What Is Pythagoras' Theorem?

In every right-angled triangle, the square on the longest side equals the sum of the squares on the other two sides. Find the hypotenuse, see the theorem in squares drawn on the sides, and write it for any lettered triangle.

Before you start

The angles of a triangle add up to $180^{\circ}$ (∠ sum of △).
Squaring a number multiplies it by itself: $5^2 = 5 \times 5 = 25$ . The square root undoes it: $\sqrt{25} = 5$ .
In $\triangle ABC$ , $AB$ is the side joining $A$ and $B$ , and $\angle C$ is the angle at $C$ . The side opposite $\angle C$ is the one that does not touch $C$ : the side $AB$ .

Right-Angled Triangles and the Hypotenuse

A right-angled triangle has one angle of $90^{\circ}$ , marked in a drawing with a small square. The other two angles add up to $180^{\circ} - 90^{\circ} = 90^{\circ}\text{,}$ so each of them is less than $90^{\circ}$ . The right angle is always the largest angle of the triangle.

The hypotenuse

In a right-angled triangle, the side opposite the right angle is called the hypotenuse.

The hypotenuse is the longest side of the triangle. This rests on a fact about every triangle, which you may take as given: the longest side is opposite the largest angle. The right angle is the largest angle, so the side opposite it is the longest.

In $\triangle ABC$ the right angle is at $C$ , so the hypotenuse is $AB$ , the side that does not touch $C$ . It is $13$ cm, longer than both $12$ cm and $5$ cm.

The hypotenuse is not always the sloping side or the side at the top. Find the right angle first; the hypotenuse is the side across from it, wherever the triangle is drawn.

Example 1: Naming the hypotenuse

Name the hypotenuse of each triangle. (a) $\triangle PQR$ with $\angle Q = 90^{\circ}$ . (b) $\triangle XYZ$ and (c) $\triangle LMN$ , drawn below.

Solution

(a) The right angle is at $Q$ . The side that does not touch $Q$ is $PR$ .

(b) The right-angle mark is at $Y$ , at the top. The side opposite $Y$ is $XZ$ , the side along the bottom.

(a) $PR$ (b) $XZ$ (c) $MN$

Squares on the Sides

Take the right-angled triangle with sides $3$ cm, $4$ cm and $5$ cm, and draw a square standing on each side. Each square is ruled in squares of $1$ cm by $1$ cm.

Count the small squares. The square on the $3$ cm side has $3^2 = 9$ , the square on the $4$ cm side has $4^2 = 16$ , and the square on the hypotenuse has $5^2 = 25$ .

And $9 + 16 = 25$ : the two smaller squares together hold exactly as many small squares as the square on the hypotenuse. In symbols, $3^2 + 4^2 = 5^2$ .

Example 2: Checking the pattern

Check that the same is true for the right-angled triangles with sides (a) $5$ cm, $12$ cm and $13$ cm, (b) $8$ cm, $15$ cm and $17$ cm.

Solution

The hypotenuse is the longest side. Add the squares of the other two sides, and compare with the square of the hypotenuse.

(a) The hypotenuse is $13$ cm. $5^2 + 12^2 = 25 + 144 = 169\text{,}$ and $13^2 = 169$ .

(b) The hypotenuse is $17$ cm. $8^2 + 15^2 = 64 + 225 = 289\text{,}$ and $17^2 = 289$ .

In both, the sum of the squares of the two shorter sides equals the square of the hypotenuse: $5^2 + 12^2 = 13^2$ and $8^2 + 15^2 = 17^2$ .

Three whole numbers that fit the pattern, such as $3$ , $4$ , $5$ and $5$ , $12$ , $13$ , are called Pythagorean triples. Multiplying a triple by the same number gives another: $3$ , $4$ , $5$ doubled is $6$ , $8$ , $10$ , and $6^2 + 8^2 = 36 + 64 = 100 = 10^2\text{.}$

Pythagoras' Theorem

The pattern holds in every right-angled triangle, whatever its size and whether its sides are whole numbers or not. It is named after the Greek mathematician Pythagoras.

Pythagoras' Theorem

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

If the hypotenuse is $c$ and the other two sides are $a$ and $b$ , then $a^2 + b^2 = c^2$ .

In areas: the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides.

Letters are not always $a$ , $b$ and $c$ . In a lettered triangle, write the theorem with the names of the sides. Start with the hypotenuse: it is the side opposite the right angle, and its square stands alone on one side of the equals sign.

The theorem for a lettered triangle

In $\triangle ABC$ with $\angle C = 90^{\circ}$ , the hypotenuse is $AB$ , so $AB^2 = AC^2 + BC^2$ .

In $\triangle PQR$ with $\angle Q = 90^{\circ}$ , the hypotenuse is $PR$ , so $PR^2 = PQ^2 + QR^2$ .

Example 3: Writing the theorem

Write Pythagoras' theorem for each triangle. (a) $\triangle XYZ$ with $\angle Y = 90^{\circ}$ . (b) $\triangle DEF$ and (c) $\triangle PQR$ , drawn below, where $p$ , $q$ and $r$ are the lengths of the sides opposite $P$ , $Q$ and $R$ .

Solution

(a) The right angle is at $Y$ , so the hypotenuse is $XZ$ . The other two sides are $XY$ and $YZ$ .

(b) The right angle is at $F$ , so the hypotenuse is $DE$ . The other two sides are $DF$ and $EF$ .

(c) The right angle is at $P$ , so the hypotenuse is the side opposite $P$ , which is $p$ . The other two sides are $q$ and $r$ .

(a) $XZ^2 = XY^2 + YZ^2$ (b) $DE^2 = DF^2 + EF^2$ (c) $p^2 = q^2 + r^2$

When Can It Be Used?

Pythagoras' theorem is true only in a right-angled triangle. In a triangle with no right angle, the square of the longest side is not the sum of the squares of the other two. For example, an equilateral triangle with sides of $2$ cm has $2^2 + 2^2 = 8$ , which is not $2^2 = 4$ .

So before using the theorem, check that the triangle has a right angle: it is marked, or stated, or the angles show it.

Example 4: Can Pythagoras' theorem be used?

Can Pythagoras' theorem be used in each triangle? (a) $\triangle ABC$ with angles $90^{\circ}$ , $45^{\circ}$ and $45^{\circ}$ . (b) $\triangle DEF$ with angles $50^{\circ}$ , $60^{\circ}$ and $70^{\circ}$ . (c) $\triangle GHI$ with angles $110^{\circ}$ , $40^{\circ}$ and $30^{\circ}$ .

Solution

(a) One angle is $90^{\circ}$ , so the triangle is right-angled. The theorem can be used, with the side opposite $A$ , which is $BC$ , as the hypotenuse.

(b) No angle is $90^{\circ}$ : all three are less than $90^{\circ}$ . The theorem cannot be used.

(c) The angles are $110^{\circ}$ , $40^{\circ}$ and $30^{\circ}$ . None is $90^{\circ}$ , and a triangle with a $110^{\circ}$ angle cannot also have a right angle, since $110^{\circ} + 90^{\circ}$ is already more than $180^{\circ}$ . The theorem cannot be used.

(a) Yes (b) No (c) No

Common Mistakes

Adding the sides instead of their squares. For the $3$ , $4$ , $5$ triangle, $3 + 4 = 7$ , not $5$ . The theorem is about the squares: $3^2 + 4^2 = 9 + 16 = 25 = 5^2\text{.}$
Squaring the sum. $(3 + 4)^2 = 7^2 = 49$ , but $3^2 + 4^2 = 25$ . Square each side first, then add.
Taking the side at the bottom as the hypotenuse. The hypotenuse is opposite the right angle, wherever the triangle is drawn. Find the right angle first.
Using the theorem without a right angle. In a triangle with angles $50^{\circ}$ , $60^{\circ}$ and $70^{\circ}$ , there is no hypotenuse and the theorem does not hold.

Practice Questions

Try each question before you open its solution. Find the right angle first, then the hypotenuse opposite it.

Question 1

In $\triangle PQR$ , $\angle Q = 90^{\circ}$ , $PQ = 5$ cm, $QR = 12$ cm and $PR = 13$ cm. (a) Which side is the hypotenuse? (b) Which angle is $PR$ opposite, and why does that make $PR$ the longest side?

ShowHide solution

(a) The right angle is at $Q$ . The side opposite $Q$ is $PR$ .

(b) $PR$ is opposite $\angle Q = 90^{\circ}$ . The other two angles add up to $90^{\circ}$ , so each is less than $90^{\circ}$ , and $\angle Q$ is the largest angle. The longest side of a triangle is opposite its largest angle, so $PR$ is the longest side.

(a) $PR$ (b) $PR$ is opposite $\angle Q$ , the right angle, which is the largest angle of the triangle; the longest side is opposite the largest angle.

Question 2

Name the hypotenuse of each triangle. (a) $\triangle DEF$ with $\angle E = 90^{\circ}$ . (b) $\triangle KLM$ with $\angle M = 90^{\circ}$ . (c) $\triangle STU$ , drawn below.

ShowHide solution

(a) The side opposite $E$ is $DF$ .

(b) The side opposite $M$ is $KL$ .

(a) $DF$ (b) $KL$ (c) $TU$

Question 3

A right-angled triangle has sides $11$ cm, $60$ cm and $61$ cm. (a) Which side is the hypotenuse? (b) Work out $11^2 + 60^2$ . (c) Work out $61^2$ . (d) Does the triangle fit the pattern of Pythagoras' theorem?

ShowHide solution

(a) The hypotenuse is the longest side: $61$ cm.

(b) $11^2 + 60^2 = 121 + 3600 = 3721\text{.}$

(d) The two answers are equal.

(a) $61$ cm (b) $3721$ (c) $3721$ (d) Yes: $11^2 + 60^2 = 61^2$ .

Question 4

Check the pattern for the right-angled triangles with sides (a) $7$ cm, $24$ cm, $25$ cm and (b) $9$ m, $12$ m, $15$ m.

ShowHide solution

(a) The hypotenuse is $25$ cm. $7^2 + 24^2 = 49 + 576 = 625\text{,}$ and $25^2 = 625$ .

(b) The hypotenuse is $15$ m. $9^2 + 12^2 = 81 + 144 = 225\text{,}$ and $15^2 = 225$ .

(a) $7^2 + 24^2 = 25^2$ (b) $9^2 + 12^2 = 15^2$ . Both fit.

Question 5

Write Pythagoras' theorem for (a) $\triangle LMN$ with $\angle N = 90^{\circ}$ , (b) $\triangle ABC$ with $\angle A = 90^{\circ}$ , (c) $\triangle UVW$ , drawn below.

ShowHide solution

(a) The hypotenuse is opposite $N$ : $LM$ . So $LM^2 = LN^2 + MN^2$ .

(b) The hypotenuse is opposite $A$ : $BC$ . So $BC^2 = AB^2 + AC^2$ .

(a) $LM^2 = LN^2 + MN^2$ (b) $BC^2 = AB^2 + AC^2$ (c) $UW^2 = UV^2 + VW^2$

Question 6

Can Pythagoras' theorem be used in each triangle? (a) Angles $30^{\circ}$ , $60^{\circ}$ and $90^{\circ}$ . (b) Angles $40^{\circ}$ , $65^{\circ}$ and $75^{\circ}$ . (c) Angles $100^{\circ}$ , $45^{\circ}$ and $35^{\circ}$ . (d) Two of the angles are $35^{\circ}$ and $55^{\circ}$ .

ShowHide solution

(a) One angle is $90^{\circ}$ : yes.

(b) No angle is $90^{\circ}$ : no.

(d) The third angle is $180^{\circ} - 35^{\circ} - 55^{\circ} = 90^{\circ}$ (∠ sum of △). The triangle is right-angled: yes.

(a) Yes (b) No (c) No (d) Yes, the third angle is $90^{\circ}$ .

Question 7

For a right-angled triangle with sides $3$ cm, $4$ cm and $5$ cm, Mei Ling writes $3^2 + 4^2 = (3 + 4)^2 = 7^2 = 49\text{.}$ Explain her mistake.

ShowHide solution

$(3 + 4)^2$ squares the sum, which is not the same as adding the squares.

Correctly, $3^2 + 4^2 = 9 + 16 = 25\text{,}$ which equals $5^2$ , the square of the hypotenuse.

She added before squaring. $3^2 + 4^2 = 9 + 16 = 25 = 5^2\text{,}$ not $49$ .

Question 8

In $\triangle ABC$ , $\angle A = 90^{\circ}$ . Raj writes $AB^2 = AC^2 + BC^2$ . What is wrong, and what should he write?

ShowHide solution

The side on its own must be the hypotenuse, which is opposite the right angle at $A$ : the side $BC$ .

$AB$ touches $A$ , so it is one of the two shorter sides.

He took $AB$ as the hypotenuse. It should be $BC^2 = AB^2 + AC^2$ .

Question 9

A garden at Punggol has a right-angled triangular lawn. Square patches of paving are laid along each side of the lawn. The patches along the two shorter sides have areas of $36$ m² and $64$ m². Find (a) the area of the patch along the longest side, (b) the length of the longest side of the lawn.

ShowHide solution

(a) By Pythagoras' theorem in areas, the square on the hypotenuse has the sum of the areas of the other two squares: $36 + 64 = 100$ m².

(b) The longest side is the side of a square of area $100$ m². Its length is $\sqrt{100} = 10$ m.

(a) $100$ m² (b) $10$ m

Question 10

Siti cuts three paper squares and lays one along each side of a right-angled triangle. The largest square has an area of $169$ cm², and one of the others has an area of $25$ cm². Find the area of the third square and the lengths of the three sides of the triangle.

ShowHide solution

The largest square stands on the hypotenuse, so its area is the sum of the other two: $169 = 25 + \text{third area}\text{.}$

Third area $= 169 - 25 = 144$ cm².

Each side is the square root of the area of its square: $\sqrt{169} = 13$ cm, $\sqrt{25} = 5$ cm and $\sqrt{144} = 12$ cm.

Check: $5^2 + 12^2 = 25 + 144 = 169 = 13^2\text{.}$

The third square is $144$ cm². The sides are $5$ cm, $12$ cm and $13$ cm.

Question 11

(a) Show that a triangle with sides $15$ cm, $20$ cm and $25$ cm fits the pattern of Pythagoras' theorem. (b) Multiply the triple $5$ , $12$ , $13$ by $3$ and check that the new numbers fit the pattern too.

ShowHide solution

(a) $15^2 + 20^2 = 225 + 400 = 625\text{,}$ and $25^2 = 625$ . (These are $3$ , $4$ , $5$ times $5$ .)

(b) $5 \times 3 = 15$ , $12 \times 3 = 36$ and $13 \times 3 = 39$ . $15^2 + 36^2 = 225 + 1296 = 1521\text{,}$ and $39^2 = 1521$ .

(a) $15^2 + 20^2 = 25^2$ (b) $15$ , $36$ , $39$ : $15^2 + 36^2 = 39^2$ .

Key Points

A right-angled triangle has one angle of $90^{\circ}$ , and it is the largest angle; the other two add up to $90^{\circ}$ .
The hypotenuse is the side opposite the right angle. It is the longest side.
Pythagoras' theorem: in a right-angled triangle, $a^2 + b^2 = c^2$ , where $c$ is the hypotenuse.
In areas: the square on the hypotenuse equals the two squares on the other sides put together.
For a lettered triangle, put the hypotenuse on its own: in $\triangle ABC$ with $\angle C = 90^{\circ}$ , $AB^2 = AC^2 + BC^2$ .
Square each side first, then add: $3^2 + 4^2 = 25$ , not $(3 + 4)^2$ .
The theorem is true only in a right-angled triangle.