Finding the Hypotenuse
The hypotenuse is the side across from the right angle, and it is always the longest side of a right-angled triangle. In NSW Year 8 maths, Pythagoras' theorem \(c^2 = a^2 + b^2\) is used to find it from the two shorter sides (NESA outcome MA4‑PYT‑C‑01).
You will name the hypotenuse in a turned triangle, square and add the shorter sides, then take the square root, giving answers as whole numbers, surds such as \(\sqrt{149}\) or decimals rounded to one place. You will change units first and spot a common mistake. A gate brace, a bushwalk east then north and the sloping side of a trapezium are all found this way. It is part of Year 8 Pythagoras.
Theory
The hypotenuse is the side opposite the right angle; it is always the longest side of a right-angled triangle. Pythagoras’ theorem says \(c^2 = a^2 + b^2\), where \(c\) is the hypotenuse. This Year 8 maths page shows how to find the hypotenuse as an exact answer, a surd or a rounded decimal.
Find the right angle first. The side that does not touch it is the hypotenuse, however the triangle is turned. It is the longest side.
Pythagoras’ theorem: the area of the square drawn on the hypotenuse equals the sum of the areas of the squares on the other two sides. With \(c\) as the hypotenuse, \(c^2 = a^2 + b^2\).
To find the hypotenuse: write \(c^2 = a^2 + b^2\), substitute the two shorter sides, add, then take the square root. A length is positive, so only the positive root is used.
Answer forms. If the root works out exactly (a whole number, or a decimal that stops) that is the exact answer. Otherwise the exact answer keeps the root sign, such as \(\sqrt{61}\), a surd; a rounded decimal is an approximation.
Estimate first. The hypotenuse is longer than each of the other two sides and shorter than the two added. Same units: change every length to one unit before you square. Scaling: sides \(15\) and \(20\) are \(5\) times \(3\) and \(4\), so the hypotenuse is \(5 \times 5 = 25\).
Not a triangle? Draw one inside the shape: a diagonal splits a rectangle in two; a line from a top corner of a right trapezium leaves a rectangle and a right-angled triangle. Where two right-angled triangles are joined, find the shared side in the first, then use it in the second.
Pythagoras' theorem (c is the hypotenuse):
| Pythagorean triad | Check |
|---|---|
| \(3, 4, 5\) | \(9 + 16 = 25\) |
| \(5, 12, 13\) | \(25 + 144 = 169\) |
| \(8, 15, 17\) | \(64 + 225 = 289\) |
| \(7, 24, 25\) | \(49 + 576 = 625\) |
How to find the hypotenuse
- Find the right angle; the side opposite it is \(c\).
- Change all lengths to the same unit.
- Write \(c^2 = a^2 + b^2\) and substitute.
- Add, then take the square root (exact, surd or rounded).
- Check: \(c\) is longer than each shorter side.
| \(c^2\) | \(=\) | \(21^2 + 28^2\) |
| \(\) | \(=\) | \(441 + 784\) |
| \(\) | \(=\) | \(1225\) |
| \(c\) | \(=\) | \(\sqrt{1225} = 35\) |
\(35\) cm. Check by scaling: \(21 = 3 \times 7\) and \(28 = 4 \times 7\), so \(c = 5 \times 7 = 35\).
| \(x^2\) | \(=\) | \(12^2 + 5^2\) |
| \(\) | \(=\) | \(144 + 25 = 169\) |
| \(x\) | \(=\) | \(\sqrt{169} = 13\) |
\(x = 13\) cm
| \(x^2\) | \(=\) | \(5^2 + 6^2\) |
| \(\) | \(=\) | \(25 + 36 = 61\) |
| \(x\) | \(=\) | \(\sqrt{61}\) |
\(x = \sqrt{61} = 7.81\ldots \approx 7.8\) cm
| \(PR^2\) | \(=\) | \(5^2 + 8^2\) |
| \(\) | \(=\) | \(25 + 64 = 89\) |
\(81 < 89 < 100\), so \(89\) is not a square number: \(PR = \sqrt{89}\) cm (about \(9.4\) cm).
| \(h^2\) | \(=\) | \(9^2 + 14^2\) |
| \(\) | \(=\) | \(81 + 196 = 277\) |
| \(h\) | \(=\) | \(\sqrt{277} = 16.6433\ldots\) |
Estimate: between \(14\) m and \(23\) m. \(h \approx 16.6\) m, which fits.
(i) Find \(AC\).
(ii) Find \(AD\) and the perimeter of \(ABCD\).
| \(AC^2\) | \(=\) | \(24^2 + 32^2\) |
| \(\) | \(=\) | \(1600\) |
| \(AD^2\) | \(=\) | \(40^2 + 9^2\) |
| \(\) | \(=\) | \(1681\) |
| \(24 + 32 + 9 + 41\) | \(=\) | \(106\) |
(i) \(AC = 40\) cm
(ii) \(AD = 41\) cm; perimeter \(106\) cm
(ii) A right trapezium has parallel sides \(21\) cm and \(14\) cm and height \(24\) cm. Find its slanted side.
(i) Same units: \(1.5\) m \(= 150\) cm.
(ii) A line down from the top corner leaves a right-angled triangle with sides \(21 - 14 = 7\) cm and \(24\) cm.
| \(\sqrt{150^2 + 80^2}\) | \(=\) | \(\sqrt{28\,900} = 170\) |
| \(\sqrt{7^2 + 24^2}\) | \(=\) | \(\sqrt{625} = 25\) |
(i) \(170\) cm
(ii) \(25\) cm
Common pitfalls
Frequently asked questions
What is the hypotenuse?
The side opposite the right angle in a right-angled triangle. It is the longest side.
What is Pythagoras' theorem?
In a right-angled triangle, \(c^2 = a^2 + b^2\), where \(c\) is the hypotenuse.
How do you find the hypotenuse?
Square the two shorter sides, add, then take the square root: \(\sqrt{5^2 + 12^2} = 13\).
When do you leave the answer as a surd?
When the question asks for an exact answer and the number under the root is not a square number, such as \(\sqrt{61}\).
How can you check your answer?
The hypotenuse must be longer than each of the other sides and shorter than the two added together.