The Pythagoras theorem and its converse

The theorem

The picture shows a right triangle with sides $3$ cm, $4$ cm and $5$ cm. The square on the longest side has area $25$ square cm, and $9 + 16 = 25$.

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

The hypotenuse is the side opposite the right angle. It is always the longest side.

A proof with similar triangles

Let $\triangle ABC$ have its right angle at B. Draw $BD \perp AC$, with D on AC.

1. First pair. $\triangle ADB$ and $\triangle ABC$ share $\angle A$, and $\angle ADB = \angle ABC = 90^\circ$. By AA they are similar. So $\dfrac{AD}{AB} = \dfrac{AB}{AC}$, which gives $AB^2 = AD \times AC$.
2. Second pair. $\triangle BDC$ and $\triangle ABC$ share $\angle C$, and $\angle BDC = \angle ABC = 90^\circ$. So they are similar, and $\dfrac{CD}{BC} = \dfrac{BC}{AC}$, which gives $BC^2 = CD \times AC$.
3. Add them. $AB^2 + BC^2 = AC \times (AD + CD) = AC \times AC = AC^2$.

That is the Pythagoras theorem.

A bonus. $\triangle ADB \sim \triangle BDC$ as well, which gives $\dfrac{AD}{BD} = \dfrac{BD}{DC}$. So $BD^2 = AD \times DC$. If $AD = 2$ cm and $DC = 8$ cm, then $BD^2 = 16$ and $BD = 4$ cm.

The converse

Converse. If in a triangle the square of one side equals the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Why? Let $AC^2 = AB^2 + BC^2$ in $\triangle ABC$.

1. Draw $\triangle PQR$ with $\angle Q = 90^\circ$, $PQ = AB$ and $QR = BC$.
2. By the theorem, $PR^2 = PQ^2 + QR^2 = AB^2 + BC^2 = AC^2$. So $PR = AC$.
3. The two triangles have three equal sides, so they are congruent by SSS.
4. Hence $\angle B = \angle Q = 90^\circ$.

Test a triangle. Are $20$ cm, $21$ cm and $29$ cm the sides of a right triangle? Compare the square of the longest side with the other two.

Yes. The right angle is opposite the side of $29$ cm.

For $6$, $8$ and $11$: $36 + 64 = 100$, which is less than $11^2 = 121$. The triangle is not right-angled. In fact, when the square of the longest side is more than the other two added, the biggest angle is more than $90^\circ$ (obtuse). When it is less, every angle is acute.

Using the theorem

A ladder. A ladder $13$ m long leans against a wall, with its foot $5$ m from the wall. How high does it reach?

The ladder is the hypotenuse: $h^2 = 13^2 - 5^2 = 169 - 25 = 144$, so $h = 12$ m.

The diagonal of a rectangle. A rectangle is $12$ cm by $16$ cm. Its diagonal is the hypotenuse of a right triangle with legs $12$ cm and $16$ cm: $d^2 = 144 + 256 = 400$, so $d = 20$ cm.

Triples. Three whole numbers that satisfy the theorem are called a Pythagorean triple. $3, 4, 5$ and $5, 12, 13$ are the two best known. Doubling or trebling every number of a triple gives another triple, such as $6, 8, 10$.

Which side is missing? Missing hypotenuse: add the squares of the legs. Missing leg: subtract the square of the known leg from the square of the hypotenuse.

Practise with new numbers

Try it in Python

The program tests four triangles. It compares the square of the longest side with the sum of the other two squares, and says whether each triangle is right-angled, acute or obtuse.

Common mistakes

• Adding when you should subtract. If the hypotenuse is given, you subtract to find a leg.
• Using the theorem without a right angle. It holds only for right triangles. Its converse tells you when there is one.
• Forgetting the square root. $c^2 = 25$ means $c = 5$.
• Choosing the wrong side as the hypotenuse. It is opposite the right angle, and it is the longest side.

Remember these

• Right angle at B: $AC^2 = AB^2 + BC^2$.
• Converse: if the squares add up in this way, the angle is a right angle.
• Draw the perpendicular from the right angle to the hypotenuse, and you get two similar triangles. Then $BD^2 = AD \times DC$.
• Check units: lengths in cm or m, areas in square cm or square m.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A right triangle has legs $6$ cm and $8$ cm. How long is the hypotenuse?

2. A right triangle has hypotenuse $26$ cm and one leg $10$ cm. Find the other leg.

3. Is a triangle with sides $9$ cm, $12$ cm and $16$ cm right-angled?

4. In $\triangle ABC$, $\angle B = 90^\circ$, $BD \perp AC$, $AD = 5$ cm and $DC = 20$ cm. Find $BD$.

5. Why is the hypotenuse the longest side of a right triangle?