Maths · 7 Trigonometry

7.1 Pythagoras’ theorem

Before you start

Squaring a number and taking a square root.
That a right angle is $90 °$ , and how it is marked on a diagram.
Reading coordinates from a grid.

Syllabus objectives covered (1)

Know and use Pythagoras’ theorem.

Part 1 of 3

The theorem, and the one thing to get right

Pythagoras' theorem applies to right-angled triangles only. If there is no right angle, it does not apply — you would need the cosine rule instead (7.5).
The hypotenuse is the longest side, always opposite the right angle. Everything depends on identifying it correctly.
$a^{2} + b^{2} = c^{2}$
In that form, $c$ is the hypotenuse and $a$ and $b$ are the two shorter sides. The hypotenuse must sit alone on one side of the equation.
The syllabus says knowledge of the theorem is required — it is not printed on the formula list, so you must recall it.
Finding the hypotenuse: square both short sides, add, take the square root. The answer must come out larger than either short side.
Finding a short side: square the hypotenuse, subtract the square of the known short side, take the square root.
$a^{2} = c^{2} - b^{2}$
Adding when you should subtract is the commonest error in this topic. Before you calculate, decide which side is missing — if it is a short side, you subtract.
The check is instant and worth doing every time: is your answer smaller than the hypotenuse? If a short side comes out longer than the hypotenuse, you added instead of subtracting.

Part 2 of 3

Distance between two points

The syllabus notes list this explicitly: finding the distance between two points given on a grid.
The trick is that any two points define a right-angled triangle if you draw a horizontal and a vertical line between them.
The horizontal side is the difference in the $x$ -coordinates; the vertical side is the difference in the $y$ -coordinates; the distance you want is the hypotenuse.
For $(1, 2)$ and $(5, 5)$ : horizontal $= 5 - 1 = 4$ , vertical $= 5 - 2 = 3$ , so distance $= \sqrt{4^{2} + 3^{2}} = \sqrt{25} = 5$ .
The differences may come out negative depending on which point you subtract from which — it does not matter, because you square them immediately. Use the positive difference and move on.
Sketching the triangle takes ten seconds and prevents the most likely error, which is subtracting an $x$ from a $y$ .

Part 3 of 3

Chords in a circle

The other two applications the syllabus names are the length of a chord and the distance of a chord from the centre of a circle.
One fact unlocks both: the perpendicular from the centre of a circle to a chord bisects the chord — it cuts it exactly in half.
So draw the radius to one end of the chord, and the perpendicular from the centre to the chord. That gives a right-angled triangle.
Radius as hypotenuse, half the chord and the perpendicular distance as the short sides.
In that triangle: the radius is the hypotenuse, the perpendicular distance from the centre is one short side, and half the chord is the other.
The word half is where marks are lost. The triangle contains half the chord, not the whole of it — so double it at the end if the question asks for the full chord length.
So to find a chord of length $2 x$ in a circle of radius $r$ at distance $d$ from the centre: $x = \sqrt{r^{2} - d^{2}}$ , then the chord is $2 x$ .
And to find the distance from the centre, rearrange the same relationship: $d = \sqrt{r^{2} - x^{2}}$ , where $x$ is half the chord.

Worked examples

Example 1. A right-angled triangle has a hypotenuse of $13$ cm and one short side of $5$ cm. Work out the length of the third side.

Show the working

Identify what is missing. The hypotenuse is known, so the missing side is a short one — which means subtracting.
Write the theorem with the hypotenuse alone on one side.
$5^{2} + b^{2} = 13^{2}$
Substitute and rearrange.
$25 + b^{2} = 169 \Rightarrow b^{2} = 144$
Take the square root.
$b = 12 cm$
Check: 12 is less than the hypotenuse of 13, as a short side must be.

Answer: $12$ cm.

Example 2. Work out the distance between the points $A (- 2, 1)$ and $B (4, 9)$ .

Show the working

Sketch the right-angled triangle with a horizontal and a vertical side joining the two points.
Horizontal difference: $4 - (- 2) = 6$ .
Vertical difference: $9 - 1 = 8$ .
The distance $A B$ is the hypotenuse, so add the squares.
$A B^{2} = 6^{2} + 8^{2} = 36 + 64 = 100$
Take the square root: $A B = 10$ . Check it is larger than both 6 and 8, which it is.

Answer: $A B = 10$ units.

Example 3. A chord of a circle of radius $10$ cm is $6$ cm from the centre. Work out the length of the chord.

Show the working

Draw the radius to one end of the chord and the perpendicular from the centre to the chord, forming a right-angled triangle.
In that triangle the radius (10 cm) is the hypotenuse and the distance from the centre (6 cm) is one short side.
The other short side is half the chord. Call it $x$ and apply the theorem.
$x^{2} + 6^{2} = 10^{2}$
Solve.
$x^{2} = 100 - 36 = 64 \Rightarrow x = 8$
Double it, because $x$ is only half the chord: the full chord is $16$ cm. Forgetting this last step is the standard error.

Answer: The chord is $16$ cm long.

Where people go wrong

Mistake: Adding the squares when finding a short side.

Instead: Subtract: $a^{2} = c^{2} - b^{2}$ when $c$ is the hypotenuse.

You only add when the hypotenuse is the unknown. Check whether your answer is shorter than the hypotenuse — if not, you added.

Mistake: Treating the longest given side as the hypotenuse without checking the right angle.

Instead: The hypotenuse is the side opposite the right angle.

In a labelled diagram the right angle tells you which side it is, and the position matters more than the length.

Mistake: Giving half the chord as the answer.

Instead: Double it — the triangle contains only half the chord.

The perpendicular from the centre bisects the chord, so the triangle you solved holds half of it.

Mistake: Using Pythagoras on a triangle with no right angle.

Instead: It applies only to right-angled triangles. Use the cosine rule otherwise.

The theorem is a statement about right angles specifically. A diagram without a right-angle mark should stop you.

Mistake: Forgetting the square root at the end.

Instead: You have found $c^{2}$ , not $c$ .

An answer of 100 cm where 10 cm was expected is a missing root, and it is easy to spot if you sanity-check the size.

Check yourself

State Pythagoras' theorem and say when it can be used.
Answer
$a^{2} + b^{2} = c^{2}$ , where $c$ is the hypotenuse. It applies only to right-angled triangles.
A right-angled triangle has short sides $9$ cm and $12$ cm. Find the hypotenuse.
Answer
$\sqrt{81 + 144} = \sqrt{225} = 15$ cm.
A right-angled triangle has hypotenuse $25$ m and one short side $7$ m. Find the other short side.
Answer
$\sqrt{625 - 49} = \sqrt{576} = 24$ m.
Find the distance between $(0, 0)$ and $(9, 12)$ .
Answer
$\sqrt{9^{2} + 12^{2}} = \sqrt{225} = 15$ .
A chord of length $24$ cm lies in a circle of radius $13$ cm. How far is it from the centre?
Answer
Half the chord is 12 cm, so the distance is $\sqrt{169 - 144} = \sqrt{25} = 5$ cm.
Why does the perpendicular from the centre of a circle to a chord create a right-angled triangle you can use?
Answer
It meets the chord at right angles and bisects it, so the radius, the perpendicular distance and half the chord form a right-angled triangle.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A ladder of length $4.5$ m leans against a vertical wall. The foot of the ladder is $1.2$ m from the base of the wall.

(a) Calculate how far up the wall the ladder reaches, correct to 3 significant figures. 3 marks

Mark scheme

M1 recognises the ladder as the hypotenuse and subtracts: 4.5² - 1.2² oe
M1 20.25 - 1.44 = 18.81 seen
A1 4.34 awrt (m)

(b) Explain how you can tell, without calculating, that the answer must be less than $4.5$ m. 1 mark No calculator

Mark scheme

B1 the wall height is a shorter side, and the hypotenuse is always the longest side oe

Question 2. The points $P (3, - 4)$ and $Q (- 5, 2)$ are plotted on a grid.

(a) Work out the length of $P Q$ . 3 marks No calculator

Mark scheme

M1 horizontal difference 8 and vertical difference 6 identified
M1 8² + 6² = 100 seen
A1 10

(b) The point $R$ has coordinates $(3, 2)$ . Show that triangle $P Q R$ has a right angle at $R$ . 2 marks No calculator

Mark scheme

M1 identifies PR as vertical and QR as horizontal, or shows PR = 6, QR = 8 and 6² + 8² = 10²
A1 correct conclusion with reasoning; nfww

Question 3. A circle has centre $O$ and radius $17$ cm. A chord $A B$ is drawn at a perpendicular distance of $8$ cm from $O$ .

(a) Calculate the length of the chord $A B$ . 3 marks

Mark scheme

M1 forms a right-angled triangle with hypotenuse 17 and short side 8
M1 √(289 - 64) = 15 seen for half the chord
A1 30 cm; dep on doubling

(b) State the property of a circle that makes this method possible. 1 mark No calculator

Mark scheme

B1 the perpendicular from the centre to a chord bisects the chord oe

Words to know

Hypotenuse: The longest side of a right-angled triangle, opposite the right angle.
Pythagoras' theorem: $a^{2} + b^{2} = c^{2}$ , where $c$ is the hypotenuse. True for right-angled triangles only.
Chord: A straight line joining two points on the circumference of a circle.
Perpendicular: At right angles. The perpendicular from the centre of a circle to a chord bisects it.

All Maths lessons Next: 7.2 Right-angled triangles