Level 4 · Mathematics M4.4

Similarity, transformations and circles

Move, turn, flip and enlarge shapes, and learn the angle rules hidden in circles.

Reflect, rotate, translate and enlarge shapes on a grid and describe each move exactly. Prove triangles congruent, use similar shapes to find missing lengths, and see how areas and volumes grow with the scale factor. Then find arc lengths and sector areas, and use the angle rules of the circle.

You are ready for this book if

You can reflect a shape in the y axis on a grid.

Written and drawn by attnom. Free to read, print and share.

M4.4.1 Chapter 1

Reflections and rotations

I flip a shape over the line y = x. Where does the corner at (1, 3) go?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

A shape on a grid can be flipped, turned, slid or made bigger. Each of these is a transformation. This chapter has the flip and the turn. The next two chapters have the slide and the enlargement.

A reflection flips a shape over a mirror line. The new shape is called the image. Each point and its image are the same distance from the mirror line, on opposite sides. The line joining them meets the mirror line at a right angle.

Draw it first

Reflect triangle A in the line x = 1

  1. A grid with an x axis and a y axis from −5 to 5. Step 1: triangle A with corners at (3, 1), (5, 2) and (3, 4), labelled A, and the dashed mirror line x = 1. Step 2: dashed lines from each corner straight across the mirror line. Step 3: the image, triangle A′, with corners at (−1, 1), (−3, 2) and (−1, 4), labelled A′.−4−224−4−2240xyx = 1A

    1 Triangle A has corners at (3, 1), (5, 2) and (3, 4). Draw the mirror line x = 1. Every point on it has x equal to 1.

  2. A grid with an x axis and a y axis from −5 to 5. Step 1: triangle A with corners at (3, 1), (5, 2) and (3, 4), labelled A, and the dashed mirror line x = 1. Step 2: dashed lines from each corner straight across the mirror line. Step 3: the image, triangle A′, with corners at (−1, 1), (−3, 2) and (−1, 4), labelled A′.−4−224−4−2240xyx = 1A

    2 From each corner, draw a line straight across the mirror. Count how far the corner is from the mirror.

  3. A grid with an x axis and a y axis from −5 to 5. Step 1: triangle A with corners at (3, 1), (5, 2) and (3, 4), labelled A, and the dashed mirror line x = 1. Step 2: dashed lines from each corner straight across the mirror line. Step 3: the image, triangle A′, with corners at (−1, 1), (−3, 2) and (−1, 4), labelled A′.−4−224−4−2240xyx = 1AA′

    3 Go the same distance past the mirror. Join the new points to make the image, A′.

The corner (3, 1) is 2 units from the line x = 1. So its image is 2 units on the other side, at (−1, 1). A point that lies on the mirror line is its own image.

Now take a mirror line that slants. The line y = x passes through (1, 1), (2, 2) and (3, 3). Every point on it has equal x and y.

A grid with an x axis and a y axis from 0 to 6. The dashed mirror line y = x slants up to the right. Triangle PQR has corners P at (1, 3), Q at (1, 5) and R at (2, 3), above the line. Its image, triangle P′Q′R′, is below the line, with P′ at (3, 1), Q′ at (5, 1) and R′ at (3, 2). The labels are P, Q, R, P′, Q′ and R′.1234561234560xyy = xPQRP′Q′R′
The mirror line y = x. Triangle PQR and its image lie on opposite sides.

The line y = x makes the same angle with both axes. Folding along it swaps the two axes, so a distance across becomes a distance up. A point (a, b) goes to the point (b, a). The numbers swap.

A grid with an x axis and a y axis from 0 to 6. The dashed mirror line y = x. Triangle A has corners (2, 4), (2, 6) and (3, 6), labelled A. Its image has corners (4, 2), (6, 2) and (6, 3), labelled A′.1234561234560xyy = xAA′
The triangle with corners (2, 4), (2, 6) and (3, 6), and its image in y = x.

Worked example

Reflect the triangle with corners (2, 4), (2, 6) and (3, 6) in the line y = x.

  1. Swap the two numbers of each corner.
  2. The corner (2, 4) goes to (4, 2).
  3. The corner (2, 6) goes to (6, 2).
  4. The corner (3, 6) goes to (6, 3).

Answer: (4, 2), (6, 2) and (6, 3)

A rotation turns a shape about one fixed point. That point is the centre of rotation. A turn can be clockwise, the way the hands of a clock move, or anticlockwise, the other way. A half turn of 180° needs no direction, because both ways end in the same place.

Draw it first

Turn triangle A by 90° clockwise about O

  1. A grid with an x axis and a y axis from −5 to 5. Step 1: triangle A with corners at (1, 1), (4, 1) and (1, 3), labelled A, and the centre O at (0, 0), a dashed line from O to the corner (4, 1), an arc with an arrowhead running clockwise down to the point (1, −4), and the label 90°. Step 2: the image, triangle A′, with corners at (1, −1), (1, −4) and (3, −1), labelled A′.−4−224−4−2240xyO90°A

    1 Triangle A has corners at (1, 1), (4, 1) and (1, 3). The centre of rotation is O at (0, 0). Take the corner (4, 1). Join it to O, then turn that line 90° clockwise. The corner lands at (1, −4).

  2. A grid with an x axis and a y axis from −5 to 5. Step 1: triangle A with corners at (1, 1), (4, 1) and (1, 3), labelled A, and the centre O at (0, 0), a dashed line from O to the corner (4, 1), an arc with an arrowhead running clockwise down to the point (1, −4), and the label 90°. Step 2: the image, triangle A′, with corners at (1, −1), (1, −4) and (3, −1), labelled A′.−4−224−4−2240xyO90°AA′

    2 Turn every corner the same way. Join the new corners to make the image, A′.

A turn keeps every point the same distance from the centre. The shape does not change size and does not flip over.

A grid with an x axis and a y axis from −5 to 5. The point P at (3, 1) and the point P′ at (1, −3). A dashed line from O at (0, 0) to each point, and an arc with an arrowhead running clockwise from P to P′. The labels are P, P′, O and 90°.−4−224−4−2240xyPP′O90°
P is 3 right and 1 up from O. After a 90° clockwise turn, P′ is 1 right and 3 down.

Look at P. Going 3 right and 1 up from O becomes going 1 right and 3 down. The right step turned into a down step, and the up step turned into a right step. So a 90° clockwise turn about O sends (x, y) to (y, −x).

A 90° anticlockwise turn about O sends (x, y) to (−y, x). A half turn of 180° about O sends (x, y) to (−x, −y). Each is the same step, turned by the angle.

A grid with an x axis and a y axis from −5 to 5. The centre C at (2, 1). Triangle A with corners at (3, 2), (5, 2) and (5, 5), labelled A. A dashed line from C to the corner (5, 2), an arc with an arrowhead running anticlockwise up to (1, 4), and the label 90°. The image, triangle A′, has corners at (1, 2), (1, 4) and (−2, 4), labelled A′.−4−224−4−2240xyC90°AA′
Triangle A turned 90° anticlockwise about C at (2, 1).

Worked example

Turn the triangle with corners (3, 2), (5, 2) and (5, 5) by 90° anticlockwise about C at (2, 1).

  1. Find how far each corner is from C. The corner (3, 2) is 1 right and 1 up. The corner (5, 2) is 3 right and 1 up. The corner (5, 5) is 3 right and 4 up.
  2. A 90° anticlockwise turn changes "right r, up u" into "left u, up r". So (3, 2) is now 1 left and 1 up from C.
  3. The corner (3, 2) goes to (1, 2). The corner (5, 2) goes to (1, 4). The corner (5, 5) goes to (−2, 4).

Answer: (1, 2), (1, 4) and (−2, 4)

To describe a rotation, say all three things: the centre, the angle and the direction. For a half turn, the direction can be left out. "90° clockwise about (0, 0)" is complete. "90° clockwise" is not.

A grid with an x axis and a y axis from −5 to 5. Triangle A with corners at (1, 1), (4, 1) and (1, 3), labelled A. A dashed straight line from the corner (4, 1) through the centre O at (0, 0) to the corner (−4, −1), with an arc running round and the label 180°. The image, triangle A′, has corners at (−1, −1), (−4, −1) and (−1, −3), labelled A′.−4−224−4−2240xyO180°AA′
A half turn about O. The same image comes from turning clockwise or anticlockwise.

See it three ways

With things

Fold a square sheet along its diagonal. Hold it to the light. A dot on one half shows through on the other half, exactly where its reflection is.

As a picture

A grid with an x axis and a y axis from 0 to 6. The dashed mirror line y = x. The point P at (1, 3) above the line and the point P′ at (3, 1) below it, joined by a dashed line that crosses the mirror line at a right angle.1234561234560xyy = xPP′

In symbols

A reflection in y = x sends (a, b) to (b, a). A 90° clockwise turn about O sends (x, y) to (y, −x).

Try it

Fold along y = x

You need: a square sheet of thin paper, a pencil, a ruler.

  1. Draw an x axis and a y axis along two edges of the paper, with the origin in one corner. Mark every 1 cm.
  2. Draw the line from the origin to the opposite corner. That is the line y = x.
  3. Mark the point (1, 3) with a dot. Fold the paper along y = x, so the two halves lie on top of each other.
  4. Hold the folded sheet up to the light. Put a second dot on the top half, exactly over the dot that shows through.

What you will see: Open the sheet. The second dot is at (3, 1). The fold swapped the two numbers.

If you cannot try it, this is what happens.
A table with the head Dot, Before and After, for the fold. Row 1: P, (1, 3) and (3, 1).DotBeforeAfterP(1, 3)(3, 1)

A wrong idea many people have

A 90° turn about (0, 0) gives the same picture as a reflection in y = x.

Why it feels right: Both moves keep the size of the shape, and both put the numbers of a point in a new order.

The test that breaks it: Take the corner (1, 3). Turn it 90° clockwise about (0, 0) and it goes to (3, −1). Reflect it in y = x and it goes to (3, 1). These are different points.

What is true: A reflection flips the shape over, so its corners run round the other way. A rotation does not flip it. They are different transformations.

Key words

transformation
a rule that moves a shape: a flip, a turn, a slide or an enlargement.
reflection
a flip of a shape over a mirror line.
mirror line
the line a shape is flipped over. A point and its image are the same distance from it.
image
the shape that a transformation makes.
rotation
a turn of a shape about a fixed point.
centre of rotation
the point that stays still while the shape turns.
clockwise
in the direction the hands of a clock move.
anticlockwise
in the opposite direction to the hands of a clock.

Back to your guess

The answer is B: To (3, 1). The two numbers swap.

The line y = x runs through every point with equal x and y, such as (2, 2). A flip over it swaps the two numbers, so (1, 3) goes to (3, 1). The other two answers are flips over the x axis and over the y axis.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. Triangle A has a corner at (2, 5). A is reflected in the line y = x. Where is that corner now?

    Tap your answer.

    1. Not this one. You changed the sign of x. That is a reflection in the y axis. A reflection in y = x swaps the two numbers. Go to M4.4.1. Try another.

    2. Yes, that is right. The mirror line y = x swaps the two numbers of every point. The corner (2, 5) goes to (5, 2).

    3. Not this one. You changed the sign of y. That is a reflection in the x axis. A reflection in y = x swaps the two numbers. Go to M3.9.3. Try another.

    See the answer

    (5, 2)

    The mirror line y = x swaps the two numbers of every point. The corner (2, 5) goes to (5, 2).

    • If you chose A: You changed the sign of x. That is a reflection in the y axis. A reflection in y = x swaps the two numbers. Go to M4.4.1.
    • If you chose C: You changed the sign of y. That is a reflection in the x axis. A reflection in y = x swaps the two numbers. Go to M3.9.3.

  2. Triangle A is reflected in the mirror line x = 3. Its corner at (1, 2) is marked. Where is the image of that corner?

    A grid with an x axis and a y axis from −5 to 5. The dashed mirror line x = 3. Triangle A with corners at (1, 2), (2, 1) and (2, 4), labelled A, on the left of the mirror line. A dot marks the corner at (1, 2).−4−224−4−2240xyx = 3A
    Tap your answer.

    1. Not this one. That is the reflection in the y axis, the line x = 0. The mirror here is the line x = 3. Go to M4.4.1. Try another.

    2. Not this one. The image is not on the mirror line. It is as far past the line as the corner is before it, which is 2 units. Go to M4.4.1. Try another.

    3. Yes, that is right. The corner (1, 2) is 2 units left of the line x = 3. Its image is 2 units right of the line, at (5, 2). The y number stays the same.

    See the answer

    (5, 2)

    The corner (1, 2) is 2 units left of the line x = 3. Its image is 2 units right of the line, at (5, 2). The y number stays the same.

    • If you chose A: That is the reflection in the y axis, the line x = 0. The mirror here is the line x = 3. Go to M4.4.1.
    • If you chose B: The image is not on the mirror line. It is as far past the line as the corner is before it, which is 2 units. Go to M4.4.1.

  3. Triangle A is turned 90° clockwise about O at (0, 0). Where does the corner (4, 1) go?

    A grid with an x axis and a y axis from −5 to 5. Triangle A with corners at (1, 1), (4, 1) and (1, 3), labelled A, and the point O at (0, 0).−4−224−4−2240xyOA
    Tap your answer.

    1. Yes, that is right. A 90° clockwise turn about O sends (x, y) to (y, −x). So (4, 1) goes to (1, −4).

    2. Not this one. You turned it the other way, anticlockwise. Clockwise is the way the hands of a clock move. Go to M4.4.1. Try another.

    3. Not this one. That is a half turn, 180°. A quarter turn of 90° is half as much. Go to M4.4.1. Try another.

    See the answer

    (1, −4)

    A 90° clockwise turn about O sends (x, y) to (y, −x). So (4, 1) goes to (1, −4).

    • If you chose B: You turned it the other way, anticlockwise. Clockwise is the way the hands of a clock move. Go to M4.4.1.
    • If you chose C: That is a half turn, 180°. A quarter turn of 90° is half as much. Go to M4.4.1.

  4. Triangle A is moved to triangle B by a rotation. Which rotation is it?

    A grid with an x axis and a y axis from −5 to 5. Triangle A with corners at (1, 1), (3, 1) and (1, 4), labelled A, and triangle B with corners at (−1, 1), (−1, 3) and (−4, 1), labelled B. The point O at (0, 0).−4−224−4−2240xyOAB
    Tap your answer.

    1. Not this one. Clockwise would put the image below the x axis. B is above it, so the turn is the other way. Go to M4.4.1. Try another.

    2. Not this one. A half turn would put B below and to the left of O, in the opposite corner of the grid. Go to M4.4.1. Try another.

    3. Yes, that is right. A turn of 90° anticlockwise about O sends (x, y) to (−y, x). The corner (1, 1) goes to (−1, 1) and (3, 1) goes to (−1, 3). Those are corners of B.

    4. Not this one. The angle and the direction are right, but the centre is not. Every point of A is turned about O, and O stays still. Go to M4.4.1. Try another.

    See the answer

    90° anticlockwise about O

    A turn of 90° anticlockwise about O sends (x, y) to (−y, x). The corner (1, 1) goes to (−1, 1) and (3, 1) goes to (−1, 3). Those are corners of B.

    • If you chose A: Clockwise would put the image below the x axis. B is above it, so the turn is the other way. Go to M4.4.1.
    • If you chose B: A half turn would put B below and to the left of O, in the opposite corner of the grid. Go to M4.4.1.
    • If you chose D: The angle and the direction are right, but the centre is not. Every point of A is turned about O, and O stays still. Go to M4.4.1.

  5. Which description of a rotation is complete?

    Tap your answer.

    1. Not this one. This gives the angle and direction but not the centre. Different centres give different images. Go to M4.4.1. Try another.

    2. Not this one. This gives the centre but not the angle or the direction. Go to M4.4.1. Try another.

    3. Not this one. This gives the angle and the centre but not the direction. A 90° turn can go either way. Go to M4.4.1. Try another.

    4. Yes, that is right. A rotation needs a centre, an angle and a direction. Only the last choice gives all three.

    See the answer

    90° clockwise about (2, 1)

    A rotation needs a centre, an angle and a direction. Only the last choice gives all three.

    • If you chose A: This gives the angle and direction but not the centre. Different centres give different images. Go to M4.4.1.
    • If you chose B: This gives the centre but not the angle or the direction. Go to M4.4.1.
    • If you chose C: This gives the angle and the centre but not the direction. A 90° turn can go either way. Go to M4.4.1.

  6. A point is reflected in the line y = x. It starts at (7, 2). What is the x number of the image?

    See the answer

    2

    A reflection in y = x swaps the two numbers. (7, 2) goes to (2, 7), so the x number is 2.

  7. A corner at (1, 4) is reflected in the line x = 3. What is the x number of the image?

    See the answer

    5

    The corner is 2 units left of the line x = 3. The image is 2 units right of it, so its x number is 3 + 2 = 5.

  8. A point is 4 units right of the centre C and 6 units above it. It makes a half turn about C. How many units below C is the image?

    See the answer

    6

    A half turn changes right into left and up into down. The image is 4 units left of C and 6 units below it.

  9. A point at (2, 6) is turned 90° clockwise about O at (0, 0). How many units below the x axis is the image?

    See the answer

    2

    A 90° clockwise turn about O sends (x, y) to (y, −x). So (2, 6) goes to (6, −2), which is 2 units below the x axis.

Keep it

Copy this into your exercise book.

A reflection flips a shape over a mirror line. In y = x the two numbers swap. A rotation needs a centre, an angle and a direction. A half turn needs no direction.

Draw: A triangle and its image in the line y = x, with the corners named.

M4.4.2 Chapter 2

Translations and combined moves

A shape slides 3 squares right and 2 squares up. How do I write that move with numbers?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

A translation slides a shape. It does not turn the shape, flip it or change its size. Every point of the shape moves the same distance in the same direction. So the image faces the same way as the shape.

Draw it first

Slide triangle A 3 right and 2 up

  1. A grid with an x axis from 0 to 8 and a y axis from 0 to 6. Step 1: triangle A with corners at (1, 1), (4, 1) and (1, 3), labelled A. Step 2: an arrow from the corner (4, 1) to the right to (7, 1), labelled 3 right, and an arrow from there up to (7, 3), labelled 2 up. Step 3: an arrow from each corner of A to the matching corner of the image. Step 4: the image, triangle A′, with corners at (4, 3), (7, 3) and (4, 5), labelled A′.123456781234560xyA

    1 Triangle A has corners at (1, 1), (4, 1) and (1, 3).

  2. A grid with an x axis from 0 to 8 and a y axis from 0 to 6. Step 1: triangle A with corners at (1, 1), (4, 1) and (1, 3), labelled A. Step 2: an arrow from the corner (4, 1) to the right to (7, 1), labelled 3 right, and an arrow from there up to (7, 3), labelled 2 up. Step 3: an arrow from each corner of A to the matching corner of the image. Step 4: the image, triangle A′, with corners at (4, 3), (7, 3) and (4, 5), labelled A′.123456781234560xy3 right2 upA

    2 Take one corner. Move it 3 squares right, then 2 squares up.

  3. A grid with an x axis from 0 to 8 and a y axis from 0 to 6. Step 1: triangle A with corners at (1, 1), (4, 1) and (1, 3), labelled A. Step 2: an arrow from the corner (4, 1) to the right to (7, 1), labelled 3 right, and an arrow from there up to (7, 3), labelled 2 up. Step 3: an arrow from each corner of A to the matching corner of the image. Step 4: the image, triangle A′, with corners at (4, 3), (7, 3) and (4, 5), labelled A′.123456781234560xy3 right2 upA

    3 Move every other corner the same way. The three arrows are parallel and equally long.

  4. A grid with an x axis from 0 to 8 and a y axis from 0 to 6. Step 1: triangle A with corners at (1, 1), (4, 1) and (1, 3), labelled A. Step 2: an arrow from the corner (4, 1) to the right to (7, 1), labelled 3 right, and an arrow from there up to (7, 3), labelled 2 up. Step 3: an arrow from each corner of A to the matching corner of the image. Step 4: the image, triangle A′, with corners at (4, 3), (7, 3) and (4, 5), labelled A′.123456781234560xy3 right2 upAA′

    4 Join the new corners. This is the image, A′.

We write a slide as a column vector: two numbers in brackets, one above the other. The top number is the move right. The bottom number is the move up. A minus sign means left, for the top number, or down, for the bottom number.

Two column vectors in brackets. The first has 3 on top and 2 below, and is labelled 3 right, 2 up. The second has −2 on top and 1 below, and is labelled 2 left, 1 up.323 right, 2 up−212 left, 1 up
The top number is the move right. The bottom number is the move up.
Two column vectors in brackets. The first has 0 on top and −4 below, and is labelled 4 down. The second has 1 on top and −1 below, and is labelled 1 right, 1 down.0−44 down1−11 right, 1 down
A zero means no move in that direction. A minus sign means left or down.

To translate a point by a vector, add the top number to x and the bottom number to y. A vector says how far to move. It does not say where to land.

A grid with an x axis from 0 to 8 and a y axis from 0 to 7. Triangle A with corners at (4, 1), (7, 1) and (7, 4), labelled A. An arrow from the corner (4, 1) to the corner (1, 3) of the image, labelled 3 left, 2 up. The image, triangle A′, with corners at (1, 3), (4, 3) and (4, 6), labelled A′.1234567812345670xy3 left, 2 upAA′
Triangle A and its image after a slide of 3 left and 2 up.

Worked example

Translate the triangle with corners (4, 1), (7, 1) and (7, 4) by the vector with −3 on top and 2 below.

  1. The top number is −3, so move every corner 3 left. The bottom number is 2, so move every corner 2 up.
  2. The corner (4, 1) goes to (1, 3).
  3. The corner (7, 1) goes to (4, 3).
  4. The corner (7, 4) goes to (4, 6).

Answer: (1, 3), (4, 3) and (4, 6)

Now go the other way. Pick one corner of the shape and its image. The move right is the new x minus the old x. The move up is the new y minus the old y.

A grid with an x axis from 0 to 10 and a y axis from 0 to 8. Triangle A with corners at (1, 7), (4, 7) and (1, 4), labelled A. An arrow from the corner (4, 7) to the right to (9, 7), labelled 5 right, and an arrow from there down to (9, 4), labelled 3 down. The image, triangle A′, has corners at (6, 4), (9, 4) and (6, 1), labelled A′.12345678910123456780xy5 right3 downAA′
The corner (4, 7) moves to (9, 4).

Worked example

The corner (4, 7) of triangle A moves to (9, 4) in the image. Write the translation as a column vector.

  1. Pick one corner and its image. The corner is (4, 7) and its image is (9, 4).
  2. The move right is the new x minus the old x: from 4 to 9 is 5.
  3. The move up is the new y minus the old y: from 7 to 4 is −3. That is 3 down.
  4. The column vector has 5 on top and −3 below.

Answer: 5 on top and −3 below

Two slides in a row make one slide. Add the top numbers and add the bottom numbers. Two moves one after the other are a combined move. Take the vector with 3 on top and 1 below, then the vector with −5 on top and 3 below. Together they make one slide of 2 left and 4 up.

A grid with an x axis from 0 to 10 and a y axis from 0 to 8. Triangle A with corners at (5, 1), (7, 1) and (5, 3), labelled A. Triangle B with corners at (8, 2), (10, 2) and (8, 4), labelled B. Triangle C with corners at (3, 5), (5, 5) and (3, 7), labelled C. An arrow from A to B, labelled first, an arrow from B to C, labelled then, and a dashed arrow from A to C, labelled total.12345678910123456780xyfirstthentotalABC
A moves to B, then B moves to C. One slide takes A straight to C.

Worked example

A shape moves by the vector with 3 on top and 1 below. Then it moves by the vector with −5 on top and 3 below. Find one translation that does the same.

  1. Add the top numbers: 3 and −5 make −2.
  2. Add the bottom numbers: 1 and 3 make 4.
  3. The one move has −2 on top and 4 below.
  4. Check with the corner (5, 1). The first move takes it to (8, 2).
  5. The second move takes (8, 2) to (3, 5). The one move takes (5, 1) to (3, 5) as well.

Answer: −2 on top and 4 below

See it three ways

With things

Slide a book across a table: 3 hand spans to the right and 2 hand spans away from you. Every corner of the book moves the same way.

As a picture

One column vector in brackets with 3 on top and 2 below, labelled 3 right, 2 up.323 right, 2 up

In symbols

The vector with 3 on top and 2 below moves the point (x, y) to (x + 3, y + 2).

Try it

Slide a tracing

You need: squared paper, thin paper to trace on, a pencil.

  1. Draw a small triangle on the squared paper. Trace it on the thin paper.
  2. Slide the tracing 3 squares right and 2 squares up. Keep its edges parallel to the grid lines. Do not turn it.
  3. Press through, or draw round the tracing, to mark the image.
  4. Join each corner of the triangle to the matching corner of the image.

What you will see: The three lines are parallel and have the same length. Every corner moved by the same vector.

If you cannot try it, this is what happens.
A grid with an x axis from 0 to 8 and a y axis from 0 to 7. Triangle A with corners at (1, 1), (4, 1) and (1, 4), labelled A. The image, triangle A′, with corners at (4, 3), (7, 3) and (4, 6), labelled A′. A dashed arrow from each corner of A to the matching corner of A′.1234567812345670xyAA′

A wrong idea many people have

The vector with 3 on top and 2 below moves a shape to the point (3, 2).

Why it feels right: It has the same two numbers as the point, and a point is written with the x number first.

The test that breaks it: Take the corner (1, 1) and the vector with 3 on top and 2 below. Add 3 to x and 2 to y. The corner goes to (4, 3), not to (3, 2).

What is true: A column vector says how far to move: right, then up. It does not say where to land. To find the image, add the vector to each corner.

Key words

translation
a slide of a shape. Every point moves the same distance in the same direction.
column vector
two numbers in brackets, one above the other. The top number is the move right and the bottom number is the move up.
combined move
two moves one after the other. For two translations it is one translation.

Back to your guess

The answer is B: With two numbers in a column: 3 on top and 2 below.

A slide has two parts, a move right and a move up, so it needs two numbers. By agreement the move right goes on top and the move up goes below. One number cannot tell the two parts apart.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. Triangle A moves to triangle B by a translation. Which column vector describes the move?

    A grid with an x axis from 0 to 9 and a y axis from 0 to 8. Triangle A with corners at (1, 4), (4, 4) and (1, 7), labelled A, and triangle B with corners at (5, 2), (8, 2) and (5, 5), labelled B.123456789123456780xyAB
    Tap your answer.

    1. Not this one. You wrote the numbers in the wrong order. The top number is the move right, which is 4. The bottom number is the move up or down. Go to M4.4.2. Try another.

    2. Yes, that is right. The corner (1, 4) moves to (5, 2). That is 4 right and 2 down, so the top number is 4 and the bottom number is −2.

    3. Not this one. You described the move from B back to A. The vector must go from the shape to its image. Go to M4.4.2. Try another.

    4. Not this one. The shape moves down, not up. A move down needs a minus sign on the bottom number. Go to M4.4.2. Try another.

    See the answer

    top number 4, bottom number −2

    The corner (1, 4) moves to (5, 2). That is 4 right and 2 down, so the top number is 4 and the bottom number is −2.

    • If you chose A: You wrote the numbers in the wrong order. The top number is the move right, which is 4. The bottom number is the move up or down. Go to M4.4.2.
    • If you chose C: You described the move from B back to A. The vector must go from the shape to its image. Go to M4.4.2.
    • If you chose D: The shape moves down, not up. A move down needs a minus sign on the bottom number. Go to M4.4.2.

  2. The corner (2, −1) is translated by the vector shown. Where does the corner go?

    One column vector in brackets with 4 on top and 3 below, labelled v.43v
    Tap your answer.

    1. Not this one. You added the numbers the wrong way round. The top number, 4, goes with x. The bottom number, 3, goes with y. Go to M4.4.2. Try another.

    2. Not this one. A vector says how far to move, not where to land. Add 4 to the x number and 3 to the y number of the corner. Go to M4.4.2. Try another.

    3. Yes, that is right. Add 4 to x: 2 + 4 = 6. Add 3 to y: −1 + 3 = 2. The corner goes to (6, 2).

    4. Not this one. You took the vector away. A translation by the vector adds its numbers. Go to M4.4.2. Try another.

    See the answer

    (6, 2)

    Add 4 to x: 2 + 4 = 6. Add 3 to y: −1 + 3 = 2. The corner goes to (6, 2).

    • If you chose A: You added the numbers the wrong way round. The top number, 4, goes with x. The bottom number, 3, goes with y. Go to M4.4.2.
    • If you chose B: A vector says how far to move, not where to land. Add 4 to the x number and 3 to the y number of the corner. Go to M4.4.2.
    • If you chose D: You took the vector away. A translation by the vector adds its numbers. Go to M4.4.2.

  3. A shape moves by the vector with 2 on top and 1 below. Then it moves by the vector with −5 on top and 2 below. Which one vector does the same?

    Tap your answer.

    1. Yes, that is right. Add the top numbers: 2 + (−5) = −3. Add the bottom numbers: 1 + 2 = 3.

    2. Not this one. You took the second move away from the first. Two moves in a row are added. Go to M4.4.2. Try another.

    3. Not this one. You multiplied. Two moves in a row are added, not multiplied. Go to M4.4.2. Try another.

    See the answer

    −3 on top, 3 below

    Add the top numbers: 2 + (−5) = −3. Add the bottom numbers: 1 + 2 = 3.

    • If you chose B: You took the second move away from the first. Two moves in a row are added. Go to M4.4.2.
    • If you chose C: You multiplied. Two moves in a row are added, not multiplied. Go to M4.4.2.

  4. A shape moves 5 units left and does not move up or down. Which column vector is that?

    Tap your answer.

    1. Not this one. A move up or down goes on the bottom. This shape moves sideways, so the 5 belongs on top. Go to M4.4.2. Try another.

    2. Not this one. A positive top number is a move right. A move left needs a minus sign. Go to M4.4.2. Try another.

    3. Yes, that is right. The top number is the move right, and left is the minus direction, so it is −5. The bottom number is 0, because there is no move up or down.

    See the answer

    −5 on top, 0 below

    The top number is the move right, and left is the minus direction, so it is −5. The bottom number is 0, because there is no move up or down.

    • If you chose A: A move up or down goes on the bottom. This shape moves sideways, so the 5 belongs on top. Go to M4.4.2.
    • If you chose B: A positive top number is a move right. A move left needs a minus sign. Go to M4.4.2.

  5. A translation moves the point (1, 1) to (4, −1). Where does the same translation move the point (0, 0)?

    Tap your answer.

    1. Not this one. That is where (1, 1) goes. A translation moves each point by the same vector, so (0, 0) lands somewhere else. Go to M4.4.2. Try another.

    2. Not this one. You used the vector the wrong way round. The vector goes from (1, 1) to (4, −1), so it is added, not taken away. Go to M4.4.2. Try another.

    3. Not this one. A translation moves every point, and (0, 0) is not a special point. The origin moves with the rest. Go to M4.4.2. Try another.

    4. Yes, that is right. From (1, 1) to (4, −1) is 3 right and 2 down, so the vector has 3 on top and −2 below. Every point moves by that vector, so (0, 0) goes to (3, −2).

    See the answer

    (3, −2)

    From (1, 1) to (4, −1) is 3 right and 2 down, so the vector has 3 on top and −2 below. Every point moves by that vector, so (0, 0) goes to (3, −2).

    • If you chose A: That is where (1, 1) goes. A translation moves each point by the same vector, so (0, 0) lands somewhere else. Go to M4.4.2.
    • If you chose B: You used the vector the wrong way round. The vector goes from (1, 1) to (4, −1), so it is added, not taken away. Go to M4.4.2.
    • If you chose C: A translation moves every point, and (0, 0) is not a special point. The origin moves with the rest. Go to M4.4.2.

  6. From an earlier chapter

    The point (4, 1) is turned 90° anticlockwise about (0, 0). What is the y number of the image?

    See the answer

    4

    A 90° anticlockwise turn about (0, 0) sends (x, y) to (−y, x). So (4, 1) goes to (−1, 4), and the y number is 4.

  7. The corner (3, 2) is translated by the vector with 4 on top and 5 below. What is the y number of the image?

    See the answer

    7

    Add the bottom number to y: 2 + 5 = 7.

  8. A corner moves from (1, 2) to (6, 9). What is the number on the bottom of the column vector?

    See the answer

    7

    The move up is the new y minus the old y: from 2 to 9 is 7.

  9. A shape moves by the vector with 2 on top and 5 below. Then it moves by the vector with −6 on top and 4 below. What is the number on the bottom of the one vector that does the same?

    See the answer

    9

    Add the bottom numbers: 5 + 4 = 9.

Keep it

Copy this into your exercise book.

A translation slides a shape. The column vector has the move right on top and the move up below. To combine two moves, add the tops and add the bottoms.

Draw: A triangle and its image with an arrow from one corner, and the vector written beside it.

M4.4.3 Chapter 3

Enlargements

A shape is enlarged by scale factor 2 about a point. Where does a corner that is 3 squares right of the point go?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

An enlargement changes the size of a shape and keeps its form. To describe one you need two things. The first is a point called the centre of enlargement. The second is a number called the scale factor. Every distance from the centre is multiplied by the scale factor.

Draw it first

Enlarge triangle A by scale factor 2 about C

  1. A grid with an x axis and a y axis from 0 to 8. Step 1: triangle A with corners at (2, 2), (4, 2) and (2, 3), labelled A, and the centre C at (1, 1). Step 2: a dashed line from C through each corner of A, ending at the image corners (3, 3), (7, 3) and (3, 5). Step 3: the image, triangle A′, with corners at (3, 3), (7, 3) and (3, 5), labelled A′.12345678123456780xyCA

    1 Triangle A has corners at (2, 2), (4, 2) and (2, 3). The centre of enlargement is C at (1, 1).

  2. A grid with an x axis and a y axis from 0 to 8. Step 1: triangle A with corners at (2, 2), (4, 2) and (2, 3), labelled A, and the centre C at (1, 1). Step 2: a dashed line from C through each corner of A, ending at the image corners (3, 3), (7, 3) and (3, 5). Step 3: the image, triangle A′, with corners at (3, 3), (7, 3) and (3, 5), labelled A′.12345678123456780xyCA

    2 Draw a line from C through each corner of A. Go on to twice the distance from C. Mark each end.

  3. A grid with an x axis and a y axis from 0 to 8. Step 1: triangle A with corners at (2, 2), (4, 2) and (2, 3), labelled A, and the centre C at (1, 1). Step 2: a dashed line from C through each corner of A, ending at the image corners (3, 3), (7, 3) and (3, 5). Step 3: the image, triangle A′, with corners at (3, 3), (7, 3) and (3, 5), labelled A′.12345678123456780xyCAA′

    3 Join the three marks. This is the image, A′. Each side is twice as long as the side it matches.

Look at the corner (2, 2). It is 1 right and 1 up from C. Its image is 2 right and 2 up from C, which is the point (3, 3). The rule is: image = C + scale factor × (corner − C).

An enlargement changes lengths and keeps angles. Each side of A′ is twice the side it matches. Every angle of A′ equals the angle it matches. That is why the shape keeps its form.

A grid with an x axis and a y axis from 0 to 8. The centre C at (1, 1). Triangle A with corners at (2, 2), (4, 2) and (2, 3), labelled A, and triangle A′ with corners at (3, 3), (7, 3) and (3, 5), labelled A′. Dashed lines run from C through the corners of A to the corners of A′.12345678123456780xyCAA′
The image corners lie on the lines from C, at twice the distance.

Worked example

Enlarge the triangle with corners (2, 2), (4, 2) and (2, 3) by scale factor 2 about C at (1, 1).

  1. Find each corner as a step from C. The corner (2, 2) is 1 right and 1 up. The corner (4, 2) is 3 right and 1 up. The corner (2, 3) is 1 right and 2 up.
  2. Multiply each step by 2. The steps become 2 right and 2 up, then 6 right and 2 up, then 2 right and 4 up.
  3. Count from C at (1, 1). The image corners are (3, 3), (7, 3) and (3, 5).

Answer: (3, 3), (7, 3) and (3, 5)

A scale factor between 0 and 1 makes the shape smaller. With scale factor 1/2 every distance from the centre is halved. The image is half as long and lies between the centre and the original.

A grid with an x axis and a y axis from 0 to 8. The centre C at (1, 1). Triangle A with corners at (3, 3), (7, 3) and (3, 7), labelled A, and the smaller triangle A′ with corners at (2, 2), (4, 2) and (2, 4), labelled A′. Dashed lines run from C through the corners of A′ to the corners of A.12345678123456780xyCAA′
Scale factor 1/2 about C. Triangle A′ is half the size of A.

Worked example

Enlarge the triangle with corners (3, 3), (7, 3) and (3, 7) by scale factor 1/2 about C at (1, 1).

  1. Find each corner as a step from C. The corner (3, 3) is 2 right and 2 up. The corner (7, 3) is 6 right and 2 up. The corner (3, 7) is 2 right and 6 up.
  2. Halve each step. The steps become 1 right and 1 up, then 3 right and 1 up, then 1 right and 3 up.
  3. Count from C at (1, 1). The image corners are (2, 2), (4, 2) and (2, 4). Each side is half as long.

Answer: (2, 2), (4, 2) and (2, 4)

Now suppose you are given a shape and its image, and you need the centre. The line through a corner and its image passes through the centre. Draw two such lines. They meet at the centre.

A grid with an x axis and a y axis from 0 to 8. Triangle A with corners at (2, 3), (4, 3) and (2, 4), labelled A, and the larger triangle A′ with corners at (3, 4), (7, 4) and (3, 6), labelled A′. Dashed lines run from each corner of A′ through the matching corner of A to the point C at (1, 2).12345678123456780xyCAA′
The lines through matching corners all meet at C.

Worked example

Triangle A has corners (2, 3), (4, 3) and (2, 4). Its image has corners (3, 4), (7, 4) and (3, 6). Find the centre and the scale factor.

  1. Join the corner (2, 3) to its image (3, 4). Draw the line on past (2, 3).
  2. Do the same for the corner (4, 3) and its image (7, 4).
  3. The two lines cross at (1, 2). That point is the centre.
  4. From C the corner (2, 3) is 1 across. The image (3, 4) is 2 across. The scale factor is 2.

Answer: centre (1, 2), scale factor 2

See it three ways

With things

Draw a small flag on squared paper. Draw it again on paper with squares 3 times as wide. Every length is 3 times as long.

As a picture

A grid with an x axis and a y axis from 0 to 8. The centre C at (1, 1). Triangle A with corners at (2, 2), (3, 2) and (2, 3), labelled A, and the larger triangle A′ with corners at (4, 4), (7, 4) and (4, 7), labelled A′. Dashed lines run from C through the corners.12345678123456780xyCAA′

In symbols

With scale factor 3 about C at (1, 1), the corner (3, 2) goes to (7, 4). Every length is 3 times as long.

Try it

Double a triangle with a ruler

You need: squared paper, a ruler, a pencil.

  1. Mark a point C near the bottom left of the paper. Draw a small triangle some way from C.
  2. Draw a line from C through each corner of the triangle. Make each line long.
  3. Measure from C to a corner. Mark a point on the line that is twice that distance from C. Do the same for each corner.
  4. Join the three new points. Measure one side of each triangle.

What you will see: Each side of the new triangle is twice the side it matches. Check the angles with a corner of the paper: they look the same.

If you cannot try it, this is what happens.
A grid with an x axis and a y axis from 0 to 8. The centre C at (1, 1). Triangle A with corners at (3, 2), (4, 2) and (3, 4), labelled A, and the larger triangle A′ with corners at (5, 3), (7, 3) and (5, 7), labelled A′. Dashed lines run from C through the corners.12345678123456780xyCAA′

A wrong idea many people have

To enlarge a shape by scale factor 2, add 2 to every length.

Why it feels right: People say "make it 2 bigger", and adding is the first way to make a number bigger.

The test that breaks it: Take a triangle with sides 3 cm and 6 cm. Add 2 to each and they become 5 cm and 8 cm. The first shape had a 1 : 2 ratio of sides. The new one does not, so it is a different shape.

What is true: An enlargement multiplies every length by the scale factor. The sides 3 cm and 6 cm become 6 cm and 12 cm. They are still in the ratio 1 : 2.

Key words

enlargement
a transformation that multiplies every distance from a centre by the same number. The shape keeps its form.
centre of enlargement
the point the distances are measured from. It does not move.
scale factor
the number every length is multiplied by. A scale factor between 0 and 1 makes the shape smaller.

Back to your guess

The answer is A: 6 squares right of the point. Every distance from the point doubles.

An enlargement multiplies every distance from the centre by the scale factor. A distance of 3 squares becomes 2 × 3 = 6 squares. Adding 2 would not keep the shape the same.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. Triangle A is enlarged by scale factor 3 about C at (1, 1). Where does the corner (3, 2) go?

    A grid with an x axis and a y axis from 0 to 8. The centre C at (1, 1) and triangle A with corners at (3, 2), (4, 2) and (3, 3), labelled A.12345678123456780xyCA
    Tap your answer.

    1. Not this one. You added 3 to each number. An enlargement multiplies the distances from the centre by the scale factor. Go to M4.4.3. Try another.

    2. Yes, that is right. The corner (3, 2) is 2 right and 1 up from C. Multiply by 3: 6 right and 3 up. Counting from C at (1, 1) gives (7, 4).

    3. Not this one. You multiplied only the right step. Both numbers of the step from the centre are multiplied. Go to M4.4.3. Try another.

    See the answer

    (7, 4)

    The corner (3, 2) is 2 right and 1 up from C. Multiply by 3: 6 right and 3 up. Counting from C at (1, 1) gives (7, 4).

    • If you chose A: You added 3 to each number. An enlargement multiplies the distances from the centre by the scale factor. Go to M4.4.3.
    • If you chose C: You multiplied only the right step. Both numbers of the step from the centre are multiplied. Go to M4.4.3.

  2. The corner (6, 4) is enlarged by scale factor 1/2 about C at (2, 2). Where does it go?

    A grid with an x axis and a y axis from 0 to 8. The centre C at (2, 2) and triangle A with corners at (4, 2), (6, 2) and (6, 4), labelled A.12345678123456780xyCA
    Tap your answer.

    1. Not this one. You halved the numbers of the corner itself. Halve the step from the centre instead, then count from the centre. Go to M4.4.3. Try another.

    2. Not this one. You doubled the step from the centre. A scale factor of 1/2 halves it. Go to M4.4.3. Try another.

    3. Yes, that is right. The corner (6, 4) is 4 right and 2 up from C. Half of that is 2 right and 1 up. Counting from C at (2, 2) gives (4, 3).

    See the answer

    (4, 3)

    The corner (6, 4) is 4 right and 2 up from C. Half of that is 2 right and 1 up. Counting from C at (2, 2) gives (4, 3).

    • If you chose A: You halved the numbers of the corner itself. Halve the step from the centre instead, then count from the centre. Go to M4.4.3.
    • If you chose B: You doubled the step from the centre. A scale factor of 1/2 halves it. Go to M4.4.3.

  3. A triangle has a side of 4 cm. It is enlarged by scale factor 3. Which statement is true?

    Tap your answer.

    1. Not this one. You added 3 cm. An enlargement multiplies each length by the scale factor. Go to M4.4.3. Try another.

    2. Not this one. The side is right, but angles do not change in an enlargement. The shape keeps its form. Go to M4.4.3. Try another.

    3. Yes, that is right. Lengths are multiplied by the scale factor: 4 × 3 = 12 cm. Angles do not change.

    See the answer

    The side is 12 cm and the angles stay the same.

    Lengths are multiplied by the scale factor: 4 × 3 = 12 cm. Angles do not change.

    • If you chose A: You added 3 cm. An enlargement multiplies each length by the scale factor. Go to M4.4.3.
    • If you chose B: The side is right, but angles do not change in an enlargement. The shape keeps its form. Go to M4.4.3.

  4. Triangle A is enlarged to triangle A′. Where is the centre of the enlargement?

    A grid with an x axis and a y axis from 0 to 8. Triangle A with corners at (3, 2), (4, 2) and (3, 3), labelled A, and triangle A′ with corners at (4, 3), (6, 3) and (4, 5), labelled A′.12345678123456780xyAA′
    Tap your answer.

    1. Not this one. This is a corner of A. The centre is where the lines through matching corners meet, and it is not a corner of A. Go to M4.4.3. Try another.

    2. Not this one. This is a corner of A′. The centre is where the lines through matching corners meet. Go to M4.4.3. Try another.

    3. Not this one. Draw the line through (3, 2) and (4, 3) and the line through (4, 2) and (6, 3). They do not cross at this point. Go to M4.4.3. Try another.

    4. Yes, that is right. The line through (3, 2) and (4, 3) and the line through (4, 2) and (6, 3) cross at (2, 1). That is the centre.

    See the answer

    (2, 1)

    The line through (3, 2) and (4, 3) and the line through (4, 2) and (6, 3) cross at (2, 1). That is the centre.

    • If you chose A: This is a corner of A. The centre is where the lines through matching corners meet, and it is not a corner of A. Go to M4.4.3.
    • If you chose B: This is a corner of A′. The centre is where the lines through matching corners meet. Go to M4.4.3.
    • If you chose C: Draw the line through (3, 2) and (4, 3) and the line through (4, 2) and (6, 3). They do not cross at this point. Go to M4.4.3.

  5. A side of 10 cm is enlarged by scale factor 1/2. How long is the image side?

    Tap your answer.

    1. Yes, that is right. Multiply the length by the scale factor: 10 × 1/2 = 5 cm.

    2. Not this one. You doubled the length. A scale factor of 1/2 is half, so the image is smaller than the original. Go to M4.4.3. Try another.

    3. Not this one. You took 1/2 away. Scale factors multiply. Half of 10 cm is 5 cm. Go to M4.4.3. Try another.

    See the answer

    5 cm

    Multiply the length by the scale factor: 10 × 1/2 = 5 cm.

    • If you chose B: You doubled the length. A scale factor of 1/2 is half, so the image is smaller than the original. Go to M4.4.3.
    • If you chose C: You took 1/2 away. Scale factors multiply. Half of 10 cm is 5 cm. Go to M4.4.3.

  6. From an earlier chapter

    The point (1, 3) is translated by the vector with 5 on top and 2 below. What is the y number of the image?

    See the answer

    5

    Add the bottom number to y: 3 + 2 = 5.

  7. The corner (2, 5) is enlarged by scale factor 3 about O at (0, 0). What is the y number of the image?

    See the answer

    15

    The corner is 2 right and 5 up from O. Multiply by 3: 6 right and 15 up. The y number is 15.

  8. The corner (5, 3) is enlarged by scale factor 2 about C at (1, 1). What is the x number of the image?

    See the answer

    9

    The corner is 4 right of C. Double it: 8 right of C. Counting from x = 1 gives 1 + 8 = 9.

  9. A side of 5 cm becomes 15 cm in an enlargement. What is the scale factor?

    See the answer

    3

    The scale factor is the new length divided by the old length: 15 ÷ 5 = 3.

Keep it

Copy this into your exercise book.

Enlarge by multiplying each distance from the centre by the scale factor. Image = centre + factor × (corner − centre). Lines through matching corners meet at the centre.

Draw: A small triangle, a bigger one, and dashed lines from the centre through matching corners.

M4.4.4 Chapter 4

Congruent triangles

Two triangles have two equal sides and one equal angle. Must they be exactly the same?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Two shapes are congruent when one fits exactly on the other. You may need to slide, turn or flip it first. Congruent shapes have the same size and the same form. Their corresponding sides and angles are equal. Those are the parts that match each other.

Two triangles drawn to the same size. Triangle ABC has sides marked with one, two and three ticks. Triangle DEF is turned and flipped but has the same ticks on the matching sides. Corners are named A, B, C, D, E and F.ABCDEF
Matching ticks mean equal sides. Matching arcs mean equal angles.

To show two triangles are congruent you do not need all six facts. Three facts, in the right pattern, are enough. There are four patterns. Each has a short name.

SSS means three sides. If all three sides of one triangle equal the three sides of another, the triangles are congruent. A triangle with sides of 4 cm, 5 cm and 6 cm can only be drawn one way.

Two triangles with sides of 4 cm, 5 cm and 6 cm, marked with one, two and three ticks. The second triangle is turned. Corners are named P, Q, R, X, Y and Z.4 cm5 cm6 cmPQR4 cm5 cm6 cmXYZ
SSS: three pairs of equal sides.

SAS means two sides and the angle between them. The angle must sit between the two sides. Then the triangle is fixed, because the two sides and the angle between them decide where the third side goes.

Two triangles. Each has a side of 5 cm with one tick, a side of 8 cm with two ticks, and a 60° angle between them marked with an arc. The second triangle is turned and flipped. Corners are named S, T, U, V, W and X.60°5 cm8 cmSTU60°5 cm8 cmVWX
SAS: two pairs of equal sides, and the equal angles are between them.

ASA means two angles and the side between them. If two angles are equal, the third angle is equal too, because the three angles of a triangle add up to 180°. So two pairs of equal angles and one pair of equal matching sides is enough, wherever that side is. This version is called AAS.

Two triangles. Each has a side of 7 cm with one tick at the bottom. The angle at the left end of that side has one arc and the angle at the right end has two arcs, in both triangles. The second triangle is turned and flipped. Corners are named J, K, L, M, N and P.7 cmJKL7 cmMNP
ASA: two equal angles and the side between them.

RHS is for right angled triangles only. It means a right angle, the hypotenuse and one other side. The hypotenuse is the longest side, opposite the right angle. A right angle, the hypotenuse and one more side fix the triangle.

Two right angled triangles. Each has a right angle mark, a hypotenuse of 5 cm with one tick, and a side of 3 cm with two ticks. The second triangle is turned. Corners are named D, E, F, G, H and J.3 cm5 cmDEF3 cm5 cmGHJ
RHS: a right angle, the hypotenuse and one other side.

One pattern does not always work. Two sides and an angle that is not between them is called SSA. Look at what can happen.

Two triangles that share a side of 5 cm with one tick, a side of 8 cm with two ticks, and a 30° angle marked with an arc at the corner A. The first triangle has a long third side. The second has a short third side. They are not the same size. Corners are named A, B, C, D, E and F.30°5 cm8 cmACB30°5 cm8 cmDFE
SSA: the same two sides and the same angle, but two different triangles.
Two triangles. Triangle ABC has a side AB of 5 cm with one tick, a side BC of 7 cm with two ticks, and an angle of 40° at B. Triangle DEF has a side DE of 5 cm with one tick, a side EF of 7 cm with two ticks, and an angle of 40° at E. The corners are named A, C, B, D, F and E.40°7 cm5 cmACB40°7 cm5 cmDFE
Triangle ABC and triangle DEF.

Worked example

Triangle ABC has AB = 5 cm, BC = 7 cm and angle B = 40°. Triangle DEF has DE = 5 cm, EF = 7 cm and angle E = 40°. Are the triangles congruent?

  1. Match the sides. AB matches DE and BC matches EF. Both pairs are equal.
  2. Angle B is between AB and BC. Angle E is between DE and EF. Both angles are 40°.
  3. Two sides and the angle between them are equal. So the triangles are congruent by SAS.

Answer: Yes, by SAS

Congruent triangles are useful because they prove things. A proof is a chain of reasons that shows a fact must be true. Take an isosceles triangle, which has two equal sides. Draw a line from the top corner to the middle of the base. It cuts the triangle into two triangles.

Two triangles drawn side by side, the two halves of an isosceles triangle. Triangle BMA has a base BM of 6 cm marked with two ticks, a side AB of 10 cm marked with three ticks, and a side AM marked with one tick. Triangle CMA is the mirror image with the same ticks. The corners are named B, M, A, C, M and A.10 cm6 cmBMA10 cm6 cmCMA
The two halves have three pairs of equal sides.

Worked example

In triangle ABC, AB = AC = 10 cm and BC = 12 cm. M is the middle of BC. Show that triangles ABM and ACM are congruent, and find BM.

  1. AB = AC = 10 cm. The problem says so.
  2. BM = MC, because M is the middle of BC. Each is half of 12 cm, which is 6 cm.
  3. AM is a side of both triangles, so it is equal to itself.
  4. Three pairs of sides are equal, so the triangles are congruent by SSS. Also BM is 6 cm.

Answer: SSS, BM is 6 cm

Because the triangles are congruent, their matching angles are equal. So angle ABM equals angle ACM. This proves that the angles at the base of any isosceles triangle are equal.

See it three ways

With things

Cut a triangle out of paper. Trace round it on a second sheet and cut out the trace. Lay one on top of the other. They match exactly.

As a picture

Two right angled triangles with sides of 3 cm, 4 cm and 5 cm. Each has a right angle mark. The second is turned. Corners are named K, L, M, N, P and Q.3 cm4 cm5 cmKLM3 cm4 cm5 cmNPQ

In symbols

Congruent triangles have equal matching sides and equal matching angles. For example KL = NP, LM = PQ and MK = QN.

Try it

Build a triangle from three strips

You need: three strips of paper or card, 3 cm, 4 cm and 5 cm long, a friend, a pencil.

  1. Lay the 3 cm and 4 cm strips so that their ends touch at a corner.
  2. Swing the 5 cm strip across to join the other two ends.
  3. Ask your friend to build a triangle from three strips of the same lengths. Lay one triangle on the other.

What you will see: The two triangles match exactly, however you place the strips. Three sides fix the triangle. That is SSS.

If you cannot try it, this is what happens.
Two triangles made from strips of 3 cm, 4 cm and 5 cm. Each has a right angle mark. One is turned. Corners are named S, T, U, V, W and X.3 cm4 cm5 cmSTU3 cm4 cm5 cmVWX

A wrong idea many people have

Two sides and an angle are always enough to show triangles are congruent.

Why it feels right: Two sides and an angle are three facts, and SAS uses exactly that.

The test that breaks it: Look at the picture of SSA. Both triangles have a side of 5 cm, a side of 8 cm and a 30° angle. But the third sides are different, so the triangles are not the same.

What is true: The angle must be between the two sides for SAS. If it is not between them, two different triangles can sometimes fit the facts, so SSA proves nothing. RHS works because the angle is 90° and the side opposite it is the longest.

Key words

congruent
having exactly the same size and form. One shape fits exactly on the other.
corresponding
matching. Corresponding sides and angles sit in the same place in each shape.
SSS
three sides of one triangle equal three sides of another.
SAS
two sides and the angle between them are equal.
ASA
two angles and the side between them are equal. If the side is not between them, the test is called AAS and it works too.
RHS
right angle, hypotenuse and one other side are equal.
proof
a chain of reasons showing a fact must be true.

Back to your guess

The answer is C: Not always. When the angle is between the two sides they must be. When it is not, they may differ.

It depends where the angle is. When the equal angle is between the two equal sides, the triangles are the same. When it is not, two different triangles can sometimes be made, so this pattern, SSA, is not a test. RHS is the one exception you will use: a right angle with the hypotenuse fixes the triangle. This chapter shows which sets of three facts are enough.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. The marks show that three pairs of sides are equal. Which test shows that the triangles are congruent?

    Two triangles with sides of 4 cm, 5 cm and 6 cm, marked with one, two and three ticks on matching sides. The corners are named A, B, C, D, E and F.4 cm5 cm6 cmABC4 cm5 cm6 cmDEF
    Tap your answer.

    1. Not this one. SAS needs an angle between two of the sides. No angle is marked, only the three sides. Go to M4.4.4. Try another.

    2. Yes, that is right. All three pairs of sides are equal, and that is the SSS test.

    3. Not this one. RHS is for right angled triangles. No right angle is marked here. Go to M4.4.4. Try another.

    See the answer

    SSS

    All three pairs of sides are equal, and that is the SSS test.

    • If you chose A: SAS needs an angle between two of the sides. No angle is marked, only the three sides. Go to M4.4.4.
    • If you chose C: RHS is for right angled triangles. No right angle is marked here. Go to M4.4.4.

  2. These two triangles have two equal sides and an equal angle. Are they congruent?

    Two triangles. Each has a side of 5 cm marked with one tick, a side of 7 cm marked with two ticks, and a 40° angle at corner A. The first triangle has a longer third side. The corners are named A, C, B, D, F and E.40°5 cm7 cmACB40°5 cm7 cmDFE
    Tap your answer.

    1. Not this one. The angle of 40° is not between the 5 cm and 7 cm sides. SAS needs the angle between the two sides. Go to M4.4.4. Try another.

    2. Not this one. SSS needs all three sides equal. The third sides are different lengths in the picture. Go to M4.4.4. Try another.

    3. Not this one. Triangles that look alike can differ in size. A reason must be one of the four tests. Go to M4.4.4. Try another.

    4. Yes, that is right. The equal angle is not between the two equal sides. That is SSA, which is not a test of congruence. The picture shows two different triangles with different third sides.

    See the answer

    No. Their third sides are different lengths. SSA cannot prove two triangles congruent.

    The equal angle is not between the two equal sides. That is SSA, which is not a test of congruence. The picture shows two different triangles with different third sides.

    • If you chose A: The angle of 40° is not between the 5 cm and 7 cm sides. SAS needs the angle between the two sides. Go to M4.4.4.
    • If you chose B: SSS needs all three sides equal. The third sides are different lengths in the picture. Go to M4.4.4.
    • If you chose C: Triangles that look alike can differ in size. A reason must be one of the four tests. Go to M4.4.4.

  3. Each triangle has a right angle. The marks show equal hypotenuses and one other equal side. Which test shows that the triangles are congruent?

    Two right angled triangles. Each has a right angle mark, a hypotenuse of 5 cm marked with one tick, and a side of 3 cm marked with two ticks. The corners are named A, B, C, D, E and F.3 cm5 cmABC3 cm5 cmDEF
    Tap your answer.

    1. Not this one. ASA needs two equal angles and a side. Here one angle is marked, and there are two sides. Go to M4.4.4. Try another.

    2. Not this one. SAS needs the angle between the two equal sides. The right angle is not between the hypotenuse and the 3 cm side. Go to M4.4.4. Try another.

    3. Yes, that is right. The triangles each have a right angle, equal hypotenuses and one more equal side. That is RHS.

    See the answer

    RHS

    The triangles each have a right angle, equal hypotenuses and one more equal side. That is RHS.

    • If you chose A: ASA needs two equal angles and a side. Here one angle is marked, and there are two sides. Go to M4.4.4.
    • If you chose B: SAS needs the angle between the two equal sides. The right angle is not between the hypotenuse and the 3 cm side. Go to M4.4.4.

  4. The marks show two equal angles and the side between them. Which test shows that the triangles are congruent?

    Two triangles. Each has a side of 7 cm with one tick. The angle at one end of that side has one arc and the angle at the other end has two arcs. The corners are named A, B, C, D, E and F.7 cmABC7 cmDEF
    Tap your answer.

    1. Yes, that is right. Two angles are equal and the side between them is equal. That is ASA.

    2. Not this one. SSS needs three equal sides. Only one side is marked here, with two angles. Go to M4.4.4. Try another.

    3. Not this one. RHS needs a right angle. None is marked here. Go to M4.4.4. Try another.

    See the answer

    ASA

    Two angles are equal and the side between them is equal. That is ASA.

    • If you chose B: SSS needs three equal sides. Only one side is marked here, with two angles. Go to M4.4.4.
    • If you chose C: RHS needs a right angle. None is marked here. Go to M4.4.4.

  5. Triangles ABM and ACM are congruent. AB = AC = 13 cm and BC = 10 cm. M is the middle of BC. How long is BM?

    Tap your answer.

    1. Not this one. That is the length of AB. BM is part of BC, the base of the triangle. Go to M4.4.4. Try another.

    2. Not this one. That is the whole base BC. M is the middle of BC, so BM is only half of it. Go to M4.4.4. Try another.

    3. Yes, that is right. M is the middle of BC, so BM is half of BC. Half of 10 cm is 5 cm.

    See the answer

    5 cm

    M is the middle of BC, so BM is half of BC. Half of 10 cm is 5 cm.

    • If you chose A: That is the length of AB. BM is part of BC, the base of the triangle. Go to M4.4.4.
    • If you chose B: That is the whole base BC. M is the middle of BC, so BM is only half of it. Go to M4.4.4.

  6. From an earlier chapter

    A side of 6 cm is enlarged by scale factor 4. How long is the image side, in centimetres?

    See the answer

    24

    An enlargement multiplies each length by the scale factor: 6 × 4 = 24 cm.

  7. Triangle ABC has sides of 5 cm, 6 cm and 9 cm. Triangle DEF is congruent to it and has sides of 6 cm, 9 cm and x cm. What is x?

    See the answer

    5

    Congruent triangles have the same three side lengths. DEF has 6 cm and 9 cm, so the third side matches the 5 cm of ABC.

  8. Triangle ABC has AB = AC and angle B = 70°. Triangles ABM and ACM are congruent, so angle C is also 70°. What is angle A, in degrees?

    See the answer

    40

    The angles of a triangle add up to 180°. Angle A = 180° − 70° − 70° = 40°.

  9. Triangle ABC has angles of 50° and 60°. Triangle DEF also has angles of 50° and 60°, and one pair of matching sides is equal. What is the third angle of each triangle, in degrees?

    See the answer

    70

    The angles of a triangle add up to 180°. The third angle is 180° − 50° − 60° = 70° in both triangles. So all three angles match, and with the equal side the triangles are congruent by AAS.

Keep it

Copy this into your exercise book.

Congruent triangles match exactly. Tests: SSS, SAS (angle between), ASA, RHS. SSA is not a test, but RHS works. Use them to prove things, such as equal base angles.

Draw: Two matching triangles with ticks on the equal sides and arcs on the equal angles.

M4.4.5 Chapter 5

Similar shapes

A 3 m pole has a 2 m shadow. At the same time a tree has a 6 m shadow. How tall is the tree?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Two shapes are similar when one is an enlargement of the other. The enlargement may also turn or flip the shape. Similar shapes have the same form but not always the same size. Each pair of matching angles is equal. Each pair of matching sides is in the same ratio.

Two similar triangles drawn to the same scale. The small triangle has sides of 3 cm, 4 cm and 5 cm. The large triangle has sides of 6 cm, 8 cm and 10 cm. The matching angles carry the same number of arcs. Corners are named A, B, C, D, E and F.3 cm4 cm5 cmABC6 cm8 cm10 cmDEF
The sides of the large triangle are 2 times the matching sides of the small one.

Take the triangles above. Divide each large side by its matching small side: 6 ÷ 3 = 2, 8 ÷ 4 = 2 and 10 ÷ 5 = 2. All three answers are 2. That number is the scale factor from the small triangle to the large one.

To show two triangles are similar, you only need their angles. The three angles of a triangle add up to 180°. If two angles match, the third is found by taking them from 180°, so it matches too.

Two triangles of different sizes. Each has a 40° angle at corner J and a 60° angle at corner K, marked with one and two arcs. The corners are named J, K, L, P, Q and R.40°60°JKL40°60°PQR
Two angles match, so the third angle is 80° in both. The triangles are similar.

This test works for triangles. For shapes with more sides, equal angles are not enough. A rectangle that is 2 cm by 4 cm and one that is 2 cm by 8 cm have only right angles. Their sides are not in the same ratio, so they are not similar.

Now use the scale factor to find a missing side. Pair up the matching sides. Divide a large side by its small match to get the scale factor. Then multiply any small side by it.

Two similar right angled triangles. Triangle ABC has sides of 3 cm and 4 cm and a hypotenuse of 5 cm. Triangle DEF has a side of 9 cm matching the 3 cm side, a side of 12 cm matching the 4 cm side, and a side marked x. The corners are named A, B, C, D, E and F.3 cm4 cm5 cmABC9 cm12 cmxDEF
Triangle DEF is similar to triangle ABC. Side x matches the 5 cm side.

Worked example

Triangle ABC has sides of 3 cm, 4 cm and 5 cm. Triangle DEF is similar. Its side of 9 cm matches the 3 cm side. How long is the side that matches 5 cm?

  1. Find the scale factor. Divide the large side by its small match: 9 ÷ 3 = 3.
  2. Multiply the 5 cm side by the scale factor: 5 × 3 = 15.
  3. The side that matches 5 cm is 15 cm long.

Answer: 15 cm

Similar triangles solve real problems. On level ground, an upright pole and an upright tree both make a right angle with the ground. At the same time the sun is at the same angle for both. So two angles match, and the shadow triangles are similar.

Two right angled triangles. The pole triangle ABC, on the left, has a pole BC of 3 m standing up and a shadow AC of 2 m lying flat. The tree triangle DEF, on the right, has a tree EF with its height marked h and a shadow DF of 6 m. The right angle is at C and at F, at the bottom left of each triangle.3 m2 mABCh6 mDEF
The pole triangle ABC is on the left and the tree triangle DEF is on the right. The sun makes the same angle with the ground for both.

Worked example

A pole 3 m high has a shadow 2 m long. At the same time a tree has a shadow 6 m long. How high is the tree?

  1. The pole and the tree are both upright, and the sun is at the same angle. The shadow triangles have equal angles, so they are similar.
  2. The scale factor from the pole shadow to the tree shadow is 6 ÷ 2 = 3.
  3. The tree is 3 times as high as the pole: 3 × 3 = 9 m.

Answer: 9 m

See it three ways

With things

Print a photo small and then large. Every person in it looks the same, only bigger. Each length in the large print is the same number of times its length in the small one.

As a picture

Two similar triangles, the large one 1.5 times the small one. The small triangle has sides of 4 cm, 5 cm and 6 cm. The large triangle has sides of 6 cm, 7.5 cm and 9 cm. Corners are named G, H, I, M, N and O.4 cm5 cm6 cmGHI6 cm7.5 cm9 cmMNO

In symbols

Scale factor = 6 ÷ 4 = 1.5. Then 5 × 1.5 = 7.5 and 6 × 1.5 = 9. Every side is multiplied by 1.5.

Try it

Draw two triangles with the same angles

You need: a ruler, a protractor, a pencil, plain paper.

  1. Draw a line 6 cm long. At one end draw an angle of 40°. At the other end draw an angle of 60°. Extend the two lines until they meet.
  2. Draw a second line, 9 cm long. Use the same two angles, 40° and 60°.
  3. Measure the sides that lie opposite the 40° angle in each triangle. Divide the larger by the smaller.
  4. Do the same for the sides opposite the 60° angle.

What you will see: Both answers are 1.5, the same as 9 ÷ 6. Triangles with the same angles have sides in the same ratio.

If you cannot try it, this is what happens.
Two triangles with a 40° angle at corner A and a 60° angle at corner B, marked with one and two arcs. The first has a base AB of 6 cm and the second a base DE of 9 cm. The corners are named A, B, C, D, E and F.40°60°6 cmABC40°60°9 cmDEF

A wrong idea many people have

If one side grows from 3 cm to 9 cm, then every side grows by 6 cm.

Why it feels right: Adding is how we often compare two numbers. The number 9 is 6 more than 3.

The test that breaks it: Add 6 to the 5 cm side and it becomes 11 cm. The sides would be 9 cm, 10 cm and 11 cm. These are not in the same ratio as 3 cm, 4 cm and 5 cm, so the shape changes.

What is true: Similar shapes grow by multiplying. The scale factor is 9 ÷ 3 = 3, so the 5 cm side becomes 5 × 3 = 15 cm.

Key words

similar
having the same form, with equal matching angles and matching sides in the same ratio. One shape is an enlargement of the other.

Back to your guess

The answer is C: 9 m, because the shadow is 3 times as long.

The pole and the tree both stand upright on level ground, and the sun is at the same angle for both. So their shadow triangles have the same shape. The tree shadow is 3 times the pole shadow, so the tree is 3 times the pole: 9 m.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. Two angles are marked in each triangle. Are the triangles similar?

    Two triangles. The first has angles of 50° and 60° marked at corners A and B. The second has angles of 60° and 70° marked at corners D and E. The corners are named A, B, C, D, E and F.50°60°ABC60°70°DEF
    Tap your answer.

    1. Yes, that is right. The first triangle has a third angle of 180° − 50° − 60° = 70°. The second has 180° − 60° − 70° = 50°. Both triangles have the angles 50°, 60° and 70°, so they are similar.

    2. Not this one. You compared only the marked angles. Work out the third angle of each triangle too. The three angles of a triangle add up to 180°. Go to M4.4.5. Try another.

    3. Not this one. Not every pair of triangles is similar. The angles must match, and being triangles is not enough. Go to M4.4.5. Try another.

    See the answer

    Yes. The missing angles are 70° and 50°, so all three angles match.

    The first triangle has a third angle of 180° − 50° − 60° = 70°. The second has 180° − 60° − 70° = 50°. Both triangles have the angles 50°, 60° and 70°, so they are similar.

    • If you chose B: You compared only the marked angles. Work out the third angle of each triangle too. The three angles of a triangle add up to 180°. Go to M4.4.5.
    • If you chose C: Not every pair of triangles is similar. The angles must match, and being triangles is not enough. Go to M4.4.5.

  2. Triangles ABC and DEF are similar. AB = 4 cm matches DE = 10 cm. BC = 6 cm. How long is EF?

    Two similar triangles. Triangle ABC has a side of 6 cm and a side AB of 4 cm. Triangle DEF has a side DE of 10 cm and a side EF marked x. The matching angles carry the same number of arcs. The corners are named A, B, C, D, E and F.6 cm4 cmABCx10 cmDEF
    Tap your answer.

    1. Not this one. You added. DE is 6 cm more than AB, so you added 6 to BC. Similar shapes grow by multiplying by the scale factor. Go to M4.4.5. Try another.

    2. Not this one. You used the scale factor the wrong way round. DEF is the larger triangle, so EF must be longer than BC. Go to M4.4.5. Try another.

    3. Yes, that is right. The scale factor is 10 ÷ 4 = 2.5. Multiply BC by it: 6 × 2.5 = 15 cm.

    See the answer

    15 cm

    The scale factor is 10 ÷ 4 = 2.5. Multiply BC by it: 6 × 2.5 = 15 cm.

    • If you chose A: You added. DE is 6 cm more than AB, so you added 6 to BC. Similar shapes grow by multiplying by the scale factor. Go to M4.4.5.
    • If you chose B: You used the scale factor the wrong way round. DEF is the larger triangle, so EF must be longer than BC. Go to M4.4.5.

  3. Triangle PQR is a smaller copy of triangle ABC. AB = 12 cm matches PQ = 8 cm. BC = 18 cm. How long is QR?

    Tap your answer.

    1. Not this one. You used the scale factor from the small triangle to the large one. PQR is smaller, so QR is shorter than BC. Go to M4.4.5. Try another.

    2. Yes, that is right. The scale factor from ABC to PQR is 8 ÷ 12 = 2/3. Multiply BC by it: 18 × 2/3 = 12 cm.

    3. Not this one. You took 4 cm away because 8 is 4 less than 12. Similar shapes shrink by multiplying, not by taking away. Go to M4.4.5. Try another.

    See the answer

    12 cm

    The scale factor from ABC to PQR is 8 ÷ 12 = 2/3. Multiply BC by it: 18 × 2/3 = 12 cm.

    • If you chose A: You used the scale factor from the small triangle to the large one. PQR is smaller, so QR is shorter than BC. Go to M4.4.5.
    • If you chose C: You took 4 cm away because 8 is 4 less than 12. Similar shapes shrink by multiplying, not by taking away. Go to M4.4.5.

  4. Which pair of shapes must be similar?

    Tap your answer.

    1. Not this one. Two rectangles have equal angles, but their sides need not be in the same ratio. A 2 cm by 4 cm rectangle and a 2 cm by 8 cm rectangle are not similar. Go to M4.4.5. Try another.

    2. Not this one. One equal angle is not enough. The other angles can differ, so the triangles can have different forms. Go to M4.4.5. Try another.

    3. Yes, that is right. Every square has four right angles and four equal sides. Any two squares have equal angles, and their sides are in the same ratio, so they are similar.

    4. Not this one. Kites can be long and thin or short and fat. Two kites need not have equal angles or sides in the same ratio. Go to M4.4.5. Try another.

    See the answer

    Two squares

    Every square has four right angles and four equal sides. Any two squares have equal angles, and their sides are in the same ratio, so they are similar.

    • If you chose A: Two rectangles have equal angles, but their sides need not be in the same ratio. A 2 cm by 4 cm rectangle and a 2 cm by 8 cm rectangle are not similar. Go to M4.4.5.
    • If you chose B: One equal angle is not enough. The other angles can differ, so the triangles can have different forms. Go to M4.4.5.
    • If you chose D: Kites can be long and thin or short and fat. Two kites need not have equal angles or sides in the same ratio. Go to M4.4.5.

  5. A 2 m stick has a shadow 3 m long. At the same time a building has a shadow 24 m long. How tall is the building?

    Tap your answer.

    1. Not this one. You used the scale factor the wrong way round, 3 ÷ 2. The building shadow is 8 times the stick shadow, so use 24 ÷ 3. Go to M4.4.5. Try another.

    2. Yes, that is right. The scale factor is 24 ÷ 3 = 8. The building is 8 times as tall as the stick: 2 × 8 = 16 m.

    3. Not this one. You added. The building shadow is 21 m longer, and you added 21 to the stick. Multiply by the scale factor instead. Go to M4.4.5. Try another.

    See the answer

    16 m

    The scale factor is 24 ÷ 3 = 8. The building is 8 times as tall as the stick: 2 × 8 = 16 m.

    • If you chose A: You used the scale factor the wrong way round, 3 ÷ 2. The building shadow is 8 times the stick shadow, so use 24 ÷ 3. Go to M4.4.5.
    • If you chose C: You added. The building shadow is 21 m longer, and you added 21 to the stick. Multiply by the scale factor instead. Go to M4.4.5.

  6. From an earlier chapter

    Triangle ABC has AB = AC and angle B = 65°. What is angle A, in degrees?

    See the answer

    50

    The base angles of an isosceles triangle are equal, so angle C is also 65°. Angle A = 180° − 65° − 65° = 50°.

  7. Triangle ABC has sides of 5 cm and 7 cm. A similar triangle has a side of 15 cm that matches the 5 cm side. How long is the side that matches 7 cm, in centimetres?

    See the answer

    21

    The scale factor is 15 ÷ 5 = 3. Multiply 7 by it: 7 × 3 = 21 cm.

  8. A triangle has sides of 12 cm and 15 cm. A similar, smaller triangle has a side of 4 cm that matches the 12 cm side. How long is the side that matches 15 cm, in centimetres?

    See the answer

    5

    The scale factor is 4 ÷ 12 = 1/3. Multiply 15 by it: 15 ÷ 3 = 5 cm.

  9. A photo is 12 cm wide and 8 cm high. It is enlarged so that it is 30 cm wide. How high is the enlarged photo, in centimetres?

    See the answer

    20

    An enlarged photo is similar to the original. The scale factor is 30 ÷ 12 = 2.5. Multiply 8 by it: 8 × 2.5 = 20 cm.

Keep it

Copy this into your exercise book.

Similar shapes have equal angles and matching sides in the same ratio. Two equal angles make triangles similar. Scale factor = a side of one ÷ the matching side of the other. Multiply.

Draw: A small and a large triangle with the same angle arcs, and the scale factor written between them.

M4.4.6 Chapter 6

Similar areas and volumes

A jar is made twice as high and twice as wide. Does it hold twice as much?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Two solids of the same shape are similar, like two jars made from the same mould. Every length of the large jar is the same number of times the matching length of the small jar. That number is the scale factor, k. Here k = 2.

Two jars of the same shape. The small jar is 8 cm high and 5 cm wide. The large jar is 16 cm high and 10 cm wide. The labels are small jar, large jar, 8 cm, 5 cm, 16 cm and 10 cm.8 cm5 cmsmall jar16 cm10 cmlarge jar
Every length of the large jar is 2 times the matching length of the small jar.

Lengths go up by k. Areas and volumes do not go up by k. Start with areas.

A square 1 cm by 1 cm with an area of 1 cm², beside a square 2 cm by 2 cm made of 4 small squares with an area of 4 cm². The labels are 1 cm, 2 cm, 1 cm² and 4 cm².1 cm1 cm²2 cm4 cm²
The sides are 2 times as long. The area is 4 times as big.

The big square has 2 rows of 2 small squares, so 2 × 2 = 4 of them. For scale factor k, the big square has k rows of k small squares, so k × k = k² of them. Any flat shape can be cut into small squares, so every area is multiplied by k². This is the area scale factor.

A cube 1 cm on each edge with a volume of 1 cm³, beside a block 2 cm on each edge made of 8 small cubes with a volume of 8 cm³. The labels are 1 cm, 2 cm, 1 cm³ and 8 cm³.1 cm1 cm³2 cm8 cm³
The edges are 2 times as long. The volume is 8 times as big.

The big block has 2 layers. Each layer has 2 rows of 2 small cubes. That is 2 × 2 × 2 = 8 cubes. For scale factor k there are k × k × k = k³ small cubes. Every volume is multiplied by k³. This is the volume scale factor.

A table with the head k, Len., Area and Vol., short for length, area and volume. Row 1: 2, × 2, × 4, × 8. Row 2: 3, × 3, × 9, × 27. Row 3: 10, × 10, × 100, × 1 000.kLen.AreaVol.2× 2× 4× 83× 3× 9× 2710× 10× 100× 1 000
Lengths multiply by k, areas by k × k, volumes by k × k × k.
A square 1 cm by 1 cm with an area of 1 cm², beside a square 3 cm by 3 cm made of 9 small squares with an area of 9 cm². The labels are 1 cm, 3 cm, 1 cm² and 9 cm².1 cm1 cm²3 cm9 cm²
With k = 3, there are 9 small squares in the big square.

Worked example

A photo frame has an area of 20 cm². A similar frame has all its lengths 3 times as long. What is the area of the larger frame?

  1. The length scale factor is k = 3.
  2. The area scale factor is k² = 3 × 3 = 9.
  3. The new area is 20 × 9 = 180 cm².

Answer: 180 cm²

Two jars of the same shape. The small jar is 12 cm high and 4 cm wide. The large jar is 24 cm high and 8 cm wide. The labels are small jar, large jar, 12 cm, 4 cm, 24 cm and 8 cm.12 cm4 cmsmall jar24 cm8 cmlarge jar
The large jar is twice as high and twice as wide.

Worked example

A small jar holds about 150 ml. A larger jar of the same shape is twice as high. About how much does the larger jar hold?

  1. Twice as high means the length scale factor is k = 2.
  2. The volume scale factor is k³ = 2 × 2 × 2 = 8.
  3. The larger jar holds about 150 × 8 = 1 200 ml.

Answer: about 1 200 ml

You can also go backwards. If you know how areas compare, take a square root to find k. If you know how volumes compare, take a cube root.

Two jars of the same shape. The small jar is 10 cm high and 6 cm wide. The large jar is 20 cm high and 12 cm wide. The labels are small jar, large jar, 10 cm, 6 cm, 20 cm and 12 cm.10 cm6 cmsmall jar20 cm12 cmlarge jar
The whole outsides of the jars have areas of about 245 cm² and 980 cm².

Worked example

Two jars have the same shape. Their whole outsides, with lid and base, have areas of about 245 cm² and 980 cm². The small jar holds about 283 ml. Find k and the volume of the large jar.

  1. The area scale factor is 980 ÷ 245 = 4.
  2. The area scale factor is k², so k is the square root of 4, which is 2.
  3. The volume scale factor is k³ = 2 × 2 × 2 = 8.
  4. The large jar holds about 283 × 8 = 2 264 ml.

Answer: k is 2 and the large jar holds about 2 264 ml

See it three ways

With things

Build a cube from 27 small cubes, 3 along each edge. Its edge is 3 times the edge of one small cube. Count the small cubes: 3 × 3 × 3.

As a picture

A cube 1 cm on each edge with a volume of 1 cm³, beside a block 3 cm on each edge made of 27 small cubes with a volume of 27 cm³. The labels are 1 cm, 3 cm, 1 cm³ and 27 cm³.1 cm1 cm³3 cm27 cm³

In symbols

Scale factor k for lengths. Scale factor k × k for areas. Scale factor k × k × k for volumes. With k = 3 these are 3, 9 and 27.

Try it

Count the squares

You need: squared paper, a pencil.

  1. Draw a square 1 cm by 1 cm. Draw a square 4 cm by 4 cm beside it.
  2. Count the 1 cm squares inside the big square.
  3. Divide the number of squares in the big square by the number in the small square.

What you will see: The big square has 16 small squares. The sides are 4 times as long, and 4 × 4 = 16.

If you cannot try it, this is what happens.
A square 1 cm by 1 cm with an area of 1 cm², beside a square 4 cm by 4 cm made of 16 small squares with an area of 16 cm². The labels are 1 cm, 4 cm, 1 cm² and 16 cm².1 cm1 cm²4 cm16 cm²

A wrong idea many people have

If a shape is enlarged by scale factor 3, its area is 3 times as big.

Why it feels right: The lengths are 3 times as long, so it seems the area should be 3 times as big too.

The test that breaks it: Take a square 1 cm by 1 cm with an area of 1 cm². Enlarge it by scale factor 3. It becomes 3 cm by 3 cm, with an area of 9 cm². That is 9 times as big, not 3 times.

What is true: Area is length times width, and both lengths are multiplied by k. So the area is multiplied by k × k = k². Volume has three lengths, so it is multiplied by k³.

Key words

area scale factor
the number every area is multiplied by. It is k², where k is the length scale factor.
volume scale factor
the number every volume is multiplied by. It is k³, where k is the length scale factor.

Back to your guess

The answer is C: It holds 8 times as much.

Doubling the height, the width and the depth makes 2 × 2 × 2 = 8 times the space inside. A jar that is twice as big in every direction holds 8 times as much. This chapter shows why.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. A small box has a volume of 40 cm³. A similar large box has every length 3 times as long. What is the volume of the large box?

    Tap your answer.

    1. Not this one. You multiplied the volume by 3. A volume has three lengths, so it is multiplied by 3 × 3 × 3. Go to M4.4.6. Try another.

    2. Not this one. You multiplied by 3 × 3 = 9, which is for an area. A volume needs three lengths, so use 3 × 3 × 3. Go to M4.4.6. Try another.

    3. Yes, that is right. The volume scale factor is k³ = 3 × 3 × 3 = 27. The large box has a volume of 40 × 27 = 1 080 cm³.

    See the answer

    1 080 cm³

    The volume scale factor is k³ = 3 × 3 × 3 = 27. The large box has a volume of 40 × 27 = 1 080 cm³.

    • If you chose A: You multiplied the volume by 3. A volume has three lengths, so it is multiplied by 3 × 3 × 3. Go to M4.4.6.
    • If you chose B: You multiplied by 3 × 3 = 9, which is for an area. A volume needs three lengths, so use 3 × 3 × 3. Go to M4.4.6.

  2. A poster has an area of 50 cm². It is enlarged by scale factor 4. What is the area of the enlarged poster?

    Tap your answer.

    1. Not this one. You multiplied the area by 4, the length scale factor. An area has two lengths, so multiply by 4 × 4. Go to M4.4.6. Try another.

    2. Yes, that is right. The area scale factor is k² = 4 × 4 = 16. The new area is 50 × 16 = 800 cm².

    3. Not this one. You multiplied by 4 × 4 × 4, which is for a volume. An area has only two lengths, so use 4 × 4. Go to M4.4.6. Try another.

    See the answer

    800 cm²

    The area scale factor is k² = 4 × 4 = 16. The new area is 50 × 16 = 800 cm².

    • If you chose A: You multiplied the area by 4, the length scale factor. An area has two lengths, so multiply by 4 × 4. Go to M4.4.6.
    • If you chose C: You multiplied by 4 × 4 × 4, which is for a volume. An area has only two lengths, so use 4 × 4. Go to M4.4.6.

  3. Two similar shapes have areas of 12 cm² and 108 cm². What is the length scale factor from the small shape to the large one?

    Tap your answer.

    1. Yes, that is right. The area scale factor is 108 ÷ 12 = 9. The length scale factor is the square root of 9, which is 3.

    2. Not this one. You divided 108 by 12. That is the area scale factor. The length scale factor is its square root. Go to M4.4.6. Try another.

    3. Not this one. You took 12 away from 108. Scale factors are found by dividing, and then a square root is needed for lengths. Go to M4.4.6. Try another.

    See the answer

    3

    The area scale factor is 108 ÷ 12 = 9. The length scale factor is the square root of 9, which is 3.

    • If you chose B: You divided 108 by 12. That is the area scale factor. The length scale factor is its square root. Go to M4.4.6.
    • If you chose C: You took 12 away from 108. Scale factors are found by dividing, and then a square root is needed for lengths. Go to M4.4.6.

  4. A small jar holds 100 ml. A similar large jar holds 800 ml. How many times as high as the small jar is the large jar?

    Tap your answer.

    1. Not this one. You divided 800 by 100. That is the volume scale factor. The length scale factor is its cube root. Go to M4.4.6. Try another.

    2. Not this one. You halved 8. Volume scales by k × k × k, so you need the number k with k × k × k = 8. Go to M4.4.6. Try another.

    3. Yes, that is right. The volume scale factor is 800 ÷ 100 = 8. The length scale factor k has k³ = 8, so k = 2, because 2 × 2 × 2 = 8.

    See the answer

    2

    The volume scale factor is 800 ÷ 100 = 8. The length scale factor k has k³ = 8, so k = 2, because 2 × 2 × 2 = 8.

    • If you chose A: You divided 800 by 100. That is the volume scale factor. The length scale factor is its cube root. Go to M4.4.6.
    • If you chose B: You halved 8. Volume scales by k × k × k, so you need the number k with k × k × k = 8. Go to M4.4.6.

  5. A shape is enlarged by scale factor 5. Which statement about its area is true?

    Tap your answer.

    1. Not this one. Five times is the scale factor for lengths. Area has two lengths, so it grows by more than that. Go to M4.4.6. Try another.

    2. Yes, that is right. The area scale factor is k² = 5 × 5 = 25.

    3. Not this one. You added 5 and 5. Areas are found by multiplying, so the factor is 5 × 5. Go to M4.4.6. Try another.

    See the answer

    The area is 25 times as big.

    The area scale factor is k² = 5 × 5 = 25.

    • If you chose A: Five times is the scale factor for lengths. Area has two lengths, so it grows by more than that. Go to M4.4.6.
    • If you chose C: You added 5 and 5. Areas are found by multiplying, so the factor is 5 × 5. Go to M4.4.6.

  6. From an earlier chapter

    Triangle ABC has sides of 4 cm and 5 cm. A similar triangle has a side of 12 cm that matches the 4 cm side. How long is the side that matches 5 cm, in centimetres?

    See the answer

    15

    The scale factor is 12 ÷ 4 = 3. Multiply 5 by it: 5 × 3 = 15 cm.

  7. A shape has an area of 6 cm². It is enlarged by scale factor 3. What is the new area, in cm²?

    See the answer

    54

    The area scale factor is 3 × 3 = 9. The new area is 6 × 9 = 54 cm².

  8. A jar holds 15 ml. A jar of the same shape has every length 2 times as long. How many ml does it hold?

    See the answer

    120

    The volume scale factor is 2 × 2 × 2 = 8. The larger jar holds 15 × 8 = 120 ml.

  9. A small jar holds 10 ml. A similar large jar holds 270 ml. How many times as high as the small jar is the large jar?

    See the answer

    3

    The volume scale factor is 270 ÷ 10 = 27. The length scale factor k has k × k × k = 27, so k = 3.

Keep it

Copy this into your exercise book.

Scale factor k: lengths × k, areas × k², volumes × k³. Doubling every length gives 4 times the area and 8 times the volume.

Draw: A 1 cm cube beside a 2 cm cube made of 8 small cubes, with the three scale factors written under them.

M4.4.7 Chapter 7

Arcs and sectors

A quarter of a round flat bread has a radius of 10 cm. How long is its curved edge?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

A circle can be cut like a flat bread. A piece of the curved edge is an arc. A slice, with two straight edges and an arc, is a sector. The straight edges of a sector are both radii. They meet at the centre, O, and make an angle there.

A circle with centre O and points A and B on it. The radius OA is labelled radius. A shaded sector runs from A anticlockwise to B. The arc from A to B is drawn heavy and labelled arc. The angle at O is marked and labelled θ.radiusarcθABO
The shaded part is a sector. The heavy curve is its arc. The angle θ is at the centre.

A full turn is 360°. A sector with an angle of θ degrees is θ out of 360 of the whole circle. If θ = 90°, the sector is 90 out of 360, which is a quarter. If θ = 180°, it is a half.

Draw it first

A sector of 90° is a quarter of the circle

  1. A circle with centre O and points A and B on it. Step 1: the radii OA and OB are drawn, with the angle of 90° at O marked and labelled 90°. Step 2: the sector between the radii is shaded. Step 3: the arc from A to B is drawn heavy and labelled arc.90°ABO

    1 Draw two radii, OA and OB, with an angle of 90° between them.

  2. A circle with centre O and points A and B on it. Step 1: the radii OA and OB are drawn, with the angle of 90° at O marked and labelled 90°. Step 2: the sector between the radii is shaded. Step 3: the arc from A to B is drawn heavy and labelled arc.90°ABO

    2 Shade the sector between the radii. It is 90 out of 360 of the circle, which is a quarter.

  3. A circle with centre O and points A and B on it. Step 1: the radii OA and OB are drawn, with the angle of 90° at O marked and labelled 90°. Step 2: the sector between the radii is shaded. Step 3: the arc from A to B is drawn heavy and labelled arc.arc90°ABO

    3 The arc is the part of the circumference inside the sector. It is a quarter of the circumference.

This gives the rules. The circumference is 2πr and the area is πr². A sector takes the fraction θ ÷ 360 of each. Use the π key on your calculator, and say how you rounded.

Arc length = (θ ÷ 360) × 2πr. Sector area = (θ ÷ 360) × πr². Perimeter of a sector = arc length + 2r. The 2r is for the two straight edges.

A circle with centre O of radius 6 cm. The radius OA is labelled 6 cm. A shaded sector runs from A anticlockwise to B, with an angle of 60° at O labelled 60°. The arc from A to B is heavy and labelled arc.6 cmarc60°ABO
A sector with a radius of 6 cm and an angle of 60°.

Worked example

Find the length of the arc of a sector with a radius of 6 cm and an angle of 60°. Give the answer to 1 decimal place.

  1. To 1 decimal place, the whole circumference is 2 × π × 6 = 37.7 cm.
  2. The angle 60° is 60 out of 360. So the arc is 60 ÷ 360 of the circumference.
  3. To 1 decimal place, the arc length is 60 ÷ 360 × 37.7 = 6.3 cm.

Answer: 6.3 cm

A circle with centre O of radius 10 cm. The radius OA is labelled 10 cm. A shaded sector runs from A anticlockwise to B, with an angle of 72° at O labelled 72°. The arc from A to B is heavy and labelled arc.10 cmarc72°ABO
A sector with a radius of 10 cm and an angle of 72°.

Worked example

Find the area of a sector with a radius of 10 cm and an angle of 72°. Give the answer to 1 decimal place.

  1. To 1 decimal place, the area of the whole circle is π × 10 × 10 = 314.2 cm².
  2. The sector is 72 out of 360 of the circle. That is 72 ÷ 360 = 0.2.
  3. To 1 decimal place, the sector area is 0.2 × 314.2 = 62.8 cm².

Answer: 62.8 cm²

To walk all the way round the edge of a sector, you go along the arc and then along both radii. So the perimeter is longer than the arc.

A circle with centre O of radius 9 cm. The radius OA is labelled 9 cm. A shaded sector runs from A anticlockwise to B, with an angle of 120° at O labelled 120°. The arc from A to B is heavy and labelled arc.9 cmarc120°ABO
A sector with a radius of 9 cm and an angle of 120°.

Worked example

Find the perimeter of a sector with a radius of 9 cm and an angle of 120°. Give the answer to 1 decimal place.

  1. To 1 decimal place, the arc length is 120 ÷ 360 × 2 × π × 9 = 18.8 cm.
  2. The sector has two straight edges. Each is a radius of 9 cm.
  3. Together the straight edges are 2 × 9 = 18 cm.
  4. Perimeter = 18.8 + 18 = 36.8 cm.

Answer: 36.8 cm

See it three ways

With things

Cut a round piece of paper. Fold it in half, then in half again. Open it. Each fold line is a radius, and a quarter of the circle lies between two of them.

As a picture

A circle with centre O. The radius OA is labelled radius. A shaded sector of a quarter of the circle runs from A anticlockwise to B, with a right angle at O labelled 90°. The arc from A to B is heavy and labelled arc.radiusarc90°ABO

In symbols

Arc length = (θ ÷ 360) × 2πr. Sector area = (θ ÷ 360) × πr². Sector perimeter = arc length + 2r.

Try it

Measure an arc with string

You need: a round tin lid or a compass, string, a ruler, a pencil.

  1. Draw a circle with a radius of 5 cm. Mark its centre.
  2. Draw two radii at right angles to each other. Check the angle with the corner of a sheet of paper.
  3. Lay a piece of string along the arc between the radii. Straighten the string and measure it.

What you will see: The string is about 7.9 cm long. That matches 90 ÷ 360 × 2 × π × 5, which is about 7.85 cm.

If you cannot try it, this is what happens.
A circle with centre O of radius 5 cm. The radius OA is labelled 5 cm. A shaded sector runs from A anticlockwise to B, with an angle of 90° at O labelled 90°. The arc from A to B is heavy and labelled arc.5 cmarc90°ABO

A wrong idea many people have

The perimeter of a sector is the length of its arc.

Why it feels right: The arc is the curved edge, and the curved edge is the part that needs the new rule.

The test that breaks it: Trace round a slice of flat bread with a finger. You go along the curve, then along one straight edge, then along the other. A sector with a radius of 9 cm has 18 cm of straight edges.

What is true: The perimeter is the whole distance round the edge. It is the arc length plus two radii: arc + 2r.

Key words

arc
a part of the circumference of a circle.
sector
a slice of a circle between two radii. It has an arc for its curved edge.

Back to your guess

The answer is B: About 15.7 cm. That is a quarter of the distance round the circle.

The curved edge is a quarter of the circumference. The circumference is about 62.8 cm, because 2 × π × 10 is about 62.8. A quarter of that is about 15.7 cm. This chapter shows how to do the same for any slice.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. A sector has a radius of 6 cm and an angle of 90°. How long is its arc, to 1 decimal place?

    A circle with centre O of radius 6 cm. The radius OA is labelled 6 cm. A shaded sector runs from A anticlockwise to B, with an angle of 90° at O labelled 90°. The arc from A to B is heavy and labelled arc.6 cmarc90°ABO
    Tap your answer.

    1. Not this one. You found the whole circumference. The arc is only a part of it, 90 out of 360. Go to M4.4.7. Try another.

    2. Yes, that is right. To 1 decimal place, the circumference is 2 × π × 6 = 37.7 cm. The sector is 90 out of 360, so the arc is 90 ÷ 360 × 37.7 = 9.4 cm.

    3. Not this one. You used the area rule, πr², and took 90 out of 360 of it. That is the area of the sector. An arc is a length, so use the circumference, 2πr. Go to M4.4.7. Try another.

    See the answer

    9.4 cm

    To 1 decimal place, the circumference is 2 × π × 6 = 37.7 cm. The sector is 90 out of 360, so the arc is 90 ÷ 360 × 37.7 = 9.4 cm.

    • If you chose A: You found the whole circumference. The arc is only a part of it, 90 out of 360. Go to M4.4.7.
    • If you chose C: You used the area rule, πr², and took 90 out of 360 of it. That is the area of the sector. An arc is a length, so use the circumference, 2πr. Go to M4.4.7.

  2. A sector has a radius of 9 cm and an angle of 40°. What is its area, to 1 decimal place?

    A circle with centre O of radius 9 cm. The radius OA is labelled 9 cm. A shaded sector runs from A anticlockwise to B, with an angle of 40° at O labelled 40°. The arc from A to B is heavy and labelled arc.9 cmarc40°ABO
    Tap your answer.

    1. Not this one. You found the area of the whole circle. The sector is only 40 out of 360 of it. Go to M4.4.7. Try another.

    2. Not this one. You used the arc length rule, with 2πr. An area uses πr², the area of the whole circle. Go to M4.4.7. Try another.

    3. Yes, that is right. The area of the whole circle is π × 9 × 9 = 254.5 cm². The sector is 40 out of 360, so its area is 40 ÷ 360 × 254.5 = 28.3 cm².

    See the answer

    28.3 cm²

    The area of the whole circle is π × 9 × 9 = 254.5 cm². The sector is 40 out of 360, so its area is 40 ÷ 360 × 254.5 = 28.3 cm².

    • If you chose A: You found the area of the whole circle. The sector is only 40 out of 360 of it. Go to M4.4.7.
    • If you chose B: You used the arc length rule, with 2πr. An area uses πr², the area of the whole circle. Go to M4.4.7.

  3. A sector has a radius of 5 cm and an angle of 90°. What is its perimeter, to 1 decimal place?

    Tap your answer.

    1. Yes, that is right. The arc is 90 ÷ 360 × 2 × π × 5 = 7.9 cm. The two radii are 2 × 5 = 10 cm. The perimeter is 7.9 + 10 = 17.9 cm.

    2. Not this one. You gave only the arc length. The perimeter goes all the way round, so add the two straight edges. Go to M4.4.7. Try another.

    3. Not this one. You added one radius. A sector has two straight edges, and both are radii. Add 2 × 5. Go to M4.4.7. Try another.

    See the answer

    17.9 cm

    The arc is 90 ÷ 360 × 2 × π × 5 = 7.9 cm. The two radii are 2 × 5 = 10 cm. The perimeter is 7.9 + 10 = 17.9 cm.

    • If you chose B: You gave only the arc length. The perimeter goes all the way round, so add the two straight edges. Go to M4.4.7.
    • If you chose C: You added one radius. A sector has two straight edges, and both are radii. Add 2 × 5. Go to M4.4.7.

  4. Which rule gives the length of an arc?

    Tap your answer.

    1. Not this one. Adding the angle to the radius mixes degrees and centimetres. An arc is a fraction of the circumference. Go to M4.4.7. Try another.

    2. Not this one. This does not give a length in centimetres. Degrees must first be turned into a fraction of 360. Go to M4.4.7. Try another.

    3. Not this one. This gives the area of a sector. An arc is a length, so it uses the circumference instead. Go to M4.4.7. Try another.

    4. Yes, that is right. An arc is a part of the circumference. The fraction of the circle is the angle ÷ 360, and the circumference is 2πr.

    See the answer

    Multiply the fraction of the circle by the circumference.

    An arc is a part of the circumference. The fraction of the circle is the angle ÷ 360, and the circumference is 2πr.

    • If you chose A: Adding the angle to the radius mixes degrees and centimetres. An arc is a fraction of the circumference. Go to M4.4.7.
    • If you chose B: This does not give a length in centimetres. Degrees must first be turned into a fraction of 360. Go to M4.4.7.
    • If you chose C: This gives the area of a sector. An arc is a length, so it uses the circumference instead. Go to M4.4.7.

  5. A sector has a radius of 4 m and an angle of 90°. What is its area, to 1 decimal place?

    Tap your answer.

    1. Not this one. You found the area of the whole circle. A sector with an angle of 90° is only a quarter of it. Go to M4.4.7. Try another.

    2. Yes, that is right. The whole circle has an area of π × 4 × 4 = 50.3 m². The sector is 90 ÷ 360 of it, which is 12.6 m².

    3. Not this one. The number is right, but the unit is not. An area is in square metres, m², and not in metres. Go to M4.4.7. Try another.

    See the answer

    12.6 m²

    The whole circle has an area of π × 4 × 4 = 50.3 m². The sector is 90 ÷ 360 of it, which is 12.6 m².

    • If you chose A: You found the area of the whole circle. A sector with an angle of 90° is only a quarter of it. Go to M4.4.7.
    • If you chose C: The number is right, but the unit is not. An area is in square metres, m², and not in metres. Go to M4.4.7.

  6. From an earlier chapter

    A shape has an area of 7 cm². It is enlarged by scale factor 3. What is the new area, in cm²?

    See the answer

    63

    The area scale factor is 3 × 3 = 9. The new area is 7 × 9 = 63 cm².

  7. A sector has a radius of 12 cm and an angle of 90°. How long is its arc, to 1 decimal place?

    See the answer

    18.8

    The circumference is 2 × π × 12 = 75.4 cm. The sector is 90 ÷ 360 of it, so the arc is 18.8 cm.

  8. A sector has a radius of 6 cm and an angle of 150°. What is its area, in cm², to 1 decimal place?

    See the answer

    47.1

    The area of the whole circle is π × 6 × 6 = 113.1 cm². The sector is 150 ÷ 360 of it, so its area is 47.1 cm².

  9. A sector has a radius of 8 cm and an angle of 45°. What is its perimeter, in cm, to 1 decimal place?

    See the answer

    22.3

    To 1 decimal place, the arc is 45 ÷ 360 × 2 × π × 8 = 6.3 cm. The two radii are 2 × 8 = 16 cm. The perimeter is 6.3 + 16 = 22.3 cm.

Keep it

Copy this into your exercise book.

Arc = (θ ÷ 360) × 2πr. Sector area = (θ ÷ 360) × πr². Sector perimeter = arc + 2r. Say how you rounded.

Draw: A circle with a shaded sector, its angle at the centre, the radius labelled, and the arc drawn heavy.

M4.4.8 Chapter 8

Angle rules in circles

Two lines from the ends of an arc meet at the centre at 100°. What is the angle when they meet at the edge instead?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

A chord is a straight line joining two points on a circle. The two points cut the circle into two arcs. A line from the centre to each end of a chord makes an angle at the centre. A line from a third point on the circle to each end makes an angle at the circumference. Both angles stand on the arc AB that does not contain C.

A circle with centre O and points A, B and C on it. Lines join O to A and to B, and C to A and to B. The angle at O between A and B is marked and labelled 150°. The angle at C between A and B is marked and labelled 75°. Both stand on the arc AB.75°150°ABCO
The angle at the centre is 150°. The angle at the circumference is 75°. They stand on the same arc AB.

The first rule. The angle at the centre is double the angle at the circumference on the same arc. Here 150° is double 75°. Measure with a protractor on a big drawing and you will see it holds for any circle.

Here is why. Draw the line from C through O to the other side of the circle, ending at D. The lines OA, OB and OC are all radii, so they are equal. That makes triangles OAC and OBC isosceles.

The same circle with a line from C through the centre O to D on the circle. At C the angle between A and D is labelled x, and the angle between D and B is labelled y. At O the angle between A and D is labelled 2x, and the angle between D and B is labelled 2y.xy2x2yABCDO
The line CD splits both angles into two parts. The angles x and y are at C. The angles 2x and 2y are at O.

Call the angle at C in triangle OAC x. Chapter 4 proved that the base angles of an isosceles triangle are equal, so the angle at A is also x. The angles on the straight line CD add to 180°, and so do the angles of triangle OAC. So the angle AOD is x + x = 2x. The same reasoning on the other side gives angle DOB = 2y. The angle at C is x + y and the angle at O is 2x + 2y. That is double. This drawing has O inside the angle at C. When O lies outside it, the same steps work with a subtraction in place of the addition.

The second rule follows at once. A diameter makes a straight angle of 180° at the centre. The angle at the circumference on the same arc is half of that, which is 90°. So the angle in a semicircle is 90°.

A circle with centre O. A diameter joins A and B. A third point C lies on the circle, and lines join C to A and to B. The angle at C is a right angle, marked with a small square and labelled 90°.90°ABCO
AB is a diameter. The angle at C is 90°, wherever C is on the circle.

The third rule is about a cyclic quadrilateral. It is a four sided shape with all four corners on a circle. Opposite angles add up to 180°. Why? The two angles at the centre on the arcs between the opposite corners add to 360°. Each angle at the edge is half of its centre angle. So the two edge angles add to half of 360°, which is 180°.

A circle with a four sided shape ABCD whose corners are on the circle. The angle at A is labelled 100°, at B 85°, at C 80° and at D 95°. The angles at A and C are opposite, and so are the angles at B and D.80°85°95°100°ABCD
Opposite corners: 100° + 80° = 180°, and 85° + 95° = 180°.

The fourth rule is about a tangent. A tangent is a straight line that touches a circle at one point. Every other point of the line lies outside the circle. So the radius to the touching point is the shortest line from the centre to the tangent. The shortest line to a line is the one at a right angle. So a tangent meets the radius at 90°.

A circle with centre O and a point P on it. A radius joins O to P. A straight line, the tangent, touches the circle at P and meets the radius at a right angle, marked with a small square and labelled 90°.tangent90°PO
The tangent touches the circle at P. The radius OP is at 90° to it.
A circle with centre O and points A, B and C on it. Lines join O to A and to B, and C to A and to B. The angle at O between A and B is labelled 140°. The angle at C between A and B is labelled x.x140°ABCO
Find the angle x.

Worked example

The angle at the centre of a circle is 140°. Find the angle x at the circumference, on the same arc.

  1. Both angles stand on the arc AB.
  2. The angle at the centre is double the angle at the circumference.
  3. So x = 140 ÷ 2 = 70°.

Answer: 70°

A circle with centre O. A diameter joins A and B. A point C lies on the circle. The angle at A between C and B is labelled 38°. The angle at C is marked as a right angle. The angle at B between A and C is labelled x.38°xABCO
Find the angle x.

Worked example

AB is a diameter and C is on the circle. Angle CAB is 38°. Find angle ABC.

  1. AB is a diameter, so the angle in a semicircle at C is 90°.
  2. The angles of triangle ABC add up to 180°.
  3. Angle ABC = 180 − 90 − 38 = 52°.

Answer: 52°

A circle with a four sided shape PQRS whose corners are on the circle. The angle at P is labelled 105° and the angle at Q is labelled 80°. The angle at R is labelled x and the angle at S is labelled y.x80°y105°PQRS
PQRS is a cyclic quadrilateral. Find x and y.

Worked example

PQRS is a cyclic quadrilateral. Angle P is 105° and angle Q is 80°. Find angle R and angle S.

  1. Opposite angles of a cyclic quadrilateral add up to 180°.
  2. Angle R is opposite angle P, so R = 180 − 105 = 75°.
  3. Angle S is opposite angle Q, so S = 180 − 80 = 100°.
  4. Check: the four angles are 105°, 80°, 75° and 100°. They add up to 360°.

Answer: R is 75° and S is 100°

A circle with centre O and a point P on it. T is a point outside the circle on the tangent at P. A radius joins O to P, and a line joins O to T. The angle at P between O and T is a right angle. The angle at O between P and T is labelled 50°. The angle at T is labelled x.x50°PTO
PT is a tangent. Find the angle x.

Worked example

PT is a tangent to a circle with centre O. Angle POT is 50°. Find angle OTP.

  1. The tangent PT meets the radius OP at 90°. So angle OPT = 90°.
  2. The angles of triangle OPT add up to 180°.
  3. Angle OTP = 180 − 90 − 50 = 40°.
  4. Check: 90° + 50° + 40° = 180°.

Answer: 40°

See it three ways

With things

Stand a coin up against the straight edge of a ruler. The ruler touches the coin at one point. Use the corner of a sheet of paper to check. The line from the middle of the coin to that point is square to the ruler.

As a picture

A circle with centre O and a point P on it. A radius joins O to P. A straight line, the tangent, touches the circle at P and meets the radius at a right angle, marked with a small square and labelled 90°.tangent90°PO

In symbols

The angle between a tangent and the radius to the touching point is 90°.

Try it

Check the angle in a semicircle

You need: a round tin lid or a compass, a ruler, the corner of a sheet of paper, a pencil.

  1. Draw round the tin lid, or use a compass. Mark the centre and draw a diameter through it.
  2. Mark any point C on the circle. Join C to both ends of the diameter.
  3. Fit the corner of a sheet of paper into the angle at C.
  4. Move C to a new place on the circle and try again.

What you will see: The corner fits exactly each time. The angle in a semicircle is 90° wherever C is.

If you cannot try it, this is what happens.
A circle with centre O. A diameter joins A and B. A point C lies on the circle and lines join C to A and to B. The angle at C is a right angle, marked with a small square and labelled 90°.90°ABCO

A wrong idea many people have

The angle at the centre equals the angle at the circumference on the same arc.

Why it feels right: Both angles are drawn from the same two points, so they look as if they should match.

The test that breaks it: Take a diameter AB. It makes a straight angle of 180° at the centre. The angle at any point C on the circle is 90°, as the tin lid test showed. They are not equal.

What is true: The angle at the centre is double the angle at the circumference on the same arc.

Key words

chord
a straight line joining two points on a circle.
stand on
an angle stands on the arc between the ends of its arms, on the side away from the angle.
semicircle
half of a circle, cut off by a diameter.
cyclic quadrilateral
a four sided shape with all four corners on a circle.
tangent
a straight line that touches a circle at one point.

Back to your guess

The answer is B: 50°, half of the angle at the centre.

The angle at the edge, on the same arc, is half the angle at the centre. So 100° at the centre makes 50° at the edge. This chapter shows why, and uses it to find angles.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. The angle at the centre is 84°. What is the angle x at the circumference on the same arc?

    A circle with centre O and points A, B and C on it. Lines join O to A and to B, and C to A and to B. The angle at O between A and B is labelled 84°. The angle at C between A and B is labelled x.x84°ABCO
    Tap your answer.

    1. Not this one. The angle at the circumference is not the same as the angle at the centre. It is half as big. Go to M4.4.8. Try another.

    2. Yes, that is right. The angle at the centre is double the angle at the circumference on the same arc. So x = 84 ÷ 2 = 42°.

    3. Not this one. You doubled the angle. It is the angle at the centre that is double, so the angle at the circumference is half. Go to M4.4.8. Try another.

    See the answer

    42°

    The angle at the centre is double the angle at the circumference on the same arc. So x = 84 ÷ 2 = 42°.

    • If you chose A: The angle at the circumference is not the same as the angle at the centre. It is half as big. Go to M4.4.8.
    • If you chose C: You doubled the angle. It is the angle at the centre that is double, so the angle at the circumference is half. Go to M4.4.8.

  2. AB is a diameter and C is on the circle. Angle CAB is 25°. What is angle ABC?

    A circle with centre O. A diameter joins A and B. A point C lies on the circle. The angle at A between C and B is labelled 25°. The angle at B between A and C is labelled x.25°xABCO
    Tap your answer.

    1. Not this one. The angle at C is 90°, but the question asks for the angle at B. Use the angles of the triangle. Go to M4.4.8. Try another.

    2. Not this one. You took 25° away from 180°. The triangle ABC also has a right angle at C, which must be taken away too. Go to M4.4.8. Try another.

    3. Yes, that is right. The angle in a semicircle at C is 90°. The angles of the triangle add up to 180°, so angle ABC = 180 − 90 − 25 = 65°.

    4. Not this one. You copied the angle at A. The two angles are not equal, because triangle ABC is not isosceles. Go to M4.4.8. Try another.

    See the answer

    65°

    The angle in a semicircle at C is 90°. The angles of the triangle add up to 180°, so angle ABC = 180 − 90 − 25 = 65°.

    • If you chose A: The angle at C is 90°, but the question asks for the angle at B. Use the angles of the triangle. Go to M4.4.8.
    • If you chose B: You took 25° away from 180°. The triangle ABC also has a right angle at C, which must be taken away too. Go to M4.4.8.
    • If you chose D: You copied the angle at A. The two angles are not equal, because triangle ABC is not isosceles. Go to M4.4.8.

  3. ABCD is a cyclic quadrilateral. Angle A is 70°. What is angle C?

    A circle with a four sided shape ABCD whose corners are on the circle. The angle at A is labelled 70° and the opposite angle at C is labelled x.70°xABCD
    Tap your answer.

    1. Yes, that is right. Opposite angles of a cyclic quadrilateral add up to 180°. Angle C = 180 − 70 = 110°.

    2. Not this one. Opposite angles of a cyclic quadrilateral add up to 180°. They are not equal unless each one is 90°. Go to M4.4.8. Try another.

    3. Not this one. You took 70° away from 360°. That rule is for all four angles together. For two opposite angles the total is 180°. Go to M4.4.8. Try another.

    See the answer

    110°

    Opposite angles of a cyclic quadrilateral add up to 180°. Angle C = 180 − 70 = 110°.

    • If you chose B: Opposite angles of a cyclic quadrilateral add up to 180°. They are not equal unless each one is 90°. Go to M4.4.8.
    • If you chose C: You took 70° away from 360°. That rule is for all four angles together. For two opposite angles the total is 180°. Go to M4.4.8.

  4. PT is a tangent to a circle with centre O. Angle POT is 48°. What is angle OTP?

    A circle with centre O and a point P on it. T is a point outside the circle on the tangent at P. A radius joins O to P and a line joins O to T. The angle at P is a right angle. The angle at O between P and T is labelled 48°. The angle at T is labelled x.x48°PTO
    Tap your answer.

    1. Not this one. You took 48° away from 180° and forgot the right angle at P. The angles of triangle OPT include the 90°. Go to M4.4.8. Try another.

    2. Not this one. You copied the angle at O. The tangent makes a right angle with the radius, so the other angle is not equal. Go to M4.4.8. Try another.

    3. Yes, that is right. The tangent meets the radius at 90°, so angle OPT = 90°. The angles of triangle OPT add up to 180°, so angle OTP = 180 − 90 − 48 = 42°.

    See the answer

    42°

    The tangent meets the radius at 90°, so angle OPT = 90°. The angles of triangle OPT add up to 180°, so angle OTP = 180 − 90 − 48 = 42°.

    • If you chose A: You took 48° away from 180° and forgot the right angle at P. The angles of triangle OPT include the 90°. Go to M4.4.8.
    • If you chose B: You copied the angle at O. The tangent makes a right angle with the radius, so the other angle is not equal. Go to M4.4.8.

  5. Which statement about a circle is true?

    Tap your answer.

    1. Not this one. They are not equal. The angle at the centre is double the angle at the circumference on the same arc. Go to M4.4.8. Try another.

    2. Yes, that is right. This is the first rule of the chapter. The angle at the centre is double the angle at the circumference when both stand on the same arc.

    3. Not this one. Opposite angles add up to 180°. They are equal only when each is 90°. Go to M4.4.8. Try another.

    See the answer

    The angle at the centre is double the angle at the circumference on the same arc.

    This is the first rule of the chapter. The angle at the centre is double the angle at the circumference when both stand on the same arc.

    • If you chose A: They are not equal. The angle at the centre is double the angle at the circumference on the same arc. Go to M4.4.8.
    • If you chose C: Opposite angles add up to 180°. They are equal only when each is 90°. Go to M4.4.8.

  6. From an earlier chapter

    A sector has a radius of 8 cm and an angle of 45°. How long is its arc, in centimetres, to 1 decimal place?

    See the answer

    6.3

    The circumference is 2 × π × 8 = 50.3 cm. The sector is 45 out of 360 of it, so the arc is 6.3 cm.

  7. The angle at the centre of a circle is 120°. What is the angle at the circumference on the same arc, in degrees?

    See the answer

    60

    The angle at the circumference is half the angle at the centre: 120 ÷ 2 = 60°.

  8. ABCD is a cyclic quadrilateral. Angle B is 83°. What is angle D, in degrees?

    See the answer

    97

    Opposite angles of a cyclic quadrilateral add up to 180°. Angle D = 180 − 83 = 97°.

  9. A tangent touches a circle at P. The centre is O and T is a point on the tangent. Angle OTP is 31°. What is angle POT, in degrees?

    See the answer

    59

    The tangent meets the radius at 90°, so angle OPT = 90°. The angles of triangle OPT add up to 180°, so angle POT = 180 − 90 − 31 = 59°.

Keep it

Copy this into your exercise book.

Angle at the centre = 2 × angle at the circumference, same arc. Angle in a semicircle = 90°. Cyclic quadrilateral: opposite angles add to 180°. A tangent is square to the radius: 90°.

Draw: A circle with a chord and the angles at the centre and the edge. Add a tangent with a right angle mark.

Key words, A to Z

Every key word in this book, what it means, and the chapter that teaches it.

anticlockwise
in the opposite direction to the hands of a clock. M4.4.1
arc
a part of the circumference of a circle. M4.4.7
area scale factor
the number every area is multiplied by. It is k², where k is the length scale factor. M4.4.6
ASA
two angles and the side between them are equal. If the side is not between them, the test is called AAS and it works too. M4.4.4
centre of enlargement
the point the distances are measured from. It does not move. M4.4.3
centre of rotation
the point that stays still while the shape turns. M4.4.1
chord
a straight line joining two points on a circle. M4.4.8
clockwise
in the direction the hands of a clock move. M4.4.1
column vector
two numbers in brackets, one above the other. The top number is the move right and the bottom number is the move up. M4.4.2
combined move
two moves one after the other. For two translations it is one translation. M4.4.2
congruent
having exactly the same size and form. One shape fits exactly on the other. M4.4.4
corresponding
matching. Corresponding sides and angles sit in the same place in each shape. M4.4.4
cyclic quadrilateral
a four sided shape with all four corners on a circle. M4.4.8
enlargement
a transformation that multiplies every distance from a centre by the same number. The shape keeps its form. M4.4.3
image
the shape that a transformation makes. M4.4.1
mirror line
the line a shape is flipped over. A point and its image are the same distance from it. M4.4.1
proof
a chain of reasons showing a fact must be true. M4.4.4
reflection
a flip of a shape over a mirror line. M4.4.1
RHS
right angle, hypotenuse and one other side are equal. M4.4.4
rotation
a turn of a shape about a fixed point. M4.4.1
SAS
two sides and the angle between them are equal. M4.4.4
scale factor
the number every length is multiplied by. A scale factor between 0 and 1 makes the shape smaller. M4.4.3
sector
a slice of a circle between two radii. It has an arc for its curved edge. M4.4.7
semicircle
half of a circle, cut off by a diameter. M4.4.8
similar
having the same form, with equal matching angles and matching sides in the same ratio. One shape is an enlargement of the other. M4.4.5
SSS
three sides of one triangle equal three sides of another. M4.4.4
stand on
an angle stands on the arc between the ends of its arms, on the side away from the angle. M4.4.8
tangent
a straight line that touches a circle at one point. M4.4.8
transformation
a rule that moves a shape: a flip, a turn, a slide or an enlargement. M4.4.1
translation
a slide of a shape. Every point moves the same distance in the same direction. M4.4.2
volume scale factor
the number every volume is multiplied by. It is k³, where k is the length scale factor. M4.4.6