Pythagoras Theorem Problems
This subtopic turns worded situations into right-angled triangles. In NSW Year 9 Core maths the step from the words to a labelled sketch is the real skill; Pythagoras' theorem then does the rest. It revises and extends the Stage 4 work under NESA outcome MA4‑PYT‑C‑01.
You will sketch and label each situation, recognise that the slanted object (a ladder, cable or slide) is the hypotenuse, solve journeys that go east and then north, find the diagonals of rectangles and squares, and work back to a perimeter or an area. You will also find the distance between two points on a map grid, the slant height of a square pyramid and how far down a wall a ladder slips when its base is pulled out. It is part of Year 9 Core Pythagoras Theorem.
Every question with a fully worked solution.
- Pythagoras Theorem Problems - Video - Pythagoras' Theorem Problem Watch
Theory
Most Pythagoras' theorem problems are written in words. This Year 9 maths page shows how to turn ladders, cables, journeys, rectangles, squares, grids and pyramids into a labelled right-angled triangle, decide whether to add or subtract, and carry an answer into a second step.
A worded problem becomes easy once you have the triangle. Sketch it, mark the right angle, write on the two lengths you know, and put a letter on the one you want.
The right angle is usually where something vertical meets something horizontal: a wall and the ground, a pole and the ground, north and east, or two edges of a rectangle.
The thing that leans across the right angle is the hypotenuse: a ladder, a cable, a ramp, a slide, a diagonal path or a straight-line distance.
Some problems take more than one step. You may need a side first and then a perimeter or an area, or one triangle first and then a second triangle that shares a side with it.
Answer the question that was asked, with units, and round only at the end.
The two forms of the theorem, with \(c\) the hypotenuse:
A square of side \(s\) and diagonal \(d\) has two equal shorter sides:
The distance between two points on a grid: count across and up, then use the theorem:
How to solve a worded Pythagoras problem
- Sketch the situation as a triangle. Nothing has to be to scale.
- Find the right angle: wall and ground, pole and ground, north and east, or the corner of a rectangle.
- Label the two known lengths and the unknown. The side that leans across the right angle is the hypotenuse.
- Add or subtract, then take the square root. In a two-triangle problem, keep the first answer exact (or as a square) for the second triangle.
- Finish the question: a perimeter, an area, a total length or a change, with units.
The diagonal cuts the square into two right-angled triangles. Both shorter sides are the side \(s\) of the square, so the unknown appears twice:
| \(s^2 + s^2\) | \(=\) | \(30^2\) |
| \(2s^2\) | \(=\) | \(900\) |
| \(s^2\) | \(=\) | \(450\) |
| \(s\) | \(=\) | \(\sqrt{450}\) |
| \(\) | \(=\) | \(21.2132\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{21.2}\ \textbf{cm}\) |
Collect \(s^2 + s^2 = 2s^2\), then divide by \(2\) before taking the root. Going the other way, a square of side \(s\) has diagonal \(\sqrt{s^2 + s^2}\).
Work through the triangles one at a time. \(AC\) belongs to both: it is the hypotenuse of \(\triangle ABC\) and a shorter side of \(\triangle ACD\).
In \(\triangle ABC\), add:
| \(AC^2\) | \(=\) | \(4^2 + 6^2\) |
| \(\) | \(=\) | \(16 + 36\) |
| \(\) | \(=\) | \(52\) |
Keep \(AC^2 = 52\). There is no need to find \(AC\) and round it. In \(\triangle ACD\), \(AD\) faces the right angle at \(C\), so add again:
| \(AD^2\) | \(=\) | \(AC^2 + CD^2\) |
| \(\) | \(=\) | \(52 + 3^2\) |
| \(\) | \(=\) | \(61\) |
| \(AD\) | \(=\) | \(\sqrt{61}\) |
| \(\) | \(\approx\) | \(\mathbf{7.8}\) |
(i) exactly
(ii) correct to one decimal place.
Draw a right-angled triangle under the flight path: across from \(P\), then up to \(Q\).
Across: from \(x = -2\) to \(x = 4\) is \(6\) units. Up: from \(y = -1\) to \(y = 2\) is \(3\) units. The flight path is the hypotenuse:
| \(PQ^2\) | \(=\) | \(6^2 + 3^2\) |
| \(\) | \(=\) | \(36 + 9\) |
| \(\) | \(=\) | \(45\) |
| \(\text{(i) } PQ\) | \(=\) | \(\boldsymbol{\sqrt{45}}\ \textbf{km}\) |
| \(\text{(ii) } PQ\) | \(\approx\) | \(\mathbf{6.7}\ \textbf{km}\) |
Count the units with care across zero: from \(-2\) to \(4\) is \(2 + 4 = 6\).
The ladder does not change length, so \(8.5\) m is the hypotenuse both times.
Step 1. How far out is the foot at the start? Subtract:
| \(b^2\) | \(=\) | \(8.5^2 - 7.5^2\) |
| \(\) | \(=\) | \(72.25 - 56.25\) |
| \(\) | \(=\) | \(16\) |
| \(b\) | \(=\) | \(4\ \text{m}\) |
Step 2. The foot moves out to \(4 + 1.1 = 5.1\) m. Subtract again for the new height:
| \(h^2\) | \(=\) | \(8.5^2 - 5.1^2\) |
| \(\) | \(=\) | \(72.25 - 26.01\) |
| \(\) | \(=\) | \(46.24\) |
| \(h\) | \(=\) | \(\sqrt{46.24}\) |
| \(\) | \(=\) | \(\mathbf{6.8}\ \textbf{m}\) |
The top slides down \(7.5 - 6.8 = 0.7\) m.
Common pitfalls
Frequently asked questions
How do you solve Pythagoras word problems?
Sketch a right-angled triangle, mark the right angle, label the two known sides and the unknown. If the unknown leans across the right angle it is the hypotenuse, so add the squares; otherwise subtract.
How do you find the diagonal of a square?
The diagonal is the hypotenuse of a right-angled triangle with two equal sides. Square the side, double it, and take the square root. For a side of \(10\) cm the diagonal is \(\sqrt{200} \approx 14.1\) cm.
How do you find the distance between two points on a grid?
Count how far across and how far up you go from one point to the other. These are the shorter sides of a right-angled triangle. Square them, add, and take the square root.
How do you find the slant height of a pyramid?
Use the right-angled triangle inside the pyramid made by the height, half a base side and the slant height. The slant height is the hypotenuse, so add the squares of the height and the half side, then take the square root.
What do you do when a problem has two right-angled triangles?
Solve the triangle with two known sides first. Its answer is usually a side of the second triangle. Keep it exact, then use it to solve the second triangle.