Pythagoras' Theorem

This lesson covers Pythagoras' Theorem for AQA GCSE Mathematics. You will learn how to find the hypotenuse, find a shorter side, and apply the theorem in real-world contexts. Higher-tier students will also study 3D Pythagoras. This is one of the most important topics in GCSE Mathematics and appears on almost every exam paper.

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What Is Pythagoras' Theorem?

Pythagoras' Theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

a squared + b squared = c squared

where c is the hypotenuse (the longest side, opposite the right angle) and a and b are the other two sides.

graph LR
    A[Right-angled triangle] --> B[Identify the hypotenuse longest side]
    B --> C{What are you finding?}
    C -->|Hypotenuse| D[c = square root of a squared + b squared]
    C -->|Shorter side| E[a = square root of c squared - b squared]

Exam Tip: The hypotenuse is ALWAYS the side opposite the right angle. It is always the longest side. If the question does not explicitly label the right angle, look for the small square symbol.

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Finding the Hypotenuse

When you know the two shorter sides and need to find the hypotenuse:

c = square root of (a squared + b squared)

Worked Example 1

Find the hypotenuse of a right-angled triangle with sides 6 cm and 8 cm.

Solution: c squared = 6 squared + 8 squared c squared = 36 + 64 c squared = 100 c = square root of 100 c = 10 cm

Worked Example 2

Find the hypotenuse of a right-angled triangle with sides 5 cm and 9 cm. Give your answer to 1 decimal place.

Solution: c squared = 5 squared + 9 squared c squared = 25 + 81 c squared = 106 c = square root of 106 c = 10.3 cm (1 d.p.)

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Finding a Shorter Side

When you know the hypotenuse and one shorter side:

a = square root of (c squared - b squared)

Worked Example 3

A right-angled triangle has a hypotenuse of 13 cm and one side of 5 cm. Find the other side.

Solution: a squared = 13 squared - 5 squared a squared = 169 - 25 a squared = 144 a = square root of 144 a = 12 cm

Worked Example 4

A right-angled triangle has a hypotenuse of 15 cm and one side of 7 cm. Find the other side to 1 decimal place.

Solution: a squared = 15 squared - 7 squared a squared = 225 - 49 a squared = 176 a = square root of 176 a = 13.3 cm (1 d.p.)

Exam Tip: When finding a shorter side, you SUBTRACT. A common mistake is to add the squares instead. Remember: hypotenuse = add, shorter side = subtract.

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Common Pythagorean Triples

A Pythagorean triple is a set of three whole numbers that satisfy Pythagoras' Theorem. Recognising these saves time.

Triple	Check
3, 4, 5	9 + 16 = 25
5, 12, 13	25 + 144 = 169
8, 15, 17	64 + 225 = 289
7, 24, 25	49 + 576 = 625

Multiples also work: (6, 8, 10), (9, 12, 15), (10, 24, 26), etc.

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Applying Pythagoras in Context

Pythagoras' Theorem is used whenever you need to find a distance in a right-angled triangle context.

Worked Example 5: Ladder Problem

A ladder of length 5 m leans against a vertical wall. The foot of the ladder is 3 m from the base of the wall. How high up the wall does the ladder reach?

Solution: The ladder is the hypotenuse (5 m), the ground distance is 3 m. h squared = 5 squared - 3 squared = 25 - 9 = 16 h = square root of 16 = 4 m

Worked Example 6: Diagonal of a Rectangle

Find the length of the diagonal of a rectangle measuring 12 cm by 5 cm.

Solution: d squared = 12 squared + 5 squared = 144 + 25 = 169 d = square root of 169 = 13 cm

Worked Example 7: Isosceles Triangle Height

An isosceles triangle has two equal sides of 10 cm and a base of 12 cm. Find the perpendicular height.

Solution: The perpendicular height bisects the base, creating two right-angled triangles each with:

• Hypotenuse = 10 cm
• Base = 12 / 2 = 6 cm

h squared = 10 squared - 6 squared = 100 - 36 = 64 h = square root of 64 = 8 cm

Exam Tip: Many questions require you to spot the right-angled triangle hidden in the problem. Diagonals of rectangles, heights of isosceles triangles, and distances across fields are all common scenarios.

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Distance Between Two Points

You can use Pythagoras to find the straight-line distance between two coordinate points.

Distance = square root of ((x2 - x1) squared + (y2 - y1) squared)

Worked Example 8

Find the distance between A(1, 2) and B(4, 6).

Solution: Horizontal distance = 4 - 1 = 3 Vertical distance = 6 - 2 = 4 Distance = square root of (3 squared + 4 squared) = square root of (9 + 16) = square root of 25 = 5 units

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3D Pythagoras [H]

In three dimensions, you can extend Pythagoras' Theorem to find the space diagonal of a cuboid or the distance between two points in 3D space.

Space diagonal of a cuboid = square root of (l squared + w squared + h squared)

This works because you apply Pythagoras twice:

1. First, find the diagonal of the base: d1 = square root of (l squared + w squared)
2. Then, use d1 and the height: space diagonal = square root of (d1 squared + h squared) = square root of (l squared + w squared + h squared)

Worked Example 9 [H]

A cuboid has dimensions 3 cm by 4 cm by 12 cm. Find the length of the space diagonal.

Solution: Space diagonal = square root of (3 squared + 4 squared + 12 squared) = square root of (9 + 16 + 144) = square root of 169 = 13 cm

Worked Example 10 [H]

A cone has radius 5 cm and slant height 13 cm. Find the vertical height.