The Theorem of Pythagoras

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\( c^2 = a^2 + b^2 \)

The Theorem of Pythagoras is the most useful single result in school geometry. It links the three sides of a right-angled triangle, so you can work out a length you cannot measure.

In Grade 8 you met the theorem. This lesson takes it further:

• State and apply the theorem in symbols
• Find the hypotenuse or a shorter side
• Recognise Pythagorean triplets and their multiples
• Classify a triangle as right, acute or obtuse
• Work through perimeters and composite figures
• Solve real-life problems and round properly

1. The Theorem

b a c

Every right-angled triangle has one angle of exactly 90°, marked with a small square. The sides have names:

Find the little square first, then look straight across from it. That is your hypotenuse — not simply the side that looks longest on a rough sketch.

\( c^2 = a^2 + b^2 \)

Learn this sentence word for word — you can be asked to state it, and it is worth two marks:

Note the first four words. In a right-angled triangle. That is a condition, not decoration — no right angle, no theorem.

\[ \begin{aligned} c^2 &= a^2 + b^2 \\ a^2 &= c^2 - b^2 \\ b^2 &= c^2 - a^2 \end{aligned} \]

The first form finds the hypotenuse. But half of all questions give you the hypotenuse and ask for a leg, so you need it rearranged.

It is the same theorem all three times — just a different letter made the subject. In a real question the sides are named after the points, so you might write \( AB^2 = BC^2 + AC^2 \). Same thing.

\[ \begin{aligned} \text{hypotenuse} &\Rightarrow \textbf{ADD} \\ \text{shorter side} &\Rightarrow \textbf{SUBTRACT} \end{aligned} \]

If you remember one thing from this lesson, make it this one.

And you get a free check. The hypotenuse is the longest side, so if you calculate a leg and get an answer bigger than the hypotenuse, you added when you should have subtracted. The triangle is telling you that you slipped.

2. Finding a Missing Side

6 cm 8 cm c

Legs of 6 cm and 8 cm. Find the hypotenuse.

Write the theorem before you substitute — it is a mark, and it stops you guessing. Then check the answer makes sense: 10 is longer than both 6 and 8, exactly as a hypotenuse should be.

b 5 cm 13 cm

Now the other direction. The hypotenuse is 13 cm and one leg is 5 cm.

The hypotenuse was given, so we subtracted. Had you added, you would have got \( \sqrt{194} \approx 13{,}9 \) — longer than the hypotenuse, which is impossible.

\( c^2 = 100 \;\; \Rightarrow \;\; c = 10 \)

This single step loses more marks in this topic than everything else put together.

Learners do all the hard work and then write 100 as the answer. The square root is always the last line.

\[ \begin{aligned} c^2 &= 10^2 + 12^2 \ &= 244 \ c &\approx 15{,}62 \end{aligned} \]

Most triangles do not give a whole number. Legs of 10 cm and 12 cm give \( c = 15{,}6204\ldots \), which rounds to \( 15{,}62 \) cm.

If you round in the middle and carry that number into another calculation, the error grows and the final answer can be wrong enough to lose the mark.

3. Pythagorean Triplets

\( 3^2 + 4^2 = 9 + 16 = 25 = 5^2 \)

Some triangles are so tidy that all three sides are whole numbers. Those side sets are called Pythagorean triplets.

Ancient builders used this: a rope with twelve equally spaced knots pulled into a 3-4-5 triangle gave a perfect right angle, every time, with no set square.

\[ \begin{aligned} 3;4;5 \quad & 5;12;13 \ 8;15;17 \quad & 7;24;25 \end{aligned} \]

Learn these four. When a question uses one, you can see the answer before you calculate — then use the working to confirm it.

\[ 6;8;10 \qquad 9;12;15 \qquad 12;16;20 \]

Any multiple of a triplet is also a triplet. Double 3-4-5 and you get 6-8-10. Triple it and you get 9-12-15.

Being whole numbers is not enough on its own, though. 6, 8 and 11 are all whole numbers, but \( 36 + 64 = 100 \) while \( 11^2 = 121 \). Not a triplet.

4. Classifying Triangles

\[ \begin{aligned} c^2 = a^2 + b^2 &\Rightarrow \text{right} \\ c^2 a^2 + b^2 &\Rightarrow \text{obtuse} \end{aligned} \]

This is the part that is new in Grade 9. The theorem also runs backwards — that is called the converse.

Instead of using the angle to find a side, you use the sides to find out about the angle. Square the longest side on its own, then compare it with the sum of the squares of the other two.

Three outcomes, and you must be able to name all three — not just say yes or no.

\[ \begin{aligned} 15^2 &= 225 \ 9^2 + 12^2 &= 225 \end{aligned} \]

Classify the triangle with sides 9; 12; 15.

This is 3-4-5 tripled, so you could have spotted it straight away.

\[ \begin{aligned} 9^2 &= 81 \ 5^2 + 7^2 &= 74 \end{aligned} \]

Classify the triangle with sides 5; 7; 9.

Larger means obtuse: one angle is over 90°.

\[ \begin{aligned} 9^2 &= 81 \ 6^2 + 7^2 &= 85 \end{aligned} \]

Classify the triangle with sides 6; 7; 9.

Compare that with 5; 7; 9, which was obtuse. One side changed by a single unit and the answer flipped. Always do the arithmetic; never judge from the picture.

5. Perimeter & Composite Figures

9 cm 14 cm ?

Perimeter questions need two skills at once. Legs of 9 cm and 14 cm — find the perimeter.

You cannot add up the sides while one is still missing. Find the third side first, then add all three.

A D C B 10 6 17

This is the Grade 9 question that separates the marks. Two right-angled triangles share the side \( AD \).

In \( \triangle ACD \) the right angle is at \( D \), with \( AC = 10 \) cm and \( CD = 6 \) cm. In \( \triangle ABD \) the right angle is also at \( D \), with \( AB = 17 \) cm.

Find the perimeter of \( \triangle ABD \).

\[ \begin{aligned} AD^2 &= 100 - 36 \ &= 64 \ AD &= 8 \text{ cm} \end{aligned} \]

You cannot start with \( \triangle ABD \) — you only know one of its sides. So start with the triangle you can finish.

In \( \triangle ACD \), the side \( AC \) is opposite the right angle, so it is the hypotenuse. Subtract.

\[ \begin{aligned} BD^2 &= 289 - 64 \ &= 225 \ BD &= 15 \text{ cm} \end{aligned} \]

Now take that \( AD = 8 \) into the second triangle.

That is the whole pattern of a composite figure: a side you find in one triangle is the key to the next one.

6. Real-Life Problems

h 2,5 m 6,5 m

A ladder 6,5 m long leans against a vertical wall, its foot 2,5 m from the wall. How high up the wall does it reach?

The wall is vertical and the ground horizontal, so the right angle is where they meet. The ladder is the sloping side.

In any leaning problem, the thing that leans is the hypotenuse — so you subtract.

24 cm 10 cm d

A rectangle is 24 cm long and 10 cm wide. Find the diagonal.

The diagonal cuts the rectangle into two right-angled triangles. A rectangle's corner is already 90°, so the theorem is allowed — and the diagonal is opposite that corner, so it is the hypotenuse. Add.

\[ \begin{aligned} 12^2 + 5^2 &= 169 \ d &= 13 \text{ cm} \end{aligned} \]

A rectangle 12 cm by 5 cm has a diagonal of 13 cm — the 5-12-13 triplet again.

Any time you see a rectangle with a diagonal drawn in, a ladder against a wall, a ramp, a roof truss, or the shortest path across a field, you are looking at a Pythagoras question. Draw the triangle, mark the right angle, and label the hypotenuse before you write anything else.

\( \text{no right angle} \Rightarrow \text{no theorem} \)

Four mistakes to avoid:

For that third one: look for the little square in the corner before you start. You can still use the converse to classify a triangle without a right angle — but you cannot use the theorem to find a side.

\[ \begin{aligned} c^2 &= a^2 + b^2 \\ a^2 &= c^2 - b^2 \end{aligned} \]

Five things to take with you:

Triplets worth knowing by heart: 3;4;5   5;12;13   8;15;17   7;24;25 — and every multiple of them.