Finding the Shorter Side
To find either of the two shorter sides in a right-angled triangle, Pythagoras' theorem is rearranged so that you subtract: \(a^2 = c^2 - b^2\). This NSW Year 9 Core maths subtopic revises Stage 4 work under NESA outcome MA4‑PYT‑C‑01, with more statements to decode and answers left in exact form.
You will decide which given length is the hypotenuse, from a diagram or from a statement such as \(\angle ABC = 90^\circ\), subtract to find the missing side, round to one or two decimal places, and simplify an exact answer such as \(6\sqrt{2}\). Problems include how far up a wall a ladder reaches, the height of a farm gate from its diagonal brace and the perimeter of a rectangle from its diagonal. It is part of Year 9 Core Pythagoras Theorem.
Every question with a fully worked solution.
- Finding the Shorter Side - Video - Pythagoras Theorem Finding Shorter Sides Watch
Theory
To find a shorter side of a right-angled triangle, square the hypotenuse, subtract the square of the known shorter side, then take the square root. This Year 9 maths page revises Pythagoras' theorem from earlier years: finding which side is the hypotenuse, working from a written statement, and giving the answer rounded or as an exact surd.
The two sides that meet at the right angle are the shorter sides. The side opposite the right angle is the hypotenuse, the longest side.
When the hypotenuse is given and a shorter side is unknown, rearrange Pythagoras' theorem. Square the hypotenuse, subtract the square of the known shorter side, then take the square root.
Always find the hypotenuse first. Look for the right angle: the hypotenuse is the side that does not touch it. With no diagram, a statement such as \(\angle ABC = 90^\circ\) tells you the right angle is at the middle letter, \(B\).
A shorter side must come out smaller than the hypotenuse. That is a quick check on every answer.
Shorter sides turn up in real life too: how high a ladder reaches up a wall, how tall a pole is from its guy rope, and how wide a rectangle is from its diagonal.
Start from \(c^2 = a^2 + b^2\), with \(c\) the hypotenuse. Subtract \(b^2\) from both sides:
The hypotenuse always goes first. In \(\triangle ABC\) with \(\angle ABC = 90^\circ\), the hypotenuse is \(AC\):
How to find a shorter side
- Find the right angle and name the hypotenuse, the side opposite it. With no diagram, sketch one.
- Write the rearranged rule with the hypotenuse first: \(a^2 = c^2 - b^2\).
- Substitute, square each number and subtract.
- Take the square root. Give it as a whole number, a surd or a rounded decimal, as the question asks.
- Check that the answer is shorter than the hypotenuse.
Find the right angle first. The side opposite it is \(25\) cm, so \(25\) cm is the hypotenuse. \(x\) touches the right angle, so it is a shorter side. Subtract, with the hypotenuse first:
| \(x^2\) | \(=\) | \(25^2 - 7^2\) |
| \(\) | \(=\) | \(625 - 49\) |
| \(\) | \(=\) | \(576\) |
| \(x\) | \(=\) | \(\sqrt{576}\) |
| \(\) | \(=\) | \(\mathbf{24}\ \textbf{cm}\) |
Check: \(24\) cm is shorter than the \(25\) cm hypotenuse, as it must be.
(i) one decimal place
(ii) two decimal places.
The middle letter of \(\angle PQR\) is \(Q\), so the right angle is at \(Q\). The hypotenuse is \(PR\), the side that does not touch \(Q\). Sketch it:
| \(QR^2\) | \(=\) | \(PR^2 - PQ^2\) |
| \(\) | \(=\) | \(9.4^2 - 4.1^2\) |
| \(\) | \(=\) | \(88.36 - 16.81\) |
| \(\) | \(=\) | \(71.55\) |
| \(QR\) | \(=\) | \(\sqrt{71.55}\) |
| \(\) | \(=\) | \(8.4587\ldots\) |
| \(\text{(i) } QR\) | \(\approx\) | \(\mathbf{8.5}\) |
| \(\text{(ii) } QR\) | \(\approx\) | \(\mathbf{8.46}\) |
| \(a^2\) | \(=\) | \(12^2 - 6^2\) |
| \(\) | \(=\) | \(144 - 36\) |
| \(\) | \(=\) | \(108\) |
| \(a\) | \(=\) | \(\sqrt{108}\) |
\(108\) is not a perfect square, so keep the root and simplify. The largest perfect square that divides \(108\) is \(36\):
| \(a\) | \(=\) | \(\sqrt{36 \times 3}\) |
| \(\) | \(=\) | \(\sqrt{36} \times \sqrt{3}\) |
| \(\) | \(=\) | \(\boldsymbol{6\sqrt{3}}\ \textbf{cm}\) |
(i) Find its width.
(ii) Find its perimeter.
The diagonal cuts the rectangle into two right-angled triangles. The diagonal is the hypotenuse, and the width \(w\) is a shorter side.
(i) Subtract, then take the square root:
| \(w^2\) | \(=\) | \(1.7^2 - 1.5^2\) |
| \(\) | \(=\) | \(2.89 - 2.25\) |
| \(\) | \(=\) | \(0.64\) |
| \(w\) | \(=\) | \(\sqrt{0.64}\) |
| \(\) | \(=\) | \(\mathbf{0.8}\ \textbf{m}\) |
(ii) The perimeter is the distance all the way round: two lengths and two widths.
| \(P\) | \(=\) | \(2 \times 1.5 + 2 \times 0.8\) |
| \(\) | \(=\) | \(3 + 1.6\) |
| \(\) | \(=\) | \(\mathbf{4.6}\ \textbf{m}\) |
Common pitfalls
Frequently asked questions
How do you find a shorter side using Pythagoras' theorem?
Square the hypotenuse, subtract the square of the known shorter side, then take the square root. For a hypotenuse of \(25\) and a side of \(7\): \(625 - 49 = 576\) and \(\sqrt{576} = 24\).
When do you add and when do you subtract in Pythagoras' theorem?
Add the squares when you want the hypotenuse. Subtract when you want a shorter side and the hypotenuse is given.
How do you know which side is the hypotenuse?
It is the side opposite the right angle, the one side that does not touch it. If \(\angle ABC = 90^\circ\), the right angle is at \(B\) and the hypotenuse is \(AC\).
Why did I get a negative number under the square root?
You subtracted the wrong way round. The hypotenuse is the longest side, so its square must come first: \(c^2 - b^2\).
How do you find the width of a rectangle from its diagonal?
The diagonal is the hypotenuse of a right-angled triangle with the length and the width as the shorter sides. Square the diagonal, subtract the square of the length, and take the square root.