Finding the Hypotenuse
When you know the two shorter sides in a right-angled triangle, Pythagoras' theorem gives the longest one: square them, add, then take the square root. This NSW Year 9 Core maths subtopic revises that Stage 4 skill (NESA outcome MA4‑PYT‑C‑01) and adds exact answers written as surds.
You will find the hypotenuse from diagrams, including triangles turned on their side, and from statements such as "the right angle is at \(B\)". You will round answers to one or two decimal places and leave answers exact, simplifying \(\sqrt{98}\) to \(7\sqrt{2}\). Real problems include the length of a ladder against a wall, the brace on a garden gate and a path straight across a rectangular park. It is part of Year 9 Core Pythagoras Theorem.
Every question with a fully worked solution.
- Finding the Hypotenuse - Video - Pythagoras Theorem Finding the Hypotenuse Watch
Theory
To find the hypotenuse of a right-angled triangle, square the two shorter sides, add them, then take the square root. This Year 9 maths page revises Pythagoras' theorem from earlier years and shows how to give the answer as a whole number, a rounded decimal or an exact surd, including the diagonal of a rectangle.
The hypotenuse is the side opposite the right angle. It is the longest side of a right-angled triangle. The other two sides are the shorter sides.
When the unknown side faces the right angle, it is the hypotenuse. Use Pythagoras' theorem as it stands: square the two shorter sides, add them, then take the square root.
The answer can come out three ways. If the total is a perfect square, the answer is a whole number (or an exact decimal). If not, the exact answer is a surd, such as \(\sqrt{50}\). If the question gives a number of decimal places, use a calculator and round.
A surd can often be simplified. Split the number under the root into a perfect square times another number, then take the root of the perfect square: \(\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\).
A diagonal of a rectangle is a hypotenuse. It cuts the rectangle into two right-angled triangles, with the length and the width as the shorter sides.
With \(c\) the hypotenuse and \(a\) and \(b\) the shorter sides:
When the sides are named by vertices, the hypotenuse is the side that does not touch the right-angle vertex. In \(\triangle XYZ\) with the right angle at \(Y\):
To simplify a surd, use the largest perfect square factor:
How to find the hypotenuse
- Find the right angle. The side opposite it is the hypotenuse. With no diagram, sketch one.
- Write the rule: \(c^2 = a^2 + b^2\), with the hypotenuse on its own.
- Substitute the two shorter sides, square each one and add.
- Take the square root. Do not stop at \(c^2\).
- Give the answer in the form asked: whole number, surd (simplified if asked) or rounded decimal, with units.
- Check that the answer is longer than both shorter sides.
The right angle is at \(Q\), so the hypotenuse is \(PR\), the side opposite \(Q\). Sketch the triangle:
Write the rule, substitute the two shorter sides, add, then take the square root:
| \(PR^2\) | \(=\) | \(PQ^2 + QR^2\) |
| \(\) | \(=\) | \(11^2 + 60^2\) |
| \(\) | \(=\) | \(121 + 3600\) |
| \(\) | \(=\) | \(3721\) |
| \(PR\) | \(=\) | \(\sqrt{3721}\) |
| \(\) | \(=\) | \(\mathbf{61}\ \textbf{cm}\) |
\(61\) cm is longer than both other sides, as a hypotenuse must be.
(i) one decimal place
(ii) two decimal places.
The triangle is turned, but the right angle is still the place to start. The side facing it is \(c\), so \(c\) is the hypotenuse. The two sloping sides are the shorter sides.
| \(c^2\) | \(=\) | \(6.1^2 + 5.4^2\) |
| \(\) | \(=\) | \(37.21 + 29.16\) |
| \(\) | \(=\) | \(66.37\) |
| \(c\) | \(=\) | \(\sqrt{66.37}\) |
| \(\) | \(=\) | \(8.1467\ldots\) |
Keep the full calculator value, then round once at the end.
| \(\text{(i) } c\) | \(\approx\) | \(\mathbf{8.1}\ \textbf{m}\) |
| \(\text{(ii) } c\) | \(\approx\) | \(\mathbf{8.15}\ \textbf{m}\) |
For two decimal places, look at the third: \(8.146\ldots\) has a \(6\) there, so round up to \(8.15\).
(i) as a surd
(ii) in simplest form.
| \(c^2\) | \(=\) | \(4^2 + 10^2\) |
| \(\) | \(=\) | \(16 + 100\) |
| \(\) | \(=\) | \(116\) |
| \(c\) | \(=\) | \(\sqrt{116}\) |
(i) \(116\) is not a perfect square (\(10^2 = 100\), \(11^2 = 121\)), so the exact answer keeps the root sign: \(\boldsymbol{c = \sqrt{116}}\) cm.
(ii) To simplify, split \(116\) into a perfect square times another number. \(116 = 4 \times 29\):
| \(c\) | \(=\) | \(\sqrt{4 \times 29}\) |
| \(\) | \(=\) | \(\sqrt{4} \times \sqrt{29}\) |
| \(\) | \(=\) | \(\boldsymbol{2\sqrt{29}}\ \textbf{cm}\) |
\(29\) has no square factor other than \(1\), so \(2\sqrt{29}\) is in simplest form.
The diagonal cuts the pool into two right-angled triangles. The length and the width are the shorter sides, and the diagonal \(d\) is the hypotenuse.
| \(d^2\) | \(=\) | \(25^2 + 10^2\) |
| \(\) | \(=\) | \(625 + 100\) |
| \(\) | \(=\) | \(725\) |
| \(d\) | \(=\) | \(\sqrt{725}\) |
| \(\) | \(=\) | \(26.9258\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{26.9}\ \textbf{m}\) |
Check: the diagonal is longer than the \(25\) m side, as it should be. To the nearest metre it is \(27\) m. A rope along both diagonals would need twice this length.
Common pitfalls
Frequently asked questions
How do you find the hypotenuse of a right-angled triangle?
Square the two shorter sides, add them, then take the square root. For shorter sides \(11\) and \(60\): \(121 + 3600 = 3721\) and \(\sqrt{3721} = 61\).
What is the formula for the hypotenuse?
\(c = \sqrt{a^2 + b^2}\), where \(c\) is the hypotenuse and \(a\) and \(b\) are the shorter sides.
Why is my hypotenuse answer smaller than one of the sides?
Something has gone wrong. The hypotenuse is always the longest side. Check that you added the squares rather than subtracting, and that you took the square root at the end.
What does exact answer mean in Pythagoras questions?
It means do not round. If the total under the square root is not a perfect square, leave the answer as a surd, such as \(\sqrt{116}\), or simplify it to \(2\sqrt{29}\).
How do you find the diagonal of a rectangle?
The diagonal is the hypotenuse of a right-angled triangle whose shorter sides are the length and the width. Square both, add, and take the square root.