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GCSE Maths · EdexcelChapter 20 · Pythagoras and trigonometry
GCSE Maths · Pythagoras and Trigonometry · Foundation · Lesson 20.1

GCSE Maths: Pythagoras’ theorem: finding the hypotenuse and a shorter side (Foundation)

Part of the GCSE Maths pythagoras and trigonometry chapter (Foundation). Chapter overview · Higher version of this lesson · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)

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20.1

Pythagoras’ theorem: finding the hypotenuse and a shorter side

Edexcel 1MA1: G20, G6
Grade 3-5

A right-angled triangle has one angle of 90° (a square corner). The longest side of a right-angled triangle is the one opposite the right angle. It has its own name: the hypotenuse. The other two sides are called the shorter sides (or the legs).

shorter sideshorter sidehypotenuse
The hypotenuse is opposite the right angle. It is always the longest side.

More than 2500 years ago, people noticed something special. Draw a square on each side of a right-angled triangle. Then the area of the square on the hypotenuse is exactly equal to the areas of the two other squares added together. In the picture below, the squares on the shorter sides have areas 9 and 16, and 9 + 16 = 25, which is the area of the biggest square.

3² = 94² = 165² = 25
Squares on the three sides of a 3, 4, 5 triangle: 3² + 4² = 5², that is 9 + 16 = 25.

This is Pythagoras’ theorem. If the shorter sides are a and b and the hypotenuse is c, then

a2 + b2 = c2

Remember from Chapter 2 that a2 means a × a. The theorem only works for right-angled triangles, and c must be the hypotenuse.

Finding the hypotenuse. If you know both shorter sides, square them, add the squares, then take the square root. The square root undoes the squaring, because we want c, not c2.

Finding a shorter side. If you know the hypotenuse and one shorter side, square them, subtract (the hypotenuse square is the big one, so you take the smaller square away), then take the square root.

Why add one time and subtract the other? Because the equation is a2 + b2 = c2. Finding c means working out the right-hand side directly. Finding a means moving b2 to the other side, which turns the plus into a minus.

A calculator gives the square root of a number that is not a perfect square as a long decimal, so we round. Always read the question to see how accurate to be. The Higher tier also needs answers left as a surd (a square root like √40 simplified to 2√10) when the question asks for an exact answer.

Key ideas

  • Hypotenuse = the side opposite the right angle = the longest side.
  • Pythagoras: a2 + b2 = c2 (c is the hypotenuse).
  • Finding the hypotenuse: add the squares, then square root.
  • Finding a shorter side: subtract the squares, then square root.
  • Check: the hypotenuse must be longer than both shorter sides.
Worked example 20.1.1
Work out the length of the hypotenuse, x.
5 cm12 cmx

Key words. The hypotenuse is the longest side of a right-angled triangle; it is always opposite the right angle. Squaring a number means multiplying it by itself. The square root (√) undoes squaring.

Method. To find the hypotenuse, square the two shorter sides, add them, then take the square root.

  1. Spot the hypotenuse. The hypotenuse is the side opposite the right angle (the little square) and it is the longest side. The side we want, x, is the one opposite the right angle, so it is the hypotenuse. We know the two shorter sides.
    Shorter sides: 5 cm and 12 cm.
  2. Write down Pythagoras’ theorem and put the numbers in. For a right-angled triangle, (shorter side)2 + (shorter side)2 = (hypotenuse)2.
    a2 + b2 = c2
    52 + 122 = x2
  3. Square each short side. Squaring means multiplying the number by itself.
    52 = 5 × 5 = 25
    122 = 12 × 12 = 144
  4. Add the two squares. This gives x2, the square of the hypotenuse.
    x2 = 25 + 144 = 169
  5. Take the square root. We want x, not x2. The square root undoes the squaring.
    x = √169 = 13 cm

Check. The hypotenuse must be the longest side: 13 is bigger than 5 and 12. Also 52 + 122 = 169 = 132. ✓

Watch out. For the hypotenuse you add the squares. Do not stop at x2: the last step is the square root.

Worked example 20.1.2
Work out the length of the hypotenuse, x. Give your answer correct to 1 decimal place.
7 cm9 cmx

Key words. The hypotenuse is the longest side of a right-angled triangle; it is always opposite the right angle. Squaring a number means multiplying it by itself. The square root (√) undoes squaring.

Method. To find the hypotenuse, square the two shorter sides, add them, then take the square root.

  1. Spot the hypotenuse. The hypotenuse is the side opposite the right angle (the little square) and it is the longest side. The side we want, x, is the one opposite the right angle, so it is the hypotenuse. We know the two shorter sides.
    Shorter sides: 7 cm and 9 cm.
  2. Write down Pythagoras’ theorem and put the numbers in. For a right-angled triangle, (shorter side)2 + (shorter side)2 = (hypotenuse)2.
    a2 + b2 = c2
    72 + 92 = x2
  3. Square each short side. Squaring means multiplying the number by itself.
    72 = 7 × 7 = 49
    92 = 9 × 9 = 81
  4. Add the two squares. This gives x2, the square of the hypotenuse.
    x2 = 49 + 81 = 130
  5. Take the square root. We want x, not x2. The square root undoes the squaring. Use the √ button on the calculator.
    x = √130 = 11.40175……
  6. Round to 1 d.p.. The question tells us how accurate to be.
    x = 11.4 cm

Check. Square the answer: 11.42 = 129.96, which is close to 130. The hypotenuse 11.4 is longer than both 7 and 9. ✓

Watch out. For the hypotenuse you add the squares. Do not stop at x2: the last step is the square root.

Worked example 20.1.3
Work out the length of the shorter side, x.
x8 cm17 cm

Key words. The hypotenuse is the longest side of a right-angled triangle; it is always opposite the right angle. Squaring a number means multiplying it by itself. The square root (√) undoes squaring.

Method. To find a shorter side, square the hypotenuse and the known side, subtract, then take the square root.

  1. Spot the hypotenuse. The hypotenuse is opposite the right angle and is the longest side. This time it is given, and we are looking for one of the two shorter sides.
    Hypotenuse = 17 cm. The other known side = 8 cm. The side we want is x.
  2. Write down Pythagoras’ theorem and put the numbers in. The hypotenuse is always the one on its own on the right-hand side: (shorter)2 + (shorter)2 = (hypotenuse)2.
    a2 + b2 = c2
    x2 + 82 = 172
  3. Square the known numbers. Squaring means multiplying the number by itself.
    82 = 8 × 8 = 64
    172 = 17 × 17 = 289
    x2 + 64 = 289
  4. Subtract 64 from both sides. We want x2 on its own. Taking 64 away from both sides keeps the equation balanced.
    x2 = 289 − 64 = 225
  5. Take the square root. We want x, not x2.
    x = √225 = 15 cm

Check. Put it back: 152 + 82 = 225 + 64 = 289 = 172. ✓ And 15 is shorter than the hypotenuse 17.

Watch out. When you are finding a shorter side you subtract. A shorter side must always be smaller than the hypotenuse; if your answer is bigger, you added by mistake.

Worked example 20.1.4
Work out the length x. Give your answer correct to 1 decimal place.
6.2 mx9.4 m

Key words. The hypotenuse is the longest side of a right-angled triangle; it is always opposite the right angle. Squaring a number means multiplying it by itself. The square root (√) undoes squaring.

Method. To find a shorter side, square the hypotenuse and the known side, subtract, then take the square root.

  1. Spot the hypotenuse. The hypotenuse is opposite the right angle and is the longest side. This time it is given, and we are looking for one of the two shorter sides.
    Hypotenuse = 9.4 m. The other known side = 6.2 m. The side we want is x.
  2. Write down Pythagoras’ theorem and put the numbers in. The hypotenuse is always the one on its own on the right-hand side: (shorter)2 + (shorter)2 = (hypotenuse)2.
    a2 + b2 = c2
    x2 + 6.22 = 9.42
  3. Square the known numbers. Squaring means multiplying the number by itself.
    6.22 = 6.2 × 6.2 = 38.44
    9.42 = 9.4 × 9.4 = 88.36
    x2 + 38.44 = 88.36
  4. Subtract 38.44 from both sides. We want x2 on its own. Taking 38.44 away from both sides keeps the equation balanced.
    x2 = 88.36 − 38.44 = 49.92
  5. Take the square root. We want x, not x2.
    x = √49.92 = 7.06540…
  6. Round to 1 d.p.. The question tells us how accurate to be.
    x = 7.1 m

Check. Put it back: 7.12 + 6.22 = 50.41 + 38.44 = 88.85, which is close to 9.42 = 88.36. ✓ The answer is shorter than the hypotenuse 9.4.

Watch out. When you are finding a shorter side you subtract. A shorter side must always be smaller than the hypotenuse; if your answer is bigger, you added by mistake.

Watch out

Do not add the two shorter sides and stop, and do not forget the last step. If c2 = 169 the answer is c = 13, not 169. Also, always check that the answer for a shorter side is smaller than the hypotenuse. If it is bigger you added when you should have subtracted.

Exercise 20A  Pythagoras with whole-number answers

Non-calculator
1
FWork out each square.
(a)42
(b)72
(c)92
(d)122
(e)152
(f)202
2
FWork out each square root.
(a)√25
(b)√49
(c)√81
(d)√144
(e)√169
(f)√225
3
FFor each right-angled triangle, write down the letters of the hypotenuse.
(a)
PQR
(b)
ABC
(c)
XYZ
4
FWork out the length of the hypotenuse, x.
3 cm4 cmx
5
FWork out the length of the hypotenuse, x, in each triangle. All the lengths are in centimetres.
(a)
6 cm8 cmx
(b)
5 cm12 cmx
(c)
8 cm15 cmx
(d)
20 cm21 cmx
6
FWork out the length of the hypotenuse, x.
24 cm7 cmx
7
FWork out the length of the shorter side, x, in each triangle. All the lengths are in centimetres.
(a)
x5 cm13 cm
(b)
6 cmx10 cm
(c)
x24 cm25 cm
(d)
15 cmx17 cm
8
FWork out the length of the shorter side, x.
x40 cm41 cm
9
PSA ladder 13 m long leans against a vertical wall. The foot of the ladder is 5 m from the wall on level ground. How high up the wall does the ladder reach?
5 mhladder 13 m
10
PSThe diagram shows a rectangle that is 24 cm long and 10 cm wide. Work out the length of the diagonal, x.
24 cm10 cmx
11
FWork out the length of the hypotenuse, x, in each triangle. Leave your answer as a square root (do not use a decimal). All lengths are in centimetres.
(a)
2 cm3 cmx
(b)
1 cm5 cmx
12
RAisha says: “The shorter sides are 6 cm and 8 cm, so the hypotenuse is 6 + 8 = 14 cm.” Explain why Aisha is wrong, and work out the correct length.
ABC6 cm8 cm14 cm
13
RKim says that the hypotenuse of this triangle is 9 cm. Without using Pythagoras’ theorem, explain why Kim must be wrong. Then work out the correct length, to 1 decimal place.
PQR10 cm6 cm
14
RA right-angled triangle has a hypotenuse of 25 cm and one shorter side of 7 cm. Show that its perimeter is 56 cm.
7 cmx25 cm

Exercise 20B  Pythagoras with a calculator

Calculator allowed
1
FWork out the length of the hypotenuse, x, in each triangle. Give your answers correct to 1 decimal place. All lengths are in centimetres.
(a)
4 cm7 cmx
(b)
6 cm9 cmx
(c)
8.5 cm5.2 cmx
(d)
12 cm12 cmx
2
FWork out the length of the shorter side, x, in each triangle. Give your answers correct to 1 decimal place. All lengths are in centimetres.
(a)
x5 cm9 cm
(b)
8.4 cmx14 cm
(c)
x6 cm7.5 cm
(d)
4.2 cmx15 cm
3
FWork out the length of the hypotenuse, x. Give your answer correct to 2 decimal places.
15.3 m8.7 mx
4
FWork out the length of the shorter side, x. Give your answer correct to 3 significant figures.
x19.4 mm23.6 mm
5
PSA television screen is a rectangle 80 cm wide and 45 cm high. Work out the length of the diagonal of the screen. Give your answer to the nearest centimetre, then say how many inches this is to the nearest inch (1 inch = 2.54 cm).
80 cm45 cmx
6
PSSam walks from one corner of a rectangular field to the opposite corner. The field is 120 m long and 90 m wide. He can walk along the two edges, or straight across the field along the diagonal. How many metres shorter is the diagonal route?
120 m90 md
7
PSA ship sails 15 km due north and then 8 km due east. How far is the ship from where it started?
15 km8 kmdStartNEnd
8
PSA vertical flagpole is 7.5 m tall. A wire runs from the top of the pole to a point on the ground 4 m from the foot of the pole. Work out the length of the wire.
4 m7.5 mx
9
PSA wheelchair ramp is 4.2 m long and rises 0.35 m. Work out the horizontal distance the ramp covers. Give your answer correct to 2 decimal places.
d0.35 mramp 4.2 m
10
RChallenge. A square has diagonals of length 10 cm. Use Pythagoras’ theorem to work out the length of one side of the square, correct to 1 decimal place. (Hint: the diagonal splits the square into two right-angled triangles whose two shorter sides are equal.)
ss10 cm

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