Finding the Hypotenuse
The hypotenuse is the longest side of a right-angled triangle, opposite the right angle. This NSW Year 9 Advanced maths subtopic revises the Stage 4 rule \(c^2 = a^2 + b^2\) (NESA outcome MA4‑PYT‑C‑01) and takes it further with exact surd answers and multi-step problems.
You will find the hypotenuse from two shorter sides, round to one or two decimal places or leave answers such as \(3\sqrt{13}\) in simplest surd form, and work with sides that are surds, like \(2\sqrt{3}\). You will also find the perimeter of a rhombus from its diagonals and the distance between two cyclists, and prove what happens to the hypotenuse when both shorter sides are doubled. It is part of Year 9 Advanced Pythagoras Theorem.
Every question with a fully worked solution.
- Finding the Hypotenuse - Video - Pythagoras Theorem Finding the Hypotenuse Watch
Theory
The hypotenuse is the side opposite the right angle, and Pythagoras' theorem gives it: \(c^2 = a^2 + b^2\). This Year 9 Advanced maths page revises the earlier theorem and takes it further: answers as a surd in simplest form or rounded, sides that are surds or pronumerals, and problems where you first build the right-angled triangle from a journey, a rectangle or a rhombus.
In a right-angled triangle, the side opposite the right angle is the hypotenuse. It is always the longest side. When the triangle is named by its vertices, find the right angle first: in \(\triangle PQR\) with \(\angle PQR = 90^\circ\), the hypotenuse is \(PR\), the side that does not touch \(Q\).
Pythagoras' theorem says the square of the hypotenuse equals the sum of the squares of the other two sides: \(c^2 = a^2 + b^2\). To find the hypotenuse, add the two squares, then take the square root.
If \(c^2\) is not a square number, \(c\) is a surd, such as \(\sqrt{58}\). An exact answer leaves it as a surd, simplified where possible: \(\sqrt{117} = 3\sqrt{13}\). A rounded answer uses the calculator value, rounded only at the end.
Many problems hide the triangle. A diagonal across a rectangle, two journeys at right angles, and the diagonals of a rhombus all make right-angled triangles. The sides can also be surds or expressions such as \(3x\).
With \(c\) the hypotenuse and \(a\) and \(b\) the shorter sides:
Squaring a surd or a pronumeral term, square every factor:
To simplify a surd answer, split off the largest square factor:
How to find the hypotenuse
- Sketch the triangle if none is given, and mark the right angle. In a word problem, find the two lengths that meet at the right angle.
- Combine first if needed: add or subtract legs of a journey that go the same way, halve diagonals, or work out distance = speed × time.
- Square the two shorter sides and add: \(c^2 = a^2 + b^2\). Square surds and bracketed terms in full.
- Square root and give the answer in the form asked: a whole number, a surd in simplest form, or a rounded decimal.
- Answer the actual question. It may want a perimeter, an area or a total length, not the hypotenuse itself.
(i) as a surd in simplest form
(ii) correct to \(2\) decimal places.
The right angle is at \(K\), so the hypotenuse is \(JL\). Add the squares of the two shorter sides.
| \(JL^2\) | \(=\) | \(JK^2 + KL^2\) |
| \(\) | \(=\) | \(6^2 + 9^2\) |
| \(\) | \(=\) | \(36 + 81\) |
| \(\) | \(=\) | \(117\) |
(i) \(117\) is not a square number, but it has the square factor \(9\).
| \(JL\) | \(=\) | \(\sqrt{117}\) |
| \(\) | \(=\) | \(\sqrt{9 \times 13}\) |
| \(\) | \(=\) | \(\sqrt{9} \times \sqrt{13}\) |
| \(\) | \(=\) | \(\mathbf{3\sqrt{13}}\ \text{cm}\) |
(ii) \(\sqrt{117} = 10.8166\ldots\), so \(JL \approx \mathbf{10.82}\) cm. Round only at the very end.
Square each side first. A surd squared gives a whole number: \((2\sqrt{5})^2 = 2^2 \times (\sqrt{5})^2\).
| \((2\sqrt{5})^2\) | \(=\) | \(4 \times 5 = 20\) |
| \(4^2\) | \(=\) | \(16\) |
| \(h^2\) | \(=\) | \(20 + 16\) |
| \(\) | \(=\) | \(36\) |
| \(h\) | \(=\) | \(\sqrt{36}\) |
| \(\) | \(=\) | \(\mathbf{6}\ \text{m}\) |
Watch out: \((2\sqrt{5})^2\) is \(20\), not \(10\). Both the \(2\) and the \(\sqrt{5}\) are squared.
Sketch the walk. Combine the east–west legs first: going west undoes part of the east leg.
| \(\text{east}\) | \(=\) | \(5 - 2 = 3\ \text{km}\) |
| \(\text{north}\) | \(=\) | \(7\ \text{km}\) |
East and north are at right angles, so the straight-line distance \(d\) is a hypotenuse.
| \(d^2\) | \(=\) | \(3^2 + 7^2\) |
| \(\) | \(=\) | \(9 + 49\) |
| \(\) | \(=\) | \(58\) |
| \(d\) | \(=\) | \(\sqrt{58}\) |
| \(\) | \(=\) | \(7.615\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{7.6}\ \text{km}\) |
(i) Show that the hypotenuse is \(x\sqrt{5}\) cm.
(ii) The hypotenuse is \(15\) cm. Find the area of the triangle.
(i) Square each side in brackets: \((2x)^2 = 4x^2\).
| \(c^2\) | \(=\) | \(x^2 + (2x)^2\) |
| \(\) | \(=\) | \(x^2 + 4x^2\) |
| \(\) | \(=\) | \(5x^2\) |
| \(c\) | \(=\) | \(\sqrt{5x^2}\) |
| \(\) | \(=\) | \(\boldsymbol{x\sqrt{5}}\) |
(ii) The area needs \(x^2\), so use \(c^2 = 5x^2\) and skip finding \(x\) itself.
| \(5x^2\) | \(=\) | \(15^2 = 225\) |
| \(x^2\) | \(=\) | \(45\) |
| \(A\) | \(=\) | \(\dfrac{1}{2} \times x \times 2x\) |
| \(\) | \(=\) | \(x^2\) |
| \(\) | \(=\) | \(\mathbf{45}\ \text{cm}^2\) |
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 \(3\) and \(7\): \(9 + 49 = 58\), so the hypotenuse is \(\sqrt{58} \approx 7.6\).
When do you leave the hypotenuse as a surd?
When the question asks for an exact answer. If \(c^2\) is not a square number, write \(c\) as a surd and simplify it, for example \(\sqrt{117} = 3\sqrt{13}\). Use a decimal only when the question asks you to round.
How do you square a surd like 2 root 5?
Square both parts and multiply: \(2^2 = 4\) and \((\sqrt{5})^2 = 5\), so \((2\sqrt{5})^2 = 20\).
How do you find the diagonal of a rectangle?
The diagonal cuts the rectangle into two right-angled triangles. The length and width are the shorter sides and the diagonal is the hypotenuse: \(d = \sqrt{l^2 + w^2}\).
Is the hypotenuse always the longest side?
Yes. \(c^2 = a^2 + b^2\), so \(c^2\) is bigger than each of the other squares, and \(c\) is longer than each shorter side. A hypotenuse answer smaller than a shorter side means a mistake.