Pythagoras Theorem Problems
Pythagoras Theorem Problems turns real situations into right-angled triangles, and the first step is always a labelled sketch. This NSW Year 9 Advanced maths subtopic builds on the Stage 4 theorem (NESA outcome MA4‑PYT‑C‑01) and moves on to problems in three dimensions.
You will solve problems about a tree, a playground slide, a TV screen, a paddock fence and a ladder that slips, find the distance between two points on a grid, and find the side of a square from its diagonal. In solids you will find the slant height of a pyramid, the height of a cone, the space diagonal of a cube and the shortest walk across a box. It is part of Year 9 Advanced Pythagoras Theorem.
Every question with a fully worked solution.
- Pythagoras Theorem Problems - Video - Find the Hypotenuse: Pythagoras Theorem Problem Watch
Theory
Harder Pythagoras' theorem problems hide the right-angled triangle, or need two of them. This Year 9 Advanced maths page revises the earlier theorem in multi-step contexts: two triangles sharing a side, sliding ladders, the space diagonal of a box, distances on a coordinate grid, cones and pyramids, and equations where a side contains \(x\).
A Pythagoras problem is any problem where a right-angled triangle can be found and two of its sides are known. The skill is spotting the triangle: walls meet the ground at right angles, north meets east, a height meets a base, and the edges of a box meet at every corner.
Many problems need two right-angled triangles joined along a shared side. The shared side is the hypotenuse of one and a shorter side of the other. Find its square in the first triangle and use that square in the second.
In three dimensions, the space diagonal of a box runs from a corner to the opposite corner. It is the hypotenuse of a triangle whose shorter sides are a face diagonal and an edge.
On a coordinate grid, the distance between two points is the hypotenuse of a triangle whose shorter sides are the horizontal and vertical steps between them.
Distance between two points \((x_1, y_1)\) and \((x_2, y_2)\):
Diagonals of a square of side \(s\), a cube of edge \(a\), and a rectangular prism \(l \times w \times h\):
How to solve a Pythagoras problem
- Sketch the situation and mark every right angle: walls and ground, north and east, heights, diagonals of rectangles, edges of boxes.
- Prepare the lengths: halve diameters and bases, add or subtract journey legs, find the steps between coordinates.
- Choose the triangle that contains the unknown. If it needs a side from another triangle, find that side squared first and keep it exact.
- Apply Pythagoras: add the squares for a hypotenuse, subtract for a shorter side, or form an equation if a side contains \(x\).
- Finish the question: a perimeter, area, total length, change in distance or rounded answer with units.
\(PR\) is the hypotenuse of \(\triangle PQR\) and a shorter side of \(\triangle PRS\). Find \(PR^2\) and keep it exact.
| \(PR^2\) | \(=\) | \(4^2 + 7^2 = 65\) |
| \(PS^2\) | \(=\) | \(PR^2 + RS^2\) |
| \(\) | \(=\) | \(65 + 6^2 = 101\) |
| \(PS\) | \(=\) | \(\sqrt{101} = 10.049\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{10.05}\) |
\(PR = \sqrt{65}\) is never worked out. Using \(PR^2 = 65\) directly avoids a rounding error.
The ladder is the hypotenuse in both positions. Find the distance of the foot from the wall before and after.
| \(\text{before: } b_1^2\) | \(=\) | \(6.5^2 - 6^2 = 6.25\) |
| \(b_1\) | \(=\) | \(2.5\ \text{m}\) |
| \(\text{after: } b_2^2\) | \(=\) | \(6.5^2 - 5.6^2 = 10.89\) |
| \(b_2\) | \(=\) | \(3.3\ \text{m}\) |
| \(d\) | \(=\) | \(3.3 - 2.5 = \mathbf{0.8}\ \text{m}\) |
The new height is \(6 - 0.4 = 5.6\) m. The foot moves \(0.8\) m, twice as far as the top slides, so the change is not the same at both ends.
The longest rod runs along the space diagonal. First find the diagonal of the base, then use it with the height.
| \(\text{base diagonal}^2\) | \(=\) | \(12^2 + 4^2 = 160\) |
| \(\text{space diagonal}^2\) | \(=\) | \(160 + 3^2\) |
| \(\) | \(=\) | \(169\) |
| \(\text{space diagonal}\) | \(=\) | \(\mathbf{13}\ \text{cm}\) |
The vertical edge meets the base diagonal at right angles, so they are the shorter sides of the second triangle.
For each side, the horizontal and vertical steps are the shorter sides of a right-angled triangle.
| \(PQ^2\) | \(=\) | \(2^2 + 4^2 = 20\) |
| \(QR^2\) | \(=\) | \(4^2 + 2^2 = 20\) |
| \(PR^2\) | \(=\) | \(6^2 + 2^2 = 40\) |
\(PQ = QR\), so it is isosceles. \(PQ^2 + QR^2 = 40 = PR^2\), so the angle at \(Q\) is a right angle.
| \(A\) | \(=\) | \(\dfrac{1}{2} \times PQ \times QR\) |
| \(\) | \(=\) | \(\dfrac{1}{2} \times \sqrt{20} \times \sqrt{20}\) |
| \(\) | \(=\) | \(\mathbf{10}\ \text{square units}\) |
Common pitfalls
Frequently asked questions
How do you solve a Pythagoras word problem?
Sketch the situation and find the right angle. Decide which side is the hypotenuse, prepare the lengths, then add the squares for the hypotenuse or subtract for a shorter side. Finish by answering what the question asks.
How do you find the diagonal of a box?
Find the diagonal of the base first, then use it with the height. In one step, \(D^2 = l^2 + w^2 + h^2\).
How do you find the distance between two points using Pythagoras?
Find the horizontal step and the vertical step between the points. These are the shorter sides, so the distance is \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
How do you find the height of a cone?
Halve the diameter to get the radius. The slant height is the hypotenuse, so \(h^2 = (\text{slant height})^2 - r^2\).
What do you do when a problem has two right-angled triangles?
Find the side they share using the first triangle, keeping its square exact, then use it in the second triangle to find the unknown.