Year 8AC9M8M06

Pythagoras' Theorem

Use Pythagoras' theorem to solve problems involving the side lengths of right-angled triangles.

One rule, two versions. Which one you use depends on whether you want the longest side.

Builds onYr 7 · Square numbers and square rootsYr 7 · Solving equations

Why this works

The theorem is a statement about areas, not lengths — and that is why the squares appear.

Draw a square on each side of a right-angled triangle. The two smaller squares have areas

a^{2}

and

b^{2}

; the big one has area

c^{2}

The theorem says the two smaller square areas add up to exactly the large one.

a^{2} + b^{2} = c^{2}

Check it with the 3-4-5 triangle:

9 + 16 = 25

. The two small squares genuinely do tile the large one.

3^{2} + 4^{2} = 9 + 16 = 25 = 5^{2}

Because it is about areas, you must square first, add or subtract second, and only take the root at the very end.

The hypotenuse is always the side opposite the right angle, and always the longest. Identify it before you write anything down — nearly every Pythagoras error traces back to getting that wrong.

Worked example

e.g.

a = 6,\ b = 8,\ \text{find the hypotenuse } c

Identify the hypotenuse — the side opposite the right angle. Here we are looking for it, so we are adding. $a^{2} + b^{2} = c^{2}$
Substitute the known sides. $6^{2} + 8^{2} = c^{2}$
Square each one. $36 + 64 = c^{2}$
Add. $100 = c^{2}$
Take the square root of both sides. Only now, at the very end. $c = \sqrt{100} = 10$
Check it. The hypotenuse must be the longest side. $10 > 8$ ✓ $c = 10$

Common mistakes

Wrong

\text{Finding a short side using } a^{2} + b^{2} = c^{2}

Right

a^{2} = c^{2} - b^{2}

The single biggest Pythagoras error. If you are looking for a short side, you must subtract. Ask first: is the missing side the longest one? If not, subtract.

Wrong

6 + 8 = c \ \text{ so } \ c = 14

Right

6^{2} + 8^{2} = c^{2} \ \text{ so } \ c = 10

Adding the sides instead of the squares. The theorem is about areas — square first, always.

Wrong

c^{2} = 100 \ \text{ so } \ c = 50

Right

c^{2} = 100 \ \text{ so } \ c = 10

Halving instead of taking the square root. The inverse of squaring is rooting.

Wrong

\text{A hypotenuse shorter than a given side}

Right

\text{Hypotenuse} > \text{either short side}

Always sanity-check the answer. If your hypotenuse comes out shorter than one of the other sides, you have subtracted when you should have added.

Practice

FluencyGet quick and accurate at the method.

a = 3,\ b = 4,\ \text{find } c

Answer

c = 5

a = 5,\ b = 12,\ \text{find } c

Answer

c = 13

a = 8,\ b = 15,\ \text{find } c

Answer

c = 17

a = 9,\ b = 12,\ \text{find } c

Answer

c = 15

c = 13,\ b = 5,\ \text{find } a

Answer

a = 12

c = 25,\ b = 24,\ \text{find } a

Answer

a = 7

c = 10,\ a = 6,\ \text{find } b

Answer

b = 8

a = 7,\ b = 24,\ \text{find } c

Answer

c = 25

c = 41,\ a = 9,\ \text{find } b

Answer

b = 40

a = 20,\ b = 21,\ \text{find } c

Answer

c = 29

ReasoningExplain why. Say it in your own words.

How can you tell, before doing any calculation, whether to add or subtract?

AnswerAsk whether the side you want is the hypotenuse. If it is, add the two squares. If you already know the hypotenuse and want a shorter side, subtract. The hypotenuse is always opposite the right angle and always the longest side.

Someone finds a hypotenuse of 6 for a triangle with sides 8 and 10. Without redoing the calculation, how do you know this is wrong?

AnswerThe hypotenuse must be the longest side, but 6 is shorter than both 8 and 10. They have almost certainly subtracted when they should have added — or misidentified which side is the hypotenuse.

Is a triangle with sides 5, 6 and 8 right-angled? Justify your answer.

AnswerNo. If it were, the two shorter squares would total the longest:

5^{2} + 6^{2} = 25 + 36 = 61

, but

8^{2} = 64

. Since

61 \neq 64

, it is not right-angled — though it is close.

Why does the theorem involve squares rather than the side lengths themselves?

AnswerBecause it is a statement about the areas of squares drawn on each side. The area of a square built on a side of length

a

a^{2}

, and it is those areas that add exactly. The lengths themselves do not.

AppliedThe same maths, inside a real question.

A 5 m ladder rests against a wall with its foot 3 m from the base. How far up the wall does it reach?

Answer4 m. The ladder is the hypotenuse, so subtract:

h^{2} = 5^{2} - 3^{2} = 25 - 9 = 16

, giving

h = 4

A rectangular gate is 120 cm wide and 90 cm tall. How long is the diagonal brace?

Answer150 cm. The diagonal is the hypotenuse:

\sqrt{120^{2} + 90^{2}} = \sqrt{14400 + 8100} = \sqrt{22500} = 150

Ana walks 9 km north then 12 km east. How far is she from her starting point in a straight line?

Answer15 km. The two legs are at right angles, so

\sqrt{9^{2} + 12^{2}} = \sqrt{81 + 144} = \sqrt{225} = 15

A TV is advertised as 40 inches, measured diagonally. Its height is 24 inches. How wide is it?

Answer32 inches. The diagonal is the hypotenuse:

w^{2} = 40^{2} - 24^{2} = 1600 - 576 = 1024

, so

w = 32

A ramp rises 1.5 m over a horizontal run of 3.6 m. How long is the ramp surface?

Answer3.9 m.

\sqrt{1.5^{2} + 3.6^{2}} = \sqrt{2.25 + 12.96} = \sqrt{15.21} = 3.9