Testing Right-Angled Triangles
The converse of Pythagoras' theorem tests whether a triangle is right-angled: if the square of the longest side equals the sum of the squares of the other two, the angle opposite the longest side is \(90^\circ\). This NSW Year 9 Advanced maths subtopic revises the Stage 4 test (NESA outcome MA4‑PYT‑C‑01).
You will test sides given as whole numbers, decimals and surds, name the right angle, and spot a student who compared the wrong squares. You will check a concrete slab and a window frame, decide whether an angle is more or less than \(90^\circ\), prove that every \(3k, 4k, 5k\) triangle is right-angled, and find both possible third sides when two sides are known. It is part of Year 9 Advanced Pythagoras Theorem.
Every question with a fully worked solution.
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Theory
The converse of Pythagoras' theorem tests a triangle: if the squares of the two shorter sides add to the square of the longest side, the triangle is right-angled. This Year 9 Advanced maths page revises the earlier test and extends it to naming the right angle, deciding whether a triangle is acute or obtuse, checking corners in practice and finding both possible third sides.
Pythagoras' theorem says that in a right-angled triangle, \(a^2 + b^2 = c^2\). The converse runs the other way: if a triangle's sides satisfy \(a^2 + b^2 = c^2\), with \(c\) the longest side, then the triangle is right-angled.
The right angle is always opposite the longest side. In \(\triangle PQR\) with longest side \(QR\), the right angle is at \(P\), written \(\angle QPR\).
If the two sides are not equal, the triangle is not right-angled. A close result, such as \(676\) against \(681.21\), still fails. The comparison also says what kind of angle is opposite the longest side: acute if \(c^2\) is smaller than \(a^2 + b^2\), obtuse if it is larger.
When a question says a triangle is right-angled but not which side is the hypotenuse, both possibilities must be checked.
Let \(c\) be the longest side and \(a\), \(b\) the other two.
Acute or obtuse describes the angle opposite \(c\), which is the largest angle.
How to test whether a triangle is right-angled
- Find the longest side. With surds or decimals, compare carefully: \(\sqrt{11} \approx 3.3\) is shorter than \(4\).
- Square the two shorter sides and add them.
- Square the longest side.
- Compare. Equal: right-angled, with the right angle opposite the longest side. Not equal: not right-angled, even if the numbers are close.
- Name the right angle with the vertex that is not on the longest side as the middle letter.
The longest side is \(QR = 6.5\). Square the other two and add.
| \(PQ^2 + PR^2\) | \(=\) | \(3.3^2 + 5.6^2\) |
| \(\) | \(=\) | \(10.89 + 31.36\) |
| \(\) | \(=\) | \(42.25\) |
| \(QR^2\) | \(=\) | \(6.5^2 = 42.25\) |
They are equal, so the triangle is right-angled. The right angle is opposite \(QR\), at the vertex not on \(QR\): it is \(\boldsymbol{\angle QPR}\).
(i) sides \(6\), \(7\) and \(9\)
(ii) sides \(5\), \(7\) and \(9\).
Compare the square of the longest side with the sum of the other two squares.
(i)
| \(6^2 + 7^2\) | \(=\) | \(36 + 49 = 85\) |
| \(9^2\) | \(=\) | \(81\) |
\(81 \lt 85\): the longest side is shorter than it would be for a right angle, so the largest angle is acute.
(ii)
| \(5^2 + 7^2\) | \(=\) | \(25 + 49 = 74\) |
| \(9^2\) | \(=\) | \(81\) |
\(81 > 74\): the longest side is longer than it would be for a right angle, so the largest angle is obtuse.
For a right angle, the diagonal squared must equal \(1.2^2 + 0.9^2\).
| \(1.2^2 + 0.9^2\) | \(=\) | \(1.44 + 0.81 = 2.25\) |
| \(1.48^2\) | \(=\) | \(2.1904\) |
\(2.1904 \ne 2.25\), so the corner is not a right angle. A right-angled corner would need a diagonal of \(\sqrt{2.25} = 1.5\) m. The diagonal is shorter, so the angle between the width and the height is smaller than \(90^\circ\).
The question does not say which side is the hypotenuse, so test both cases.
Case 1: the third side is the hypotenuse.
| \(c^2\) | \(=\) | \(8^2 + 10^2 = 164\) |
| \(c\) | \(=\) | \(\sqrt{164} = \mathbf{2\sqrt{41}}\ \text{cm}\) |
Case 2: the \(10\) cm side is the hypotenuse.
| \(b^2\) | \(=\) | \(10^2 - 8^2 = 36\) |
| \(b\) | \(=\) | \(\mathbf{6}\ \text{cm}\) |
The \(8\) cm side cannot be the hypotenuse, because it is shorter than \(10\) cm. So the third side is \(2\sqrt{41}\) cm or \(6\) cm.
Common pitfalls
Frequently asked questions
How do you prove a triangle is right-angled?
Square the two shorter sides and add them. If the total equals the square of the longest side, the triangle is right-angled by the converse of Pythagoras' theorem.
What is the converse of Pythagoras' theorem?
If the sides of a triangle satisfy \(a^2 + b^2 = c^2\), where \(c\) is the longest side, then the angle opposite \(c\) is a right angle.
Which angle is the right angle?
The right angle is opposite the longest side. It sits at the vertex that is not on the longest side.
How do you tell if a triangle is acute or obtuse from its sides?
Compare the square of the longest side with the sum of the squares of the other two. If it is smaller, the largest angle is acute. If it is larger, the largest angle is obtuse.
How do builders check a right angle?
They mark \(1.5\) m along one edge and \(2\) m along the other, then measure between the marks. If it is exactly \(2.5\) m, the corner is a right angle, because \(1.5^2 + 2^2 = 2.5^2\).