Pythagoras' theorem

What Pythagoras' theorem says

In a right-angled triangle, the side opposite the right angle is called the hypotenuse. It is always the longest side. The other two sides are the shorter sides, and they meet at the right angle.

Pythagoras' theorem: in any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

$$a^2+b^2=c^2$$

Here $c$ is the hypotenuse and $a$ and $b$ are the two shorter sides. It does not matter which shorter side you call $a$ and which you call $b$, but it matters very much which side is $c$.

"Square" can be read literally. Draw a square on each side of the triangle, facing outwards. The area of the square on the hypotenuse is exactly the total area of the squares on the other two sides. For the triangle with sides 3, 4 and 5: $3^2+4^2=9+16=25=5^2$.

The theorem is true only for right-angled triangles. If there is no right angle, you cannot use it.

The hypotenuse $c$ is opposite the right angle. Find it first in every question: it is the side that does not touch the right-angle mark.

Why the theorem is true

Take four copies of a right-angled triangle with shorter sides $a$ and $b$ and hypotenuse $c$. Arrange them inside a large square of side $a+b$, one in each corner, as in the diagram below.

The shape left in the middle has four sides, each of length $c$. Each of its corners is a right angle, because the angles on a straight line add up to ${180}^{°}$, and the two acute angles of a right-angled triangle add up to ${90}^{°}$, leaving ${180}^{°}-{90}^{°}={90}^{°}$. So the middle shape is a square of side $c$.

Now work out the area of the large square in two ways.

• As one square: $(a+b{)}^2=a^2+2ab+b^2$.
• As four triangles plus the middle square: $4\times \frac{1}{2}ab+c^2=2ab+c^2$.

These are the same area, so $a^2+2ab+b^2=2ab+c^2$. Take $2ab$ from both sides and you are left with $a^2+b^2=c^2$.

You will not be asked to prove the theorem, but knowing why it works helps you remember the most important fact about it: the hypotenuse is on its own on one side of the equation.

The large square has area $(a+b{)}^2$. It is also four triangles, each of area $\frac{1}{2}ab$, plus the shaded square of area $c^2$.

Finding the hypotenuse

When you know the two shorter sides and want the hypotenuse:

1. Identify the hypotenuse: the side opposite the right angle.
2. Square each of the two shorter sides and add.
3. Take the square root.
4. Check that your answer is longer than each of the other two sides, but shorter than the two of them added together.

Write the working as an equation, $c^2=a^2+b^2$, then $c=\sqrt{\dots }$. An examiner gives a method mark for the squares added, even if the final answer goes wrong.

Finding a shorter side

When you know the hypotenuse and one shorter side, rearrange the theorem: $a^2=c^2-b^2$.

1. Square the hypotenuse and square the side you know.
2. Subtract: the hypotenuse squared minus the other side squared.
3. Take the square root.

Why subtract? The square on the hypotenuse is the total of the other two squares. If you know the total and one of the parts, the other part is what is left when you take the known part away.

A quick check catches the most common error: a shorter side must come out shorter than the hypotenuse. If your answer is longer, you added instead of subtracting.

Pythagorean triples and working without a calculator

On the non-calculator papers the numbers are chosen so that the squares are easy and the square root comes out exactly. Sets of three whole numbers that fit $a^2+b^2=c^2$ are called Pythagorean triples. The ones worth knowing are:

• 3, 4, 5
• 5, 12, 13
• 8, 15, 17
• 7, 24, 25

Any multiple of a triple is also a triple, because multiplying every side by the same number gives a triangle of the same shape. So 6, 8, 10 and 30, 40, 50 and even 1.5, 2, 2.5 are all right-angled triangles.

Spotting a triple tells you the answer quickly, but still show the working. Write ${10}^2+{24}^2=100+576=676$ and $\sqrt{676}=26$; a bare 26 risks losing the method mark if it is wrong.

Without a calculator you need your squares up to ${15}^2=225$, and a way of finding a square root by trying: $\sqrt{576}$ ends in 4 or 6 (because $4^2$ and $6^2$ end in 6) and lies between $20$ and $25$ because ${20}^2=400$ and ${25}^2=625$. Try ${24}^2=576$. It works.

Is a triangle right-angled?

The theorem also works backwards. If the square of the longest side equals the sum of the squares of the other two sides, then the triangle has a right angle, and the right angle is opposite the longest side. This is called the converse of Pythagoras' theorem.

To test a triangle:

1. Square the longest side.
2. Square the other two sides and add.
3. Compare. Equal means right-angled. Not equal means not right-angled.

For example, a triangle with sides 9 cm, 40 cm and 41 cm: ${41}^2=1681$ and $9^2+{40}^2=81+1600=1681$. They are equal, so the triangle is right-angled, with the right angle opposite the 41 cm side.

A triangle with sides 7 cm, 9 cm and 12 cm: ${12}^2=144$ but $7^2+9^2=49+81=130$. They are not equal, so there is no right angle.

In a "show that" question you must write both numbers and a conclusion in words: "$841=841$, so the triangle is right-angled." Builders use exactly this idea: a rope with 12 equally spaced knots pegged out as a 3, 4, 5 triangle gives a true right angle for the corner of a building.

Spotting the right-angled triangle in a problem

Exam questions rarely hand you a bare triangle. Look for the right angle in these places:

• Rectangles and squares: a diagonal splits the shape into two right-angled triangles, with the diagonal as the hypotenuse.
• Isosceles triangles: the line of symmetry from the top vertex meets the base at right angles and cuts the base in half. Use half the base, not the whole base.
• Journeys: "due north" then "due east" makes a right angle at the turning point.
• Walls, poles and ladders: a wall is vertical and the ground is horizontal, so they meet at a right angle. The ladder is the hypotenuse.
• Two triangles sharing a side: find the shared side from the first triangle, then use it in the second. Keep its full calculator value; do not round it.

Always draw a sketch if there is no diagram, mark the right angle, and label the hypotenuse before you write any numbers.