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\( c^2 = a^2 + b^2 \)
Pythagoras' Theorem is one of the most famous results in mathematics. It connects the three sides of any right-angled triangle and lets us find any missing side when the other two are known.
In this lesson you will learn to:
A right-angled triangle has one angle equal to exactly 90°. The three sides have special names:
\[ c^2 = a^2 + b^2 \]
Pythagoras' Theorem: In any right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.
The squares proof shows why the theorem works: the area of the square on the hypotenuse (\( c^2 \)) equals the combined area of the squares on the two legs (\( a^2 + b^2 \)).
This is not just a formula to memorise — it is a geometric truth about areas.
\[ c^2 = a^2 + b^2 \quad \Rightarrow \quad c = \sqrt{a^2+b^2} \]
Key facts to remember:
\[ c = \sqrt{a^2 + b^2} \]
To find the hypotenuse, square both legs, add them, then take the square root.
Steps:
\[ c^2 = 3^2 + 4^2 = 9 + 16 = 25 \]
Example: A right-angled triangle has legs \( a = 3 \) cm and \( b = 4 \) cm. Find \( c \).
\[ c = \sqrt{5^2+12^2} = \sqrt{169} = 13 \text{ cm} \]
Example: Legs are \( 5 \) cm and \( 12 \) cm. Find the hypotenuse.
\[ c = \sqrt{a^2+b^2} \]
Finding the hypotenuse — checklist:
\[ a = \sqrt{c^2 - b^2} \]
When you know the hypotenuse and one leg, rearrange the theorem to find the missing leg.
Start from \( c^2 = a^2 + b^2 \), then subtract \( b^2 \) from both sides:
\[ a = \sqrt{10^2-6^2} = \sqrt{64} = 8 \text{ cm} \]
Example: Hypotenuse \( c = 10 \) cm, one leg \( b = 6 \) cm. Find the other leg.
\[ a = \sqrt{15^2-9^2} = \sqrt{144} = 12 \text{ m} \]
Example: \( c = 15 \) m, \( b = 9 \) m. Find \( a \).
\[ \text{Hypotenuse} \to \text{subtract} \quad\quad \text{Hypotenuse} \leftarrow \text{add} \]
Key distinction:
\[ \text{If } c^2 = a^2+b^2 \Rightarrow \text{right-angled} \]
The converse of Pythagoras lets you test whether a triangle is right-angled — without measuring angles.
Given three sides, label the longest side \( c \) and the others \( a \) and \( b \), then check the relationship.
\[ c^2 = a^2+b^2 \Rightarrow \text{right-angled} \\[6pt] c^2 > a^2+b^2 \Rightarrow \text{obtuse-angled} \\[6pt] c^2 < a^2+b^2 \Rightarrow \text{acute-angled} \]
Three possible results:
\[ 25^2 = 625 \quad 7^2+24^2 = 625 \quad \checkmark \]
Example: Sides are 7, 24, 25. What type of triangle?
\[ 8^2 = 64 > 61 = 5^2+6^2 \Rightarrow \text{obtuse} \]
Example: Sides are 5, 6, 8. What type of triangle?
\[ a^2 + b^2 = c^2 \quad a,b,c \in \mathbb{Z}^+ \]
A Pythagorean triple is a set of three positive whole numbers \( (a, b, c) \) that satisfy Pythagoras' theorem exactly.
Knowing common triples lets you solve problems instantly — no calculator needed!
\[ 3\text{-}4\text{-}5 \qquad 5\text{-}12\text{-}13 \qquad 8\text{-}15\text{-}17 \qquad 7\text{-}24\text{-}25 \]
Memorise these four families:
\[ 3\text{-}4\text{-}5 \xrightarrow{\times 2} 6\text{-}8\text{-}10 \xrightarrow{\times 3} 9\text{-}12\text{-}15 \]
Multiples of a triple are also triples.
Multiply every number in a triple by the same integer \( k \):
\[ \text{Spot the triple} \Rightarrow \text{no calculator needed!} \]
Exam strategy: Always check if the given numbers form a triple or a multiple of one.
\[ \text{Draw} \to \text{Label} \to \text{Apply} \to \text{Answer} \]
4-step approach for any word problem:
\[ h = \sqrt{13^2 - 5^2} = \sqrt{144} = 12 \text{ m} \]
Problem: A 13 m ladder leans against a wall. Its foot is 5 m from the wall. How high up the wall does it reach?
\[ d = \sqrt{9^2+12^2} = \sqrt{225} = 15 \text{ cm} \]
Problem: A rectangle is 9 cm wide and 12 cm long. Find the diagonal.
\[ d = \sqrt{8^2+6^2} = \sqrt{100} = 10 \text{ km} \]
Problem: A hiker walks 8 km east, then 6 km north. How far from the start?
\[ c = \sqrt{a^2+b^2} \quad\text{or}\quad a = \sqrt{c^2-b^2} \]
Word problems to watch for:
\[ c^2 = a^2 + b^2 \]
Pythagoras' Theorem — complete summary: