Ext Maths · 6 Trigonometry

6.1 Pythagoras’ theorem

Before you start

Squaring a number and finding a square root, with and without a calculator.
The area of a triangle, $\frac{1}{2} \times base \times height$ .
Rounding to 3 significant figures.

Syllabus objectives covered (1)

Know and use Pythagoras’ theorem.

Part 1 of 6

Find the right angle, then find the hypotenuse

Pythagoras' theorem is about right-angled triangles only. Before you use it, point to the right angle. If there isn't one, this is the wrong tool.
The hypotenuse is the side opposite the right angle. It never touches the right angle, and it is always the longest side.
The theorem says: the square of the hypotenuse equals the sum of the squares of the other two sides. With $c$ as the hypotenuse:
$a^{2} + b^{2} = c^{2}$
Learn it in words, not just letters. On the paper the sides will be called $P Q$ , $x$ or 'the ladder', never conveniently $a$ , $b$ and $c$ .
It is not on the List of formulas at the front of the paper, so you must know it by heart.

Part 2 of 6

Where the theorem comes from

Take a right-angled triangle with shorter sides $3$ and $4$ and draw a square on each of its three sides.
The two small squares together have exactly the area of the square on the hypotenuse: 9 + 16 = 25.
The small squares have areas $3^{2} = 9$ and $4^{2} = 16$ . The square on the longest side has area $5^{2} = 25$ .
And $9 + 16 = 25$ . That is the whole theorem: the two smaller squares exactly fill the big one. It is true for every right-angled triangle, not just this one.
Seeing it as areas tells you why you square the sides first. You are adding areas, not lengths, so $3 + 4 \neq 5$ but $9 + 16 = 25$ .

Part 3 of 6

Finding the hypotenuse: square, add, square root

When you know the two shorter sides and want the hypotenuse, add their squares, because the hypotenuse's square is the total.
Then take the square root. The equation gives you $c^{2}$ , and the question wants $c$ .
$c = \sqrt{a^{2} + b^{2}}$
Check: your answer must be longer than both of the sides you started with. If it isn't, you have subtracted by mistake.

Part 4 of 6

Finding a shorter side: square, subtract, square root

When you know the hypotenuse and one shorter side, subtract. The hypotenuse's square is the total, so taking away one piece leaves the other piece.
$a^{2} = c^{2} - b^{2}$
Always subtract the smaller square from the hypotenuse's square. Doing it the other way round gives a negative number, and you cannot square-root that.
The trap is adding out of habit. Before you press anything, ask: is the side I want the hypotenuse? If no, subtract.
Check: a shorter side must come out shorter than the hypotenuse.

Part 5 of 6

Spotting the right-angled triangle inside a problem

Exam questions rarely hand you a bare triangle. Sketch the situation, then look for a right angle that is already there.
A ladder against a wall: the wall is vertical and the ground horizontal, so the right angle is at the foot of the wall. The ladder is the hypotenuse.
The diagonal of a rectangle: every corner is a right angle, so the diagonal is the hypotenuse of a triangle made of a length and a width.
The height of an isosceles triangle: the line from the top vertex straight down to the base cuts the base in half at a right angle. Use half the base, not the whole base.
The distance between two points on a grid: go across, then up. The horizontal and vertical steps are the two shorter sides.
Without a calculator, look for Pythagorean triples, whole-number sides that fit exactly: $3, 4, 5$ and $5, 12, 13$ and $8, 15, 17$ , and their multiples, such as $6, 8, 10$ .
If a non-calculator answer does not come out whole, leave it as a square root such as $\sqrt{52}$ cm rather than guessing a decimal.
With a calculator, give lengths to 3 significant figures. If the length is used again later, keep the full value on your calculator and round only the final answer.

Part 6 of 6

Using the theorem backwards: is there a right angle?

The theorem also works as a test. If the squares of the two shorter sides add up to the square of the longest side, the triangle is right-angled.
If they don't add up, the triangle has no right angle at all.
The right angle is always opposite the longest side, because that side is the hypotenuse.
In a 'Show that' question, write out both numbers and say what they prove. Two matching numbers with no sentence is not a finished answer.

Worked examples

Example 1. A right-angled triangle has shorter sides of $7.2$ cm and $9.6$ cm. Calculate the length of the hypotenuse.

Show the working

You want the hypotenuse, so add the squares.
Square each side and add.
$c^{2} = {7.2}^{2} + {9.6}^{2} = 51.84 + 92.16 = 144$
Square-root, because the line above gives $c^{2}$ , not $c$ .
$c = \sqrt{144} = 12 cm$
The answer is exact, so there is nothing to round. Check: $12$ is longer than both $7.2$ and $9.6$ .

Answer: $12$ cm.

Example 2. A ladder $6.5$ m long leans against a vertical wall. Its foot is on horizontal ground, $1.8$ m from the wall. Calculate how far up the wall the ladder reaches.

Show the working

Sketch it. The wall meets the ground at a right angle, and the ladder is opposite that right angle, so the ladder is the hypotenuse.
You want a shorter side (the height up the wall), so subtract.
$h^{2} = {6.5}^{2} - {1.8}^{2} = 42.25 - 3.24 = 39.01$
Square-root.
$h = \sqrt{39.01} = 6.2457 \dots$
Round to 3 significant figures: $6.25$ m. Check: it is shorter than the $6.5$ m ladder, as it must be.

Answer: $6.25$ m (3 s.f.).

Example 3. Without a calculator: an isosceles triangle has two sides of $13$ cm and a base of $10$ cm. Work out its height and its area.

Show the working

Draw the height from the top vertex down to the base. It meets the base at a right angle and cuts it in half, so each half is $5$ cm.
That makes a right-angled triangle with hypotenuse $13$ cm (a sloping side) and one short side $5$ cm. You want the other short side, so subtract.
$h^{2} = 13^{2} - 5^{2} = 169 - 25 = 144$
So $h = 12$ cm. This is the $5, 12, 13$ triple, which is why it came out whole.
Now the area uses the whole base with this height.
$Area = \frac{1}{2} \times 10 \times 12 = 60 {cm}^{2}$

Answer: Height $12$ cm, area $60$ cm².

Example 4. Without a calculator: work out the distance between the points $(1, 2)$ and $(7, 10)$ .

Show the working

Across: from $x = 1$ to $x = 7$ is $6$ units. Up: from $y = 2$ to $y = 10$ is $8$ units.
The across and up steps meet at a right angle, and the distance between the points is the hypotenuse.
$d^{2} = 6^{2} + 8^{2} = 36 + 64 = 100$
So $d = \sqrt{100} = 10$ . It is the $3, 4, 5$ triple doubled.

Answer: $10$ units.

Where people go wrong

Mistake: Adding the squares when the side you want is not the hypotenuse.

Instead: Add only to find the hypotenuse. To find a shorter side, subtract from the hypotenuse's square.

The hypotenuse's square is the total of the other two. Adding two pieces gives the total; you subtract to find a missing piece.

Mistake: Stopping at $c^{2}$ , for example giving the ladder height as $39.01$ m.

Instead: Take the square root as the last step.

The theorem is about the squares of the sides. An answer far bigger than any side in the problem is the giveaway.

Mistake: Adding the sides before squaring, as in ${(7.2 + 9.6)}^{2}$ .

Instead: Square each side separately, then add.

The theorem adds the areas of separate squares. ${(a + b)}^{2}$ is not $a^{2} + b^{2}$ .

Mistake: Using the whole base of an isosceles triangle when finding its height.

Instead: Use half the base.

The height cuts the base in half, so the right-angled triangle has only half the base as one of its sides.

Mistake: Rounding a length early and then using it again.

Instead: Keep the full calculator value until the final answer.

The rounding error is carried into the next step, and the final answer can then miss the accuracy mark.

Check yourself

State Pythagoras' theorem in words.
Answer
In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.
How do you recognise the hypotenuse?
Answer
It is the side opposite the right angle, and it is always the longest side.
Without a calculator, work out the hypotenuse of a right-angled triangle with shorter sides $5$ cm and $12$ cm.
Answer
$\sqrt{25 + 144} = \sqrt{169} = 13$ cm.
Without a calculator, a right-angled triangle has hypotenuse $10$ cm and one side $6$ cm. Work out the third side.
Answer
$\sqrt{100 - 36} = \sqrt{64} = 8$ cm.
Calculate the hypotenuse when the shorter sides are $4.3$ cm and $6.1$ cm.
Answer
$\sqrt{18.49 + 37.21} = \sqrt{55.7} = 7.46$ cm (3 s.f.).
A rectangle has a diagonal of $25$ cm and a width of $7$ cm. Work out its length.
Answer
$\sqrt{625 - 49} = \sqrt{576} = 24$ cm.
A triangle has sides $9$ cm, $12$ cm and $16$ cm. Is it right-angled? Explain.
Answer
No. $9^{2} + 12^{2} = 81 + 144 = 225$ , but $16^{2} = 256$ . The squares of the shorter sides do not add up to the square of the longest side.

Exam questions

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

Question 1. A rectangular television screen is $104$ cm wide. The diagonal of the screen is $120$ cm.

(a) Calculate the height of the screen. 3 marks

Mark scheme

M1 120² − 104² oe, or 3584 seen
M1 √(their 3584)
A1 59.9 awrt (cm)

(b) Calculate the area of the screen. 2 marks

Mark scheme

M1 104 × their (a)
A1 6230 awrt (cm²); FT their (a)

Question 2. Do not use a calculator in this question. Triangle $P Q R$ has $P Q = 20$ cm, $Q R = 21$ cm and $P R = 29$ cm.

(a) Show that triangle $P Q R$ is right-angled. 2 marks No calculator

Mark scheme

M1 20² + 21² = 841 (or 400 + 441) and 29² = 841 both seen
A1 statement that these are equal so the triangle is right-angled

(b) Write down which angle of the triangle is $90 °$ . 1 mark No calculator

Mark scheme

B1 angle PQR (the angle opposite PR) cao

(c) $S$ is the point on $Q R$ with $Q S = 15$ cm. Work out the length of $P S$ . 2 marks No calculator

Mark scheme

M1 20² + 15² oe, or 625 seen
A1 25 cao (cm)

Question 3. A rectangular field measures $85$ m by $42$ m. Jo wants to get from one corner to the opposite corner. She can walk along two edges of the field, or straight along the diagonal.

(a) Calculate how much shorter the diagonal route is than the route along the edges. 3 marks

Mark scheme

M1 85² + 42² oe, or 8989 seen
M1 85 + 42 − √(their 8989); dep on first M1
A1 32.2 awrt (m)

How this is examined

Deciding whether to add or subtract the squares before you calculate anything: A 2- or 3-mark Calculate, often as the first part of a longer question about a real object: a ladder, a screen, a ramp, a field. The method mark is for the correct combination written with the numbers in, such as $120^{2} - 104^{2}$ . A second method mark goes to the square root of that. The accuracy mark needs the final length to 3 s.f., for example $59.9$ cm.
Finding the hidden right-angled triangle in a shape that is not a right-angled triangle: An isosceles triangle, a rectangle's diagonal, a kite or two points on a grid, where you have to draw in the height or the diagonal yourself before Pythagoras can start. Using half the base in an isosceles triangle, or the across and up steps between two points, as the two shorter sides. A correct combination of squares from that triangle earns the method mark even if the arithmetic slips afterwards.
Proving a triangle is right-angled from its three sides: A 2-mark Show (that), usually on the non-calculator paper, with whole-number sides. Both values written, for example $20^{2} + 21^{2} = 841$ and $29^{2} = 841$ , followed by a statement that they are equal so the triangle is right-angled.
A Pythagoras length feeding a later part: Part (a) finds a length and part (b) uses it for an area, a perimeter or an angle. Using the unrounded value from (a) in (b). Follow-through marks usually protect you if (a) was wrong, but only if your working for (b) is shown.

Words to know

Hypotenuse: The side of a right-angled triangle opposite the right angle; always the longest side.
Pythagoras' theorem: In a right-angled triangle with hypotenuse $c$ , $a^{2} + b^{2} = c^{2}$ .
Pythagorean triple: Three whole numbers that fit the theorem exactly, such as $3, 4, 5$ or $5, 12, 13$ .
Isosceles triangle: A triangle with two equal sides; the height from the top vertex cuts the base in half at a right angle.

All Ext Maths lessons Next: 6.2 Right-angled triangles