Pythagoras' Theorem

Pythagoras' Theorem is one of the most important results in geometry. It links the three sides of a right-angled triangle and appears throughout Edexcel GCSE Mathematics — from straightforward calculations to multi-step problem solving. Higher tier extends to 3D applications.

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Key Vocabulary

Term	Meaning
Right-angled triangle	A triangle containing one 90° angle
Hypotenuse	The longest side, opposite the right angle
Pythagorean triple	A set of three positive integers that satisfy $a^2 + b^2 = c^2$a2+b2=c2 (e.g. 3, 4, 5)

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The Theorem

In any right-angled triangle with hypotenuse c and shorter sides a and b:

$a^2 + b^2 = c^2$a2+b2=c2

This means: the square of the hypotenuse equals the sum of the squares of the other two sides.

Edexcel Formula Sheet: Pythagoras' Theorem is NOT on the Edexcel formula sheet — you must memorise it.

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Understanding the Derivation

Consider a right-angled triangle with sides a, b and hypotenuse c. If you draw a square on each side:

• Area of square on side a = a²
• Area of square on side b = b²
• Area of square on hypotenuse = c²

Pythagoras' Theorem states that $a^2 + b^2 = c^2$a2+b2=c2.

This can be proved by arranging four copies of the triangle inside a large square, or by algebraic methods. You are not required to prove it for GCSE, but understanding where it comes from helps you remember and apply it.

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Common Pythagorean Triples

Memorising these saves time in the exam:

Triple	Multiples
3, 4, 5	6, 8, 10 and 9, 12, 15 and 15, 20, 25
5, 12, 13	10, 24, 26
8, 15, 17	—
7, 24, 25	—

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Finding the Hypotenuse

When you need the longest side, add the squares.

c = $\sqrt{a^{2} + b^{2}}$a2+b2​

Worked Example 1

A right-angled triangle has shorter sides 6 cm and 8 cm. Find the hypotenuse.

$c^2 = 6^2 + 8^2$c2=62+82 = 36 + 64 = 100 c = $\sqrt{100}$100​ = 10 cm

Worked Example 2

A right-angled triangle has shorter sides 5 cm and 9 cm. Find the hypotenuse to 1 d.p.

$c^2 = 5^2 + 9^2$c2=52+92 = 25 + 81 = 106 c = $\sqrt{106}$106​ = 10.3 cm (1 d.p.)

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Finding a Shorter Side

When you know the hypotenuse and one shorter side, subtract the squares.

a = $\sqrt{c^{2} - b^{2}}$c2−b2​

Worked Example 3

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

$a^2 = 13^2 - 5^2$a2=132−52 = 169 - 25 = 144 a = $\sqrt{144}$144​ = 12 cm

Worked Example 4

A right-angled triangle has hypotenuse 15 cm and one side 7 cm. Find the other side to 1 d.p.

$a^2 = 15^2 - 7^2$a2=152−72 = 225 - 49 = 176 a = $\sqrt{176}$176​ = 13.3 cm (1 d.p.)

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Problem-Solving Applications

Diagonals of Rectangles

Worked Example 5

A rectangle is 12 cm long and 5 cm wide. Find the length of the diagonal.

The diagonal forms the hypotenuse of a right-angled triangle. $d^2 = 12^2 + 5^2$d2=122+52 = 144 + 25 = 169 d = $\sqrt{169}$169​ = 13 cm

Isosceles Triangles

Worked Example 6

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

The perpendicular height bisects the base, giving two right-angled triangles with hypotenuse 10 and base 6. $h^2 = 10^2 - 6^2$h2=102−62 = 100 - 36 = 64 h = $\sqrt{64}$64​ = 8 cm

Distance Between Two Points

The distance between points (x₁, y₁) and (x₂, y₂) is:

d = $\sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2}}$(x2​−x1​)2+(y2​−y1​)2​

This is just Pythagoras applied to a coordinate grid.

Worked Example 7

Find the distance between (1, 3) and (4, 7).

d = $\sqrt{(4 - 1)^{2} + (7 - 3)^{2}}$(4−1)2+(7−3)2​ = $\sqrt{9 + 16}$9+16​ = $\sqrt{25}$25​ = 5

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3D Pythagoras [H]

In 3D problems, apply Pythagoras' Theorem twice.

Worked Example 8 [H]

A cuboid has length 6 cm, width 4 cm and height 3 cm. Find the length of the space diagonal (from one corner to the opposite corner).

Step 1: Find the diagonal of the base. d₁² = 6² + 4² = 36 + 16 = 52 d₁ = $\sqrt{52}$52​

Step 2: The space diagonal forms a right-angled triangle with d₁ and the height. d² = d₁² + 3² = 52 + 9 = 61 d = $\sqrt{61}$61​ = 7.8 cm (1 d.p.)

Shortcut for 3D: For a cuboid with dimensions a, b, c, the space diagonal = $\sqrt{a^{2} + b^{2} + c^{2}}$a2+b2+c2​.

Worked Example 9 [H]

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

$h^2 = 13^2 - 5^2$h2=132−52 = 169 - 25 = 144 h = $\sqrt{144}$144​ = 12 cm

Worked Example 10 [H]

A square-based pyramid has base side 8 cm and slant height 10 cm (from the midpoint of a base edge to the apex). Find the perpendicular height.

Half the base side = 4 cm. $h^2 = 10^2 - 4^2$h2=102−42 = 100 - 16 = 84 h = $\sqrt{84}$84​ = 9.2 cm (1 d.p.)

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Checking if a Triangle is Right-Angled

If $a^2 + b^2 = c^2$a2+b2=c2 (where c is the longest side), the triangle is right-angled. If $a^2 + b^2 > c^2$a2+b2>c2, the triangle is acute. If $a^2 + b^2 < c^2$a2+b2<c2, the triangle is obtuse.

Worked Example 11

Is a triangle with sides 7, 10 and 12 right-angled?

7² + 10² = 49 + 100 = 149 12² = 144 Since 149 > 144, the triangle is not right-angled (it is acute).

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Common Mistakes and Misconceptions