Pythagoras' Theorem

Use Pythagoras' theorem to find the longest or a shorter side of a right-angled triangle, split isosceles triangles down the middle, chain two triangles through a shared side, and add up perimeters and costs.

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, Cambridge O Level Maths 4024, Edexcel Further Pure Maths 4PM1 · 64 real past-paper questions below

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The Longest Side Squared

In a right-angled triangle, the square on the longest side equals the two other squares added together.

Formula

a^{2} + b^{2} = c^{2}

$c$ is the hypotenuse, across from the right angle. $a$ and $b$ are the two shorter sides.

Worked example

1
$d^{2} = 9. 2^{2} + 6. 5^{2}$ — The diagonal of a rectangle is across from the right angle, so it is the longest side. Square the two shorter sides and add.
2
$d^{2} = 84.64 + 42.25$ — Square each side: 9.2 × 9.2 = 84.64 and 6.5 × 6.5 = 42.25.
3
$d^{2} = 126.89$ — Add the two squares.
4
$d = 126.89 = 11.26 \dots$ — We have d squared, so take the square root to undo the squaring.
5
$d = 11.3 cm$ — Round to 3 significant figures and add the unit.

Common mistake

Adding the sides: $9.2 + 6.5 = 15.7$ . Square them first, then add, then square root.

Key idea

About 1800 BC, long before Pythagoras, a scribe in Babylonia (today's Iraq) made a clay tablet now called Plimpton 322. It lists 15 rows of right-triangle number sets, including $1270 9^{2} + 1350 0^{2} = 1854 1^{2}$ .

Finding a Shorter Side

The 3-mark question can give the hypotenuse and ask for a shorter side. Then you subtract. The hypotenuse is 15.4 cm and one side is 8.1 cm.

Worked example

1
$x^{2} = 15. 4^{2} - 8. 1^{2}$ — 15.4 is the hypotenuse. From $a^{2} + b^{2} = c^{2}$ , take $b^{2}$ from both sides, so a shorter side squared is $c^{2}$ minus the other short side squared.
2
$x^{2} = 237.16 - 65.61$ — Square each side: 15.4 × 15.4 = 237.16 and 8.1 × 8.1 = 65.61.
3
$x^{2} = 171.55$ — Subtract the two squares.
4
$x = 171.55 = 13.098 \dots$ — We have x squared, so take the square root.
5
$x = 13.1 cm$ — Round to 3 significant figures. It is shorter than 15.4, as a shorter side should be.

Common mistake

Adding the squares and getting 17.4 cm. That is longer than the hypotenuse, which is impossible.

Key idea

Write the Pythagoras line first, before any numbers. A correct line earns a method mark even if the arithmetic slips later.

Isosceles Triangles: Split Down the Middle

The 4 or 5-mark question: an isosceles triangle. The line from the top to the middle of the base is the height, and it makes two right-angled triangles.

The base is 16 cm, the two equal sides are unknown and the area is $72 cm^{2}$ . Find the perimeter.

Worked example

1
$\frac{1}{2} \times 16 \times h = 72$ — Area of a triangle is half the base times the height. The base is 16 and the area is 72, so h is the only unknown.
2
$8 h = 72$ — Half of 16 is 8.
3
$h = 9$ — Divide both sides by 8 to leave h alone.
4
$half the base = 16 \div 2 = 8$ — The height cuts the base in half and makes two right-angled triangles. Each one has a base of 8, not 16.
5
$x^{2} = 8^{2} + 9^{2}$ — x is an equal side, across from the right angle, so it is the hypotenuse. Square the two shorter sides and add.
6
$x^{2} = 64 + 81$ — Square each side: 8 × 8 = 64 and 9 × 9 = 81.
7
$x^{2} = 145$ — Add the two squares.
8
$x = 145 = 12.04 \dots$ — Take the square root. This is the length of one equal side.
9
$16 + 2 \times 12.04 \dots$ — The perimeter is the base plus the two equal sides.
10
$perimeter = 40.1 cm$ — Round to 3 significant figures and add the unit.

Common mistake

Using the whole base, 16, in Pythagoras. The right angle is in the middle, so each half has base 8.

Two Triangles: Find the Shared Side First

The 5 or 6-mark question has two right-angled triangles joined along one side. Work out the shared side in the first, then use it in the second.

Triangle $A B C$ has a right angle at $B$ , with $A B = 12$ and $B C = 7$ . Triangle $A C D$ has a right angle at $C$ , with $C D = 10$ . Find $A D$ .

Worked example

1
$A C^{2} = 1 2^{2} + 7^{2}$ — The right angle of triangle ABC is at B, so AC is its hypotenuse. AC is also the side the two triangles share.
2
$A C^{2} = 144 + 49$ — Square each side: 12 × 12 = 144 and 7 × 7 = 49.
3
$A C^{2} = 193$ — Add the two squares. Keep 193, do not take the square root yet.
4
$A D^{2} = A C^{2} + 1 0^{2}$ — The right angle of triangle ACD is at C, so AD is its hypotenuse. Its shorter sides are AC and CD = 10.
5
$A D^{2} = 193 + 100$ — Put in the exact $A C^{2} = 193$ and square 10 to get 100. We need $A C^{2}$ , not AC, so no square root is needed here.
6
$A D^{2} = 293$ — Add the two numbers.
7
$A D = 293 = 17.117 \dots$ — Take the square root to undo the squaring.
8
$A D = 17.1 cm$ — Round to 3 significant figures and add the unit.

Common mistake

Rounding $A C$ to 14 first. Then $296 = 17.2$ , which loses the mark. Keep the exact value.

Perimeters and Costs

The 4-mark question: a frame or a shape made of right-angled triangles. Find the missing edge, add up all the edges, then answer the money part.

A triangular frame has sides of 3.6 metres and 2.7 metres at the right angle. Steel costs $15 per metre and is sold in whole metres.

Worked example

1
$c^{2} = 3. 6^{2} + 2. 7^{2}$ — The sides 3.6 and 2.7 meet at the right angle, so the missing edge c is the hypotenuse.
2
$c^{2} = 12.96 + 7.29$ — Square each side: 3.6 × 3.6 = 12.96 and 2.7 × 2.7 = 7.29.
3
$c^{2} = 20.25$ — Add the two squares.
4
$c = 20.25 = 4.5 m$ — Take the square root to undo the squaring.
5
$3.6 + 2.7 + 4.5 = 10.8 m$ — The frame has three edges, so add them all to get the length of steel needed.
6
$buy 11 m$ — Steel is sold in whole metres. 10.8 is not a whole number, so round up to 11. Rounding down would leave you short.
7
$11 \times 15 = $165$ — Each metre costs 15 dollars, so multiply the 11 metres by 15.

Common mistake

Working out $10.8 \times 15 = $162$ . You cannot buy 10.8 metres, so you must buy 11.

Is It a Right Angle? The Converse

Some questions give three sides and ask whether the triangle has a right angle. Pythagoras works backwards: if the numbers fit, the angle is 90°.

Rule

Square the two shorter sides and add them. Square the longest side. If the two answers are equal, the triangle is right-angled. If they are different, it is not.

Worked example

1
$sides 9, 12, 15 : test 15$ — The longest side must be the hypotenuse, so test with 15. The other two, 9 and 12, are the shorter sides.
2
$9^{2} + 1 2^{2} = 225$ — Square 9 to get 81 and 12 to get 144, then add.
3
$1 5^{2} = 225$ — Now square the longest side.
4
$9^{2} + 1 2^{2} = 1 5^{2}$ — The two answers are equal, so the sides fit Pythagoras.
5
$right-angled$ — Write this conclusion. The angle opposite the 15 is 90°.

Now a triangle that fails the test. The sides are 6, 7 and 10.

Worked example

1
$6^{2} + 7^{2} = 36 + 49 = 85$ — Square the two shorter sides and add.
2
$1 0^{2} = 100$ — Square the longest side.
3
$85 \neq = 100$ — The answers are different, so the sides do not fit Pythagoras.
4
$not right-angled$ — Write this conclusion. The longest side is longer than a right angle would give.

Common mistake

Testing with the wrong side as the hypotenuse. Always square the longest side by itself and add the other two. Otherwise the answers will never match.

Exact Answers, Surds (Extended/Higher)

Some questions say “give your answer as a surd” or “show that the height is 5√3”. Then leave the square root unevaluated and simplify it. An equilateral triangle has side 10 cm. Find its height as a surd.

Formula

a \times b = a \times b

Here a and b are positive numbers. Pick a to be the biggest square number that divides the number under the root.

Worked example

1
$half the base = 10 \div 2 = 5$ — The height of an equilateral triangle cuts the base in half and makes a right-angled triangle. The sloping side, 10, is the hypotenuse.
2
$h^{2} = 1 0^{2} - 5^{2}$ — h is a shorter side, so take the square of the other short side away from the square of the hypotenuse.
3
$h^{2} = 100 - 25$ — Square each side: 10 × 10 = 100 and 5 × 5 = 25.
4
$h^{2} = 75$ — Subtract.
5
$h = 75$ — Take the square root. Leave it as a root, because the question asks for a surd.
6
$75 = 25 \times 3$ — Find the biggest square number that divides 75. It is 25, because 25 × 3 = 75.
7
$25 \times 3 = 25 \times 3$ — Split the root into two roots using the formula above.
8
$h = 53 cm$ — $25 = 5$ . $3$ cannot be simplified, so stop. Check: 5 × 1.732… = 8.66…, the same as $75$ .

Common mistake

Writing 8.66 when the question asks for a surd. A decimal is not an exact answer, so it loses the final mark.

Key idea

Simplifying surds has its own lesson: see Surds.

Which Method? Read the Question

Pythagoras needs a right angle and two known sides.

You see	You do
Find the longest side	square, add, square root
Find a shorter side	square, subtract, square root
Isosceles triangle	split it: half the base and the height
Is it a right-angled triangle?	square the shorter two and add; compare with the longest squared
Exact answer or surd (Extended/Higher)	leave the root, take out the biggest square number
An angle is given or wanted	Choosing the Right Trigonometric Ratio
A shape needing a side and an angle	Using Trigonometry and Pythagoras Together

Key idea

Check before you finish: the hypotenuse must be the longest side.

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