order gives each new shape. The picture shows the whole journey.
B (5, 1), (3, 1), (5, 4) then C (3, 1), (5, 1), (3, 4)
Decide which type of transformation it is. A translation only slides the shape. If the shape had turned, its sides would no longer be parallel to the originals.
Compare a side: it is 3 long in A and 3 long in C.
The size is unchanged, C faces the same way as A, and every side is parallel to its original. So the transformation is a translation.
Pick a vertex and its image. Every point moves by the same amount, so one vertex is enough to find the move.
(−3, 1) → (3, 1)
Find the move across and the move up or down. Subtract the x-coordinates, then the y-coordinates (image − object). Positive means right/up, negative means left/down.
Across: 3 − −3 = 6 (6 right)
Up/down: 1 − 1 = 0 (no move)
Write the move as a column vector. The across move goes on top and the up/down move goes underneath.
Translation by 6
0
Write the full description. We have found every piece of information needed for this transformation.
Translation by 6
0

Check. Test another vertex: (−1, 1) + 6
0 = (5, 1), which is the matching vertex of C. ✓

Watch out. Write the vector as across on top, up/down underneath, and take image − object, not the other way round.

Worked example 21.6.2

A is mapped to B by a reflection in the line y = 0. B is mapped to C by a reflection in the line x = 0. Describe fully the single transformation that maps A onto C. A has vertices at (1, 1), (4, 1), (1, 3).

Key words. A combination of transformations means doing one transformation and then another. A point that does not move is an invariant point.

Do the first transformation: reflection in the line y = 0. Reflecting in a horizontal line y = 0: the x-coordinate stays the same and the new y is 2 × 0 − y. Do this to every vertex of shape A.
Shape B:
(1, 1) → (1, 2 × 0 − 1) = (1, −1)
(4, 1) → (4, 2 × 0 − 1) = (4, −1)
(1, 3) → (1, 2 × 0 − 3) = (1, −3)
Then do the second transformation: reflection in the line x = 0. Reflecting in a vertical line x = 0: the y-coordinate stays the same and the new x is 2 × 0 − x (the same distance on the other side of the line). Do this to every vertex of shape B.
Shape C:
(1, −1) → (2 × 0 − 1, −1) = (−1, −1)
(4, −1) → (2 × 0 − 4, −1) = (−4, −1)
(1, −3) → (2 × 0 − 1, −3) = (−1, −3)
Plot the shapes. Joining the image points in order gives each new shape. The picture shows the whole journey.
B (1, −1), (4, −1), (1, −3) then C (−1, −1), (−4, −1), (−1, −3)
Decide which type of transformation it is. A rotation turns the shape. Compare with a translation, where all sides stay parallel to the originals.
Compare a side: it is 3 long in A and 3 long in C.
The size is unchanged, C is not flipped, but its sides are not parallel to the originals: the shape has turned. So the transformation is a rotation.
Find the centre of rotation. In a half turn, the centre is exactly half-way between a point and its image. Find the midpoint for two different vertices: they should be the same point.
(1, 1) and (−1, −1): midpoint (0, 0)
(4, 1) and (−4, −1): midpoint (0, 0)
Both midpoints are the same, so the centre is (0, 0).
Find the angle and direction. From C the moves to the vertex and to its image point in exactly opposite directions, so it is a half turn. For a half turn the direction does not matter.
From C to (1, 1): (1, 1); from C to (−1, −1): (−1, −1). Both signs have changed, so the angle is 180°.
Write the full description. We have found every piece of information needed for this transformation.
Rotation, 180°, centre (0, 0)

Check. Rotate (4, 1) about (0, 0) by 180°: we get (−4, −1), which matches the vertex (−4, −1) of C. ✓

Watch out. Give three things for a rotation: the angle, the direction (clockwise or anticlockwise) and the centre. A 180° turn does not need a direction.

Worked example 21.6.3

A is mapped to B by a rotation of 90° clockwise about (0, 0). B is mapped to C by a reflection in the line y = 0. Describe fully the single transformation that maps A onto C. A has vertices at (1, 1), (4, 1), (1, 3).

Key words. A combination of transformations means doing one transformation and then another. A point that does not move is an invariant point.

Do the first transformation: rotation of 90° clockwise about (0, 0). Rotate about (0, 0): write each point as a move from the centre, turn the move using (a, b) → (b, −a), then add the centre back on. Do this to every vertex of shape A.
Shape B:
(1, 1): move from centre (1, 1) → turned (1, −1) → (1, −1)
(4, 1): move from centre (4, 1) → turned (1, −4) → (1, −4)
(1, 3): move from centre (1, 3) → turned (3, −1) → (3, −1)
Then do the second transformation: reflection in the line y = 0. Reflecting in a horizontal line y = 0: the x-coordinate stays the same and the new y is 2 × 0 − y. Do this to every vertex of shape B.
Shape C:
(1, −1) → (1, 2 × 0 − (−1)) = (1, 1)
(1, −4) → (1, 2 × 0 − (−4)) = (1, 4)
(3, −1) → (3, 2 × 0 − (−1)) = (3, 1)
Plot the shapes. Joining the image points in order gives each new shape. The picture shows the whole journey.
B (1, −1), (1, −4), (3, −1) then C (1, 1), (1, 4), (3, 1)
Decide which type of transformation it is. A reflection flips the shape over. All other transformations of this type keep the shape facing the same way.
Compare a side: it is 3 long in A and 3 long in C.
The size is unchanged, but C is the mirror image of A: if you go round the corners in the same order, one shape is clockwise and the other is anticlockwise. So the transformation is a reflection.
Compare the coordinates of a vertex and its image. For the diagonal mirror lines y = x and y = −x the coordinates change in a special way.
The coordinates have been swapped: (4, 1) became (1, 4).
Use the midpoint as a check. The mirror line must pass through the midpoint of a point and its image.
Midpoint of (4, 1) and (1, 4) = (2.5, 2.5), which lies on y = x.
Write the equation of the mirror line. A reflection that swaps x and y is a reflection in the line y = x.
Mirror line: y = x
Write the full description. We have found every piece of information needed for this transformation.
Reflection in the line y = x

Check. Reflect (1, 3) in y = x and we get (3, 1), which matches (3, 1) in C. ✓

Watch out. A reflection needs the equation of the mirror line. Writing just “reflection” or “reflect in the x-axis” for a line like x = 3 is wrong.

Worked example 21.6.4

A is mapped to B by a translation by 2
1. B is mapped to C by a reflection in the line y = 0. Write down the coordinates of the vertices of C. A has vertices at (1, 2), (3, 2), (1, 5).

Key words. A combination of transformations means doing one transformation and then another. A point that does not move is an invariant point.

Method. Carry out the transformations one at a time, in the order given, using the image of one as the object for the next.

Do the first transformation: translation by 2
1. Add 2 to every x and 1 to every y. Do this to every vertex of shape A.
Shape B:
(1, 2) → (1 + 2, 2 + 1) = (3, 3)
(3, 2) → (3 + 2, 2 + 1) = (5, 3)
(1, 5) → (1 + 2, 5 + 1) = (3, 6)
Then do the second transformation: reflection in the line y = 0. Reflecting in a horizontal line y = 0: the x-coordinate stays the same and the new y is 2 × 0 − y. Do this to every vertex of shape B.
Shape C:
(3, 3) → (3, 2 × 0 − 3) = (3, −3)
(5, 3) → (5, 2 × 0 − 3) = (5, −3)
(3, 6) → (3, 2 × 0 − 6) = (3, −6)
Plot the shapes. Joining the image points in order gives each new shape. The picture shows the whole journey.
B (3, 3), (5, 3), (3, 6) then C (3, −3), (5, −3), (3, −6)

Check. Shapes A and C are congruent (same size), and the vertex (1, 2) has gone to (3, −3): the transformations only move or flip the shape. ✓

Watch out. Do the transformations in the order given. Doing them the other way round usually gives a different answer.

Worked example 21.6.5

Shape A has vertices at (1, 1), (3, 1), (1, 4). Shape B is found by doing a rotation of 90° clockwise about (0, 0) and then a translation by 4
0. Shape C is found by doing the translation by 4
0 first and then the rotation of 90° clockwise about (0, 0). Is shape C in the same place as shape B? Show your working.

Key words. A combination of transformations means doing one transformation and then another. A point that does not move is an invariant point.

Method. Work out the final image for each order, then compare them.

Do transformation 1 then transformation 2. First rotation of 90° clockwise about (0, 0), then translation by 4
0.
After the first: (1, −1), (1, −3), (4, −1)
After the second: (5, −1), (5, −3), (8, −1)
Call this shape B.
Now do them in the opposite order. First translation by 4
0, then rotation of 90° clockwise about (0, 0).
After the first: (5, 1), (7, 1), (5, 4)
After the second: (1, −5), (1, −7), (4, −5)
Call this shape C.
Compare B and C. If the order did not matter, B and C would be in exactly the same place.
B: (5, −1), (5, −3), (8, −1)
C: (1, −5), (1, −7), (4, −5)
The vertices are different, so the shapes are in different places.

Check. Look at the picture: B and C are not on top of each other. ✓

Watch out. Many pairs of transformations give different results when the order is swapped. (Two translations are one exception: they always give the same result.)

Worked example 21.6.6

Reflect a point in the x-axis and then in the y-axis. Write down the invariant point(s) of this combination.

Key words. A combination of transformations means doing one transformation and then another. A point that does not move is an invariant point.

Method. Think about what the transformation does to each point, then find which points stay where they are.

Say what an invariant point is. An invariant point is a point that does not move when the transformation is done.
Invariant = unchanged
Think about what a rotation does. Every point is moved (turned or stretched) about the centre, except the centre itself: it is at distance 0 from the centre, so it cannot move.
The centre is (0, 0).
Test the centre. Apply the transformation to the centre.
(0, 0) → (0, 0)
Write the answer. Only the centre is invariant (the scale factor is not 1, and the turn is not 0° or 360°).
Invariant point: (0, 0) only.

Check. Test another point, (1, 1): it goes to (−1, −1), which is different, so it is not invariant. ✓

Watch out. Do not confuse “invariant point” with “image”. An invariant point is its own image.

Watch out

Do the transformations in the order stated, and always use the image of the first transformation as the object for the second, not the original shape. And a “single transformation” must be a reflection, rotation, translation or enlargement with all its detail.

21.7

Congruent shapes and triangles

Edexcel 1MA1: G5
Grade 3-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Congruent shapes and trianglesComing soon

Two shapes are congruent if they are exactly the same size and shape. You can place one exactly on top of the other, perhaps after turning it, sliding it or flipping it over. Reflections, rotations and translations all give congruent images. For congruent shapes, all matching (corresponding) sides are equal and all matching angles are equal.

For triangles we do not need to check all six facts (three sides and three angles). If three particular facts are equal, the rest are forced to be equal as well. There are four conditions to remember. For triangles ABC and PQR:

SSS: all three sides are equal.
SAS: two sides and the angle between them (the included angle) are equal.
ASA: two angles and the side between them are equal. (If the equal side is not between the angles it is AAS, which also works because the third angle is then known too.)
RHS: both are right-angled, the hypotenuses are equal and one other side is equal.

SSS: the three sides are 5 cm, 6 cm and 7 cm in both triangles, so they are congruent (one is flipped and turned).

SAS: two sides (5 cm and 7 cm) and the angle between them (50°).

ASA: two angles (40° and 70°) and the side between them (6 cm).

RHS: a right angle, the hypotenuse (10 cm) and one other side (6 cm).

What does not work. Three equal angles (AAA) are not enough: triangles with the same angles can be different sizes. They are similar, not congruent (see section 21.8). Also, two sides and an angle that is not between them (SSA) is not enough. The picture below shows two different triangles that both have a 30° angle, a side of 10 cm next to it and a side of 6 cm opposite it.

SSA does not prove congruence. Triangles AB₁C and AB₂C both have A = 30°, AC = 10 cm and a side of 6 cm opposite the angle, but they are different triangles.

Writing a proof. In a proof you must give a reason for every fact. Typical reasons are “given”, “common side” (shared by both triangles), “vertically opposite angles are equal”, “alternate angles are equal” (parallel lines) and “base angles of an isosceles triangle are equal”. Finish by naming the condition, for example “triangle ABC is congruent to triangle CDA (SAS)”, and write the vertices in matching order.

Key ideas

Congruent shapes are exactly the same size and shape. Reflection, rotation and translation give congruent images.
Triangles are congruent if: SSS, SAS (included angle), ASA (or AAS), or RHS.
AAA is not enough (similar only). SSA is not enough (unless the angle is a right angle, which is RHS).
In a proof, give a reason for each fact and name the condition. Write the vertices in matching order.

Worked example 21.7.1

Are triangles ABC and PQR congruent? Give a reason.

Key words. Two shapes are congruent if they are exactly the same size and shape (they may be turned or flipped). The included angle is the angle between two given sides. The hypotenuse is the side opposite the right angle.

Method. Triangles are congruent if one fits exactly on the other. For triangles, we only need to check one of four conditions: SSS, SAS, ASA or RHS.

Write down what is known about each triangle. We can only use information that is given in the question or can be worked out from it.
Triangle 1: AB = 6 cm, BC = 4 cm, CA = 5 cm
Triangle 2: PQ = 6 cm, QR = 4 cm, RP = 5 cm
Match the equal parts. Parts in the same position (corresponding parts) must be equal in both triangles.
AB = PQ = 6 cm (side)
BC = QR = 4 cm (side)
CA = RP = 5 cm (side)
Decide which condition is satisfied. SSS (side, side, side): all three sides of one triangle equal the three sides of the other triangle.
The three equal parts make SSS.
State the conclusion with its reason. In an exam, name the condition as your reason.
The triangles are congruent (SSS).

Check. Put the three sides in order: 4, 5, 6 in both triangles. If you cut one out and flip it over, it fits exactly. ✓

Watch out. Match the sides by length, not by where they appear on the page: the second triangle has been turned over.

Worked example 21.7.2

Are triangles ABC and PQR congruent? Give a reason.

Method. Triangles are congruent if one fits exactly on the other. For triangles, we only need to check one of four conditions: SSS, SAS, ASA or RHS.

Write down what is known about each triangle. We can only use information that is given in the question or can be worked out from it.
Triangle 1: AB = 6 cm, AC = 8 cm, angle A = 40°
Triangle 2: PQ = 6 cm, PR = 8 cm, angle P = 40°
Match the equal parts. Parts in the same position (corresponding parts) must be equal in both triangles.
AB = PQ = 6 cm (side)
angle A = angle P = 40° (angle, between the two sides)
AC = PR = 8 cm (side)
Decide which condition is satisfied. SAS (side, angle, side): two sides and the angle between them (the included angle) are equal.
The three equal parts make SAS.
State the conclusion with its reason. In an exam, name the condition as your reason.
The triangles are congruent (SAS).

Check. Use the cosine rule to find BC and QR: both are √(6² + 8² − 2 × 6 × 8 × cos 40°) = 5.14 cm, so all three sides match. ✓

Watch out. The angle must be between the two sides. A 40° angle in another position would not be enough.

Worked example 21.7.3

Are these two triangles congruent? Give a reason.

Key words. Two shapes are congruent if they are exactly the same size and shape. The included angle is the angle between two given sides.

Method. Check the known facts against the four conditions for congruent triangles. If none is satisfied, we cannot conclude that the triangles are congruent.

Write down what is known about each triangle. We can only use information that is given in the question or can be worked out from it.
Triangle 1: angles 40°, 60°, 80° (no sides given)
Triangle 2: angles 40°, 60°, 80° (no sides given)
Compare with the four conditions. To be sure, the facts must match SSS, SAS, ASA/AAS or RHS.
Three equal angles is AAA, which is not one of the conditions: no side is known to be equal.
State the conclusion. If none of the conditions is met, we cannot say that the triangles are congruent.
The triangles are not necessarily congruent.

Check. Draw a triangle with angles 40°, 60° and 80° with a base of 5 cm, then another with a base of 10 cm: both have the same angles but different sizes. ✓

Watch out. AAA proves the triangles are similar (same shape) but not congruent (same size).

Worked example 21.7.4

Are triangles ABC and PQR congruent? Give a reason.

Method. Triangles are congruent if one fits exactly on the other. For triangles, we only need to check one of four conditions: SSS, SAS, ASA or RHS.

Write down what is known about each triangle. We can only use information that is given in the question or can be worked out from it.
Triangle 1: right angle at B, AC = 13 cm (hypotenuse), BC = 5 cm
Triangle 2: right angle at Q, PR = 13 cm (hypotenuse), QR = 5 cm
Match the equal parts. Parts in the same position (corresponding parts) must be equal in both triangles.
angle B = angle Q = 90° (right angle)
AC = PR = 13 cm (hypotenuse)
BC = QR = 5 cm (side)
Decide which condition is satisfied. RHS (right angle, hypotenuse, side): both triangles have a right angle, the hypotenuses are equal and one other pair of sides is equal.
The three equal parts make RHS.
State the conclusion with its reason. In an exam, name the condition as your reason.
The triangles are congruent (RHS).

Check. Pythagoras gives the missing side in each triangle: √(13² − 5²) = √144 = 12 cm, so all three sides match. ✓

Watch out. The side must be a different side from the hypotenuse, and the equal angle must be the right angle.

Worked example 21.7.5

ABCD is a parallelogram. The diagonal AC is drawn. Prove that triangles ABC and CDA are congruent.

Key words. A proof is a chain of reasons that shows a statement is always true. Each line must have a reason, such as “given”, “common side”, “alternate angles” or “vertically opposite angles”.

Method. To prove triangles are congruent, state three pairs of equal parts, each with a reason, name the condition (SSS, SAS, ASA or RHS), and then write the conclusion.

AB = CD Opposite sides of a parallelogram are equal (marked with one tick).
BC = DA Opposite sides of a parallelogram are equal (marked with two ticks).
AC = CA AC is a common side, shared by both triangles.
Conclusion. All the facts needed for SSS are shown.
Triangle ABC is congruent to triangle CDA (SSS).

Check. The matching vertices are A ↔ C, B ↔ D, C ↔ A, and the pairs of sides match: AB = CD, BC = DA, AC = CA. ✓

Watch out. Write the vertices in matching order: triangle ABC is congruent to triangle CDA, not triangle ACD.

Watch out

Check that the equal parts are in the correct positions. For SAS the angle must be between the two equal sides. AAA and SSA do not prove congruence. In a proof, write the vertices in matching order and give a reason for every statement.

21.8

Similar shapes and scale factors

Edexcel 1MA1: G7, G19, R12
Grade 3-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Similar shapes and scale factorsComing soon

Two shapes are similar if one is an enlargement of the other. They have exactly the same shape but can be different sizes (one might also be turned or flipped). For similar shapes:

all corresponding angles are equal;
all corresponding sides are in the same ratio. This ratio, large ÷ small, is the scale factor.

Corresponding sides are the ones in the same position in both shapes. For example, the shortest side of one triangle matches the shortest side of the similar triangle. Remember the difference: congruent shapes are the same size and shape; similar shapes are the same shape but not necessarily the same size.

Triangle PQR is similar to triangle ABC with scale factor 1.5: 6 × 1.5 = 9, 4 × 1.5 = 6, 5 × 1.5 = 7.5. The angles are equal.

Finding a missing length. (1) Find a pair of corresponding sides that you know both lengths of. (2) Work out the scale factor: larger ÷ smaller. (3) Use it on the side you want: multiply if you are going from the small shape to the large shape, divide if you are going from the large shape to the small one.

Are two shapes similar? Check that all pairs of corresponding sides are in the same ratio. A rectangle 4 cm by 6 cm and a rectangle 6 cm by 9 cm are similar: 6 ÷ 4 = 1.5 and 9 ÷ 6 = 1.5. A 4 cm by 6 cm rectangle and a 6 cm by 8 cm rectangle are not similar, since 6 ÷ 4 = 1.5 but 8 ÷ 6 = 1.33…. (For triangles, equal angles are enough: you do not need to check the sides.)

Similar triangles from parallel lines. Parallel lines create similar triangles. If DE is parallel to BC in triangle ABC, then triangle ADE is similar to triangle ABC. The angle at A is shared, and the corresponding angles ADE and ABC are equal, and so are AED and ACB. Note that the matching side to AD is the whole of AB, not DB. Two triangles that meet at a point (a “bow-tie” shape) with a pair of parallel sides are also similar, because the alternate angles are equal and the vertically opposite angles are equal.

DE is parallel to BC. Triangle ADE is similar to triangle ABC with scale factor AB ÷ AD = 10 ÷ 4 = 2.5.

AB is parallel to CD, so triangles XAB and XCD are similar with scale factor 1.5.

Proving two triangles are similar (Higher). Show that two angles are equal (the third then must be equal as well), and give a reason for each pair: parallel lines (corresponding or alternate angles), vertically opposite angles, a common angle, or angles in a triangle that add up to 180°. Then state: “the triangles are similar (equal angles)”.

Key ideas

Similar shapes: equal corresponding angles and corresponding sides in the same ratio. The ratio is the scale factor.
Scale factor = length on the larger shape ÷ matching length on the smaller shape.
Small to large: multiply by the scale factor. Large to small: divide.
Parallel lines make similar triangles; look for a common angle, alternate or corresponding angles, or vertically opposite angles.
Congruent = same size and shape. Similar = same shape, any size.

Worked example 21.8.1

The two shapes are mathematically similar. The diagrams are drawn to scale. Work out the value of x.

Key words. Two shapes are similar if one is an enlargement of the other: same shape, different size. Corresponding sides are in the same position in both shapes. The scale factor is the number you multiply lengths by to go from one shape to the other.

Method. Similar shapes have corresponding sides in the same ratio. Find the scale factor from a pair of matching sides we know, then use it on the side we want.

Match the corresponding sides. In similar shapes the sides that are in the same position match. Their lengths are in the same ratio, and that ratio is the scale factor.
Known pair: 6 cm (small shape) and 9 cm (large shape).
The side we want is the one marked x.
Find the scale factor from the pair of sides we know. Scale factor = length in the large shape ÷ matching length in the small shape.
9 cm ÷ 6 cm = 32
So the large shape is 32 times as big as the small shape.
Use the scale factor to find x. x is on the large shape and we know the matching side on the small shape. Going from small to large we multiply by the scale factor.
x = 5 cm × 32 = 7.5 cm
Write the answer with its unit. Lengths need a unit.
x = 7.5 cm

Check. Ratio check: 7.5 cm ÷ 5 cm = 32, the same scale factor as before. ✓

Watch out. Make sure you pair the matching sides (same position in both shapes). Multiply when going from the small shape to the large one and divide when going from large to small.

Worked example 21.8.2

The two shapes are mathematically similar. The diagrams are drawn to scale. Work out the value of x.

Method. Similar shapes have corresponding sides in the same ratio. Find the scale factor from a pair of matching sides we know, then use it on the side we want.

Match the corresponding sides. In similar shapes the sides that are in the same position match. Their lengths are in the same ratio, and that ratio is the scale factor.
Known pair: 4 cm (small shape) and 10 cm (large shape).
The side we want is the one marked x.
Find the scale factor from the pair of sides we know. Scale factor = length in the large shape ÷ matching length in the small shape.
10 cm ÷ 4 cm = 52
So the large shape is 52 times as big as the small shape.
Use the scale factor to find x. x is on the small shape and we know the matching side on the large shape. Going from large to small we divide by the scale factor.
x = 15 cm ÷ 52 = 6 cm
Write the answer with its unit. Lengths need a unit.
x = 6 cm

Check. Ratio check: 15 cm ÷ 6 cm = 52, the same scale factor as before. ✓

Watch out. Make sure you pair the matching sides (same position in both shapes). Multiply when going from the small shape to the large one and divide when going from large to small.

Worked example 21.8.3

The two shapes are mathematically similar. The diagrams are drawn to scale. Work out the value of x.

Method. Similar shapes have corresponding sides in the same ratio. Find the scale factor from a pair of matching sides we know, then use it on the side we want.

Match the corresponding sides. In similar shapes the sides that are in the same position match. Their lengths are in the same ratio, and that ratio is the scale factor.
Known pair: 6 m (small shape) and 12 m (large shape).
The side we want is the one marked x.
Find the scale factor from the pair of sides we know. Scale factor = length in the large shape ÷ matching length in the small shape.
12 m ÷ 6 m = 2
So the large shape is 2 times as big as the small shape.
Use the scale factor to find x. x is on the large shape and we know the matching side on the small shape. Going from small to large we multiply by the scale factor.
x = 5 m × 2 = 10 m
Write the answer with its unit. Lengths need a unit.
x = 10 m

Check. Ratio check: 10 m ÷ 5 m = 2, the same scale factor as before. ✓

Watch out. Make sure you pair the matching sides (same position in both shapes). Multiply when going from the small shape to the large one and divide when going from large to small.

Worked example 21.8.4

In triangle ABC, D is on AB and E is on AC. DE is parallel to BC. The diagram is drawn to scale. Work out the value of x.

Method. Parallel lines make triangles ADE and ABC similar. Find the scale factor from a pair of sides we know, then use it on the side we need.

Explain why the triangles are similar. We may only use scale factors if the triangles are similar. Parallel lines give equal corresponding angles.
DE is parallel to BC, so angle ADE = angle ABC (corresponding angles) and angle AED = angle ACB (corresponding angles). Angle A is common to both triangles. All three angles match, so triangles ADE and ABC are similar.
Find AB. AB is the whole side. The small triangle ADE uses AD, and the large triangle ABC uses AB = AD + DB.
AB = AD + DB = 4 cm + 6 cm = 10 cm
Find the scale factor from the small triangle to the large triangle. Scale factor = length in ABC ÷ matching length in ADE. AB matches AD.
AB ÷ AD = 10 cm ÷ 4 cm = 52
Use it to find BC. BC matches DE. Going from the small triangle to the large one, multiply.
BC = DE × 52 = 8 cm × 52 = 20 cm
Write the answer with its unit.
x = 20 cm

Check. Ratio check: AD : AB = 4 : 10 and DE : BC = 8 : 20. Both equal 25 when written as a fraction. ✓

Watch out. The side AB of the large triangle is the whole side AD + DB. Do not use DB alone as the matching side for AD.

Worked example 21.8.5

AC and BD cross at X. AB is parallel to CD. The diagram is drawn to scale. Work out the value of x.

Method. The crossing lines make triangles XAB and XCD similar. Find the scale factor from a pair of sides we know, then use it on the side we need.

Explain why the triangles are similar. Parallel lines give equal alternate angles, and crossing lines give equal vertically opposite angles.
AB is parallel to CD, so angle XAB = angle XCD (alternate angles) and angle XBA = angle XDC (alternate angles). Angle AXB = angle CXD (vertically opposite angles). All three angles match, so triangles XAB and XCD are similar.
Find the scale factor from the small triangle to the large triangle. XA matches XC. Scale factor = XC ÷ XA.
4.5 cm ÷ 3 cm = 32
Use it to find CD. CD matches AB. Multiply by the scale factor.
CD = AB × 32 = 5 cm × 32 = 7.5 cm
Write the answer with its unit.
x = 7.5 cm

Check. Ratio check: CD ÷ AB = 7.5 cm ÷ 5 cm = 32. ✓

Watch out. Match the sides by position: XA matches XC, XB matches XD and AB matches CD. The small triangle is the one at the top.

Worked example 21.8.6

A man who is 1.8 m tall casts a shadow 2.4 m long. At the same time a tree casts a shadow 10 m long. Work out the height of the tree.

Method. The man and the tree make similar right-angled triangles with the sun’s rays, because the sun’s rays are parallel and both stand vertically. Use the scale factor from the shadows.

Explain why the triangles are similar. The man and the tree are both vertical, the ground is horizontal and the rays of the sun are parallel. So the triangles have equal angles.
Man’s triangle and tree’s triangle are similar.
Find the scale factor from the man’s shadow to the tree’s shadow. The shadows match each other. Scale factor = tree shadow ÷ man shadow.
10 ÷ 2.4 = 25 ÷ 6 = 256
Use it to find the height of the tree. The heights match each other. Multiply the man’s height by the scale factor.
1.8 × 256 = 45 ÷ 6 = 7.5 m

Check. Check the ratio the other way: 1.8 ÷ 2.4 = 0.75 and 7.5 ÷ 10 = 0.75. ✓

Watch out. Match shadow to shadow and height to height. Do not match the height of one with the shadow of the other.

Worked example 21.8.7

Triangles ABC and PQR are similar. AB = 6 cm and PQ = 9 cm. BC = x cm and QR = (x + 4) cm. Work out x.

Method. Write the scale factor two ways: from the lengths we know and from the lengths with x. They must be equal, so we get an equation to solve.

Find the scale factor from AB to PQ. Scale factor = PQ ÷ AB.
9 ÷ 6 = 1.5
Write QR in terms of BC using the scale factor. QR matches BC, and every length on PQR is 1.5 times the matching length on ABC.
x + 4 = 1.5 × x
Solve the equation. Subtract x from both sides, then divide by 0.5.
4 = 1.5x − x = 0.5x
x = 4 ÷ 0.5 = 8

Check. BC = 8 cm and QR = 12 cm, and 12 ÷ 8 = 1.5. ✓

Watch out. Write the scale factor as a number first, then form the equation. Do not mix the two triangles up.

Watch out

Match the corresponding sides carefully: the side opposite the same angle, or in the same position. In the nested triangle, the side matching AD is AB (the whole side), not DB. Also remember: from the small shape to the large shape multiply; from the large shape to the small shape divide.

21.9

Area and volume scale factors

Edexcel 1MA1: G19, R12
Grade 4-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Area and volume scale factorsComing soon

When a shape is enlarged by a scale factor k, every length is multiplied by k. What happens to the area and the volume? Look at squares. A square with side 1 has area 1. If the side is doubled to 2, the square holds 2 × 2 = 4 of the little squares, so the area is multiplied by 4 (not 2). If the side is trebled to 3, the area is 3 × 3 = 9: multiplied by 9.

Doubling the side of a square multiplies its area by 2² = 4. Trebling multiplies it by 3² = 9.

For cubes it is the same idea in three directions. A cube with side 2 is made of 2 × 2 × 2 = 8 little cubes, so doubling the lengths multiplies the volume by 8. Trebling the lengths gives 3 × 3 × 3 = 27 little cubes.

Cubes with side 1, 2 and 3 have volumes 1, 8 and 27: the volume is multiplied by k³.

This works for any similar shapes, not only squares and cubes. If the length scale factor is k:

Quantity	Scale factor
Lengths (sides, perimeters, heights, radii)	k
Areas (surface area, cross-section area)	k²
Volumes (also capacity and, for the same material, mass)	k³

Using ratio notation. If the lengths of two similar shapes are in the ratio a : b, then their areas are in the ratio a² : b² and their volumes in the ratio a³ : b³. For example, if two similar vases have heights in the ratio 2 : 3, their surface areas are in the ratio 4 : 9 and their volumes in the ratio 8 : 27.

Working backwards. If you are given the area (or volume) scale factor, take the square root (or cube root) to find the length scale factor. For example, if the areas are in the ratio 9 : 25, the lengths are in the ratio 3 : 5, and the volumes are in the ratio 27 : 125.

Key ideas

If the length scale factor is k, the area scale factor is k² and the volume scale factor is k³.
Length ratio a : b gives area ratio a² : b² and volume ratio a³ : b³.
Higher: to find k from areas take the square root; from volumes take the cube root.
This works for similar shapes only.
Surface areas use k² and capacities use k³.

Worked example 21.9.1

Two similar triangles have a length scale factor of 3. The area of the smaller triangle is 12 cm². Work out the area of the larger triangle.

Key words. The length scale factor k says how many times longer every length becomes. The area scale factor is k² and the volume scale factor is k³. These work only for similar shapes or solids.

Method. For similar shapes, if the lengths are multiplied by k, the areas are multiplied by k².

Write down the length scale factor. It is the number that every length is multiplied by.
k = 3
Work out the area scale factor. Area has two lengths in it (length × width), so it is multiplied by k twice: k².
area scale factor = 3² = 3 × 3 = 9
Multiply the original area by the area scale factor. The new value is the old value multiplied by the scale factor.
12 × 9 = 108 cm²

Check. Divide back: 108 ÷ 9 = 12. ✓

Watch out. Do not multiply the area or volume by k. Area uses k², volume uses k³.

Worked example 21.9.2

Two similar solids have a length scale factor of 2. The volume of the smaller solid is 15 cm³. Work out the volume of the larger solid.

Method. For similar shapes, if the lengths are multiplied by k, the volumes are multiplied by k³.

Write down the length scale factor. It is the number that every length is multiplied by.
k = 2
Work out the volume scale factor. Volume has three lengths in it (length × width × height), so it is multiplied by k three times: k³.
volume scale factor = 2³ = 2 × 2 × 2 = 8
Multiply the original volume by the volume scale factor. The new value is the old value multiplied by the scale factor.
15 × 8 = 120 cm³

Check. Divide back: 120 ÷ 8 = 15. ✓

Watch out. Do not multiply the area or volume by k. Area uses k², volume uses k³.

Worked example 21.9.3

The lengths of two similar shapes are in the ratio 2 : 5. Write down the ratio of (a) their areas, (b) their volumes.

Method. If the lengths of two similar shapes are in the ratio a : b, then the areas are in the ratio a² : b² and the volumes are in the ratio a³ : b³.

Write the length ratio in simplest form. Divide both numbers by their highest common factor.
Lengths: 2 : 5
Square both numbers for the area ratio. The area scale factor is the square of the length scale factor.
Areas: 2² : 5² = 4 : 25
Cube both numbers for the volume ratio. The volume scale factor is the cube of the length scale factor.
Volumes: 2³ : 5³ = 8 : 125

Check. Test with a = 2, b = 5: a cube of side 2 has volume 8 and a cube of side 5 has volume 125. ✓

Watch out. Square (or cube) each number in the ratio, not just one of them.

Worked example 21.9.4

Two similar shapes have areas 20 cm² and 80 cm². A side of the smaller shape is 7 cm. Work out the length of the matching side of the larger shape.

Method. The areas are in the ratio k². Find the ratio of the areas, then take the square root to get the length scale factor k.

Find the area scale factor. Area scale factor = new area ÷ old area.
80 ÷ 20 = 4
Take the square root to find the length scale factor. The area scale factor is k², so k is its square root.
k = √4 = 2
Use k to find the new length. Lengths are multiplied by the length scale factor k.
7 × 2 = 14 cm

Check. Put it back: k² = 2² = 4, and 20 × 4 = 80. ✓

Watch out. The square root goes from area to length. Do not use the area ratio as the length scale factor.

Worked example 21.9.5

The areas of two similar shapes are in the ratio 9 : 25. Work out the ratio of their volumes.

Method. Go from the area ratio to the length ratio, then to the volume ratio.

Find the length ratio. The area ratio is the length ratio raised to the power 2, so take the square root of each number.
Areas 9 : 25
Lengths: √9 : √25 = 3 : 5
Raise the length ratio to the power 3. The volume ratio is the length ratio raised to the power 3.
3³ : 5³ = 27 : 125
Write the answer. Name what the ratio is for.
Volumes: 27 : 125

Check. Go back: 27 : 125 with the cube root gives 3 : 5, the length ratio. ✓

Watch out. You cannot go straight from area to volume by multiplying: always go through the length ratio.

Watch out

Do not multiply the area or the volume by the length scale factor. If the lengths double, the area goes up by a factor of 4 and the volume by a factor of 8. And when you are given an area ratio, take the square root to get the length ratio, not half of it.

Chapter summary

Skill	Method
Reflection	Mirror line, same distance on the other side. Describe with the equation of the mirror line. y = x swaps the coordinates; y = −x swaps and changes both signs.
Rotation	Angle + direction + centre. Write each vertex as a move from the centre and turn it: 90° clockwise (a, b) → (b, −a); 90° anticlockwise (−b, a); 180° (−a, −b).
Translation	Column vector: top = across (right +), bottom = up/down (up +). Vector = image − object.
Enlargement	Scale factor + centre. Multiply the distance from the centre by the scale factor. A fraction makes the shape smaller; a negative scale factor puts the image on the other side of the centre (Higher).
Describe fully	Name the type and give its detail: line / angle, direction, centre / vector / scale factor, centre.
Combinations (Higher)	Do them in order. Two reflections in parallel lines = translation; in crossing lines = rotation. Invariant point: unchanged by the transformation.
Congruent triangles	SSS, SAS (included angle), ASA / AAS, RHS. AAA and SSA are not enough.
Similar shapes	Equal angles, sides in the same ratio. Scale factor = large ÷ small. Small to large: multiply. Large to small: divide.
Area and volume	Length scale factor k: area × k², volume × k³. Ratio a : b gives a² : b² and a³ : b³. Take roots to go back.

Can you…?

I can reflect a shape in x = a, y = b, y = x and y = −x.
I can rotate a shape about any centre through 90° or 180° and describe a rotation fully.
I can translate a shape using a column vector and find the vector of a translation.
I can enlarge a shape with a whole-number or fractional scale factor from any centre, and describe an enlargement fully.
I can describe fully any single transformation, and I know which are congruent and which are similar.
Higher: I can use negative scale factors, combine transformations and find invariant points.
I can decide whether two triangles are congruent using SSS, SAS, ASA or RHS and give a reason.
I can find a missing length in similar shapes using a scale factor, including similar triangles from parallel lines.
I can use k, k² and k³ to compare lengths, areas and volumes of similar shapes.
Higher: I can prove that two triangles are congruent or similar.

Key terms

Transformation: A change to a shape: reflection, rotation, translation or enlargement.
Object and image: The object is the original shape; the image is the shape after the transformation.
Mirror line: The line a shape is reflected in. Each point and its image are the same distance from it.
Centre of rotation: The fixed point a shape is turned about.
Column vector: Two numbers written one above the other. The top number is the move across and the bottom number is the move up or down.
Scale factor: The number every length is multiplied by in an enlargement or between similar shapes.
Centre of enlargement: The fixed point that distances are measured from in an enlargement.
Invariant point: A point that does not move when a transformation is carried out.
Congruent: Exactly the same size and shape (one can be placed on top of the other).
Similar: The same shape but not necessarily the same size: equal angles and sides in the same ratio.
Corresponding: In the same position in two shapes (sides or angles that match).
Included angle: The angle between two given sides.
Hypotenuse: The longest side of a right-angled triangle, opposite the right angle.

Total 46 marks. Allow about 46 minutes.

← Chapter 20: Pythagoras and trigonometry Chapter 22: Constructions, loci, bearings and vectors →

GCSE Maths Transformations, Congruence and Similarity (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

You should already know

Where this leads

Exercise P Do you remember?

Reflection

Key ideas

Watch out

Exercise 21A Reflection

Rotation

Key ideas

Watch out

Exercise 21B Rotation

Translation

Key ideas

Watch out

Exercise 21C Translation

Enlargement

Key ideas

Watch out

Exercise 21D Enlargement

Describing and identifying single transformations

Key ideas

Watch out

Exercise 21E Describing single transformations

Combinations of transformations and invariant points

Key ideas

Watch out

Exercise 21F Combinations and invariant points

Congruent shapes and triangles

Key ideas

Watch out

Exercise 21G Congruent triangles

Exercise 21H Congruence: more practice and proof

Similar shapes and scale factors

Key ideas

Watch out

Exercise 21I Similar shapes and missing lengths

Area and volume scale factors

Key ideas

Watch out

Exercise 21J Area and volume scale factors

Exercise 21K Area and volume: problem solving

Chapter summary

Can you…?

Key terms

Exercise 21Z Check-up (non-calculator)

Exercise 21X Exam-style questions