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Maths · Angles and trigonometry

Pythagoras' theorem 2

Pythagoras hidden inside other problems: the distance between two points, the height of an isosceles triangle, the slant side of a trapezium, and two triangles joined together.

  • 4 key terms
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Teacher resources

The teacher copies: slides with the questions built in, the answers, and anything else attached to this lesson for whoever is teaching it.

Student handouts

The same files the students see, to print or hand out.

Last Lesson and Before

Answer each one, then check.

  1. 1

    Last lesson: a right-angled triangle has shorter sides 6 and 8. How long is the hypotenuse?

    Show answerHide answer

    10

  2. 2

    What is the formula for the area of a triangle?

    Show answerHide answer

    \(\frac{1}{2} \times \text{base} \times \text{height}\)

  3. 3

    Work out \(4 - (-2)\).

    Show answerHide answer

    6

  4. 4

    Is 5, 12, 13 a Pythagorean triple?

    Show answerHide answer

    Yes: \(25 + 144 = 169\)

Learning Objectives

  1. 1Find the distance between two points on a coordinate grid.
  2. 2Find the height of an isosceles triangle, and use it to find the area.
  3. 3Spot the right-angled triangle hidden inside a shape.
  4. 4Solve problems that need Pythagoras' theorem twice.

Finding the Right-Angled Triangle

Most problems don't hand you a right-angled triangle - you have to draw it in.

  • Isosceles triangle

    The line of symmetry cuts it into two right-angled triangles, and halves the base.

  • Rectangle

    A diagonal makes two right-angled triangles.

  • Trapezium

    Drop a vertical height from a top corner to make a right-angled triangle at the end.

  • Coordinates

    The horizontal and vertical distances between two points are the two shorter sides.

The Distance Between Two Points

Work out the length of the line joining A(1, 2) and B(10, 7). Give your answer to 1 decimal place.

Show the solutionHide the solution
  1. 1 Horizontal distance \(10 - 1 = 9\)
  2. 2 Vertical distance \(7 - 2 = 5\)
  3. 3 Pythagoras \(AB^2 = 9^2 + 5^2 = 81 + 25 = 106\)
  4. 4 Square root \(AB = \sqrt{106} = 10.295\ldots\)

Answer10.3 units

Area of an Isosceles Triangle

An isosceles triangle has sides 10 cm, 10 cm and 12 cm. Work out its area.

Show the solutionHide the solution
  1. 1 The line of symmetry halves the base Half base = 6 cm, hypotenuse 10 cm
  2. 2 Pythagoras for the height \(h^2 = 10^2 - 6^2 = 100 - 36 = 64\), so \(h = 8\)
  3. 3 Area \(\frac{1}{2} \times 12 \times 8 = 48\)

Answer48 cm²

The Slant Side of a Trapezium

An isosceles trapezium has parallel sides 10 cm and 16 cm, and a height of 8 cm. Work out its perimeter, to 1 decimal place.

Show the solutionHide the solution
  1. 1 The extra length on the longer side is shared between the two ends \((16 - 10) \div 2 = 3\) cm
  2. 2 Pythagoras for one slant side \(s^2 = 3^2 + 8^2 = 73\), so \(s = 8.544\ldots\)
  3. 3 Add the four sides \(10 + 16 + 2 \times 8.544\ldots = 43.088\ldots\)

Answer43.1 cm

Two Triangles, Two Steps

When two right-angled triangles share a side, find the shared side first.

  1. 1 Sketch

    Draw both triangles and mark the right angles.
  2. 2 Shared side

    Find the side the triangles share, using the triangle where you know two sides.
  3. 3 Keep it exact

    Keep the full value (or its square) in your calculator - don't round yet.
  4. 4 Second triangle

    Use the shared side to find the length asked for.
  5. 5 Round last

    Round only the final answer.

Shortest Route

A spider is in one corner of the floor of a room 4 m long and 3 m wide. A fly is in the opposite corner of the floor. (a) How far does the spider walk if it goes along two walls? (b) How far if it walks straight across the floor? (c) Points P(−3, 1) and Q(5, 7) are two towns on a map grid in km. How far apart are they?

1. Sketch the right-angled triangle.

2. Label the two shorter sides.

3. Use Pythagoras.

A good answer shows: (a) \(4 + 3 = 7\) m (b) \(\sqrt{16 + 9} = 5\) m (c) Horizontal 8, vertical 6, so \(\sqrt{64 + 36} = 10\) km.

Can I...?

  1. 1Find the distance between two points on a grid.
  2. 2Find the height of an isosceles triangle.
  3. 3Find the area of an isosceles triangle.
  4. 4Find the slant side of a trapezium.
  5. 5Use Pythagoras twice with two triangles.

Summary & Exam Focus

  • Distance between points: horizontal and vertical differences are the shorter sides.
  • Isosceles triangle: the line of symmetry halves the base and makes a right angle.
  • Two triangles: find the shared side first.

Exam focus

Triangle ABC is right-angled at B, with AB = 5 cm and BC = 12 cm. Triangle ACD is right-angled at D, with CD = 5 cm. Work out the length of AD. (4 marks) (4 marks)

Mark every right angle on the diagram before you start. Each Pythagoras step uses one right-angled triangle - write which one.

Key terms

The vocabulary this lesson expects you to use. Each one is linked from the first place it appears above.

Hypotenuse
The longest side of a right-angled triangle, opposite the right angle.
Line of symmetry
A line that splits a shape into two mirror-image halves.
Perpendicular height
The height measured at right angles to the base.
Coordinates
A pair of numbers \((x, y)\) giving the position of a point.

Questions and answers

7 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Non-calculator 3 marks Easier

Point A has coordinates \((-2, 3)\) and point B has coordinates \((4, 11)\). Work out the length of AB.

Mark scheme — 3 marks available

  • 6 and 8 — M1
  • \(6^2 + 8^2\) — M1
  • 10 — A1

Model answer

Horizontal \(4 - (-2) = 6\), vertical \(11 - 3 = 8\). \(AB = \sqrt{6^2 + 8^2} = \sqrt{100} = 10\)

2. Exam question Non-calculator 4 marks Easier

An isosceles triangle has sides of 13 cm, 13 cm and 10 cm. Work out the area of the triangle.

Mark scheme — 4 marks available

  • Half base 5 used in a right-angled triangle with hypotenuse 13 — P1
  • \(13^2 - 5^2\) — P1
  • Height 12 — P1
  • 60 cm² — A1

Model answer

Half the base is 5 cm. Height \(= \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\) cm. Area \(= \frac{1}{2} \times 10 \times 12 = 60\) cm².

3. Exam question Non-calculator 4 marks Easier

The diagram shows two right-angled triangles, ABC and ACD. AB = 5 cm, BC = 12 cm and CD = 5 cm. Work out the length of AD.

Two right-angled triangles sharing side AC: ABC with right angle at B, AB 5 cm and BC 12 cm, and ACD with right angle at D and CD 5 cm.

Mark scheme — 4 marks available

  • \(5^2 + 12^2\) — P1
  • AC = 13 — P1
  • \(13^2 - 5^2\) — P1
  • AD = 12 cm — A1

Model answer

Triangle ABC: \(AC = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\) cm. Triangle ACD: \(AD = \sqrt{13^2 - 5^2} = \sqrt{144} = 12\) cm.

4. Exam question Calculator 4 marks Easier

An isosceles trapezium has parallel sides of 10 cm and 16 cm. Its perpendicular height is 8 cm. Work out the perimeter of the trapezium. Give your answer to 1 decimal place.

Mark scheme — 4 marks available

  • 3 found — P1
  • \(3^2 + 8^2\) — P1
  • \(10 + 16 + 2\sqrt{73}\) — P1
  • 43.1 cm — A1

Model answer

Each end overhangs \((16 - 10) \div 2 = 3\) cm. Slant side \(= \sqrt{3^2 + 8^2} = \sqrt{73} = 8.544\ldots\) cm. Perimeter \(= 10 + 16 + 2 \times 8.544\ldots = 43.1\) cm.

5. Multiple choice 1 mark Easier

How far apart are the points \((1, 1)\) and \((4, 5)\)?

  1. A 7
  2. B 5 Correct
  3. C 25
  4. D 4.5

Why: Horizontal 3, vertical 4, so the distance is \(\sqrt{9 + 16} = 5\).

6. Multiple choice 1 mark Core

An isosceles triangle has sides 5 cm, 5 cm and 8 cm. What is its height?

  1. A 3 cm Correct
  2. B 4 cm
  3. C 6.4 cm
  4. D 9 cm

Why: Half the base is 4, so the height is \(\sqrt{25 - 16} = 3\) cm.

7. Multiple choice 1 mark Stretch

A square has a diagonal of 10 cm. What is its area?

  1. A 100 cm²
  2. B 25 cm²
  3. C 50 cm² Correct
  4. D 70.7 cm²

Why: If the side is \(s\), \(s^2 + s^2 = 100\), so \(2s^2 = 100\) and the area \(s^2 = 50\) cm².