Apply Pythagoras theorem to find an unknown side in a right-angled triangle and to test whether a triangle is right-angled

What this dot point is asking

SEAB wants you to apply Pythagoras theorem to find a missing side of a right-angled triangle, and to use its converse to test whether a triangle has a right angle. This theorem is one of the most used tools in the syllabus, appearing in mensuration, trigonometry and coordinate geometry.

The answer

The theorem

In a right-angled triangle, the square of the hypotenuse (the longest side, opposite the right angle) equals the sum of the squares of the other two sides:

where $c$ is the hypotenuse and $a$, $b$ are the two legs.

Finding the hypotenuse

When the two legs are known, square each, add, and take the square root:

The hypotenuse is always the longest side, so this answer should exceed each leg.

Finding a shorter side

When the hypotenuse and one leg are known, rearrange by subtracting:

Here you subtract because the unknown is a leg, not the hypotenuse, so the result is smaller than the hypotenuse.

The converse

If the sides of a triangle satisfy $a^2 + b^2 = c^2$ for the longest side $c$, then the triangle is right-angled, with the right angle opposite $c$. This converse is the standard way to test for a right angle from three side lengths.

Pythagoras inside three-dimensional shapes

Pythagoras extends to solids by applying it twice. To find the space diagonal of a cuboid with edges $a$, $b$ and $c$, first find the diagonal of the base, $\sqrt{a^2 + b^2}$, then use that as one leg of a second right triangle whose other leg is the height $c$. This gives the space diagonal $\sqrt{a^2 + b^2 + c^2}$. So a box measuring $3 \times 4 \times 12$ has a space diagonal of $\sqrt{9 + 16 + 144} = \sqrt{169} = 13$. Recognising that a three-dimensional length problem is two flat Pythagoras problems chained together is a frequently tested E-Maths extension.

Pythagorean triples speed up working

A Pythagorean triple is a set of three whole numbers satisfying $a^2 + b^2 = c^2$, such as $(3, 4, 5)$, $(5, 12, 13)$ and $(8, 15, 17)$. Spotting one, or a multiple of one like $(6, 8, 10)$, lets you write down the missing side without a calculator. If a right triangle has legs $9$ and $12$, recognising these as $3 \times (3, 4)$ gives the hypotenuse $3 \times 5 = 15$ instantly. Memorising the common triples and their multiples is a quick-win that saves time and provides a check on a calculated answer.

Examples in context

Example 1. Diagonal of a screen. The diagonal of a rectangular screen is the hypotenuse of a right triangle whose legs are the width and height. Pythagoras theorem converts width and height into the diagonal size quoted for televisions and phones.

Example 2. Shortest distance. The straight-line distance across a rectangular field, corner to corner, is found by Pythagoras from the two side lengths. This is always shorter than walking along two sides, a everyday use of the theorem.

Try this

Q1. A right-angled triangle has legs $9\ \text{cm}$ and $40\ \text{cm}$. Find the hypotenuse. [2 marks]

• Cue. $\sqrt{9^2 + 40^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41\ \text{cm}$.

Q2. The hypotenuse is $25\ \text{cm}$ and one leg is $7\ \text{cm}$. Find the other leg. [2 marks]

• Cue. $\sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24\ \text{cm}$.

Q3. Is a triangle with sides $6\ \text{cm}$, $8\ \text{cm}$ and $11\ \text{cm}$ right-angled? [2 marks]

• Cue. $6^2 + 8^2 = 100$ but $11^2 = 121$; they differ, so it is not right-angled.