Mixed Problems
In Mixed Problems the question no longer says which side is missing, so the first step is a decision: is the unknown the hypotenuse or one of the shorter sides? This NSW Year 9 Core maths subtopic revises Pythagoras' theorem from Stage 4 (NESA outcome MA4‑PYT‑C‑01) by mixing both kinds of calculation.
You will check whether the unknown side faces the right angle, then add or subtract the squares, work with decimals and with statements that have no diagram, give exact answers such as \(\sqrt{80}\), and carry a first answer into a second step, such as the height of an isosceles triangle. Real problems include the diagonal of a tabletop, the ground length of a wheelchair ramp and a yacht's distance from where it started. It is part of Year 9 Core Pythagoras Theorem.
Every question with a fully worked solution.
- Mixed Problems - Video - Pythagorean Theorem Word Problems Watch
Theory
In mixed Pythagoras' theorem problems the question does not say which side is missing. This Year 9 maths page shows how to decide: if the unknown is the hypotenuse, add the squares; if the hypotenuse is given, subtract. It also covers written statements with no diagram and two-step problems such as the height and area of an isosceles triangle.
Every Pythagoras question gives two sides of a right-angled triangle and asks for the third. The skill here is the decision: add or subtract?
Find the right angle, then look at the side it faces. That side is the hypotenuse.
If the unknown side faces the right angle, the unknown is the hypotenuse. Add the squares of the two known sides.
If the side facing the right angle is already known, the unknown is a shorter side. Subtract: hypotenuse squared minus the other side squared.
Some problems take two steps. Use Pythagoras' theorem to find a missing side, then use that side for something else, such as an area or a perimeter. An isosceles triangle often needs this: its height cuts it into two right-angled triangles.
With \(c\) the hypotenuse and \(a\) and \(b\) the shorter sides:
Area of a triangle, using two sides that meet at a right angle as base and height:
How to solve any Pythagoras problem
- Sketch the right-angled triangle if there is no diagram. Mark the right angle and the two known sides.
- Decide: does the unknown face the right angle? Yes: add. No (the hypotenuse is given): subtract.
- Calculate: square, add or subtract, then take the square root.
- Finish any second step, such as an area, a perimeter or a total distance.
- Check: the hypotenuse must be the longest side.
Both known sides touch the right angle. The unknown \(x\) faces it, so \(x\) is the hypotenuse. Add the squares:
| \(x^2\) | \(=\) | \(20^2 + 48^2\) |
| \(\) | \(=\) | \(400 + 2304\) |
| \(\) | \(=\) | \(2704\) |
| \(x\) | \(=\) | \(\sqrt{2704}\) |
| \(\) | \(=\) | \(\mathbf{52}\ \textbf{m}\) |
Check: \(52\) m is longer than both known sides, as a hypotenuse must be.
Sketch it. The wall meets the ground at a right angle. The ladder faces that right angle, so the ladder is the hypotenuse, and it is given. The height is a shorter side. Subtract:
| \(h^2\) | \(=\) | \(7.3^2 - 4.8^2\) |
| \(\) | \(=\) | \(53.29 - 23.04\) |
| \(\) | \(=\) | \(30.25\) |
| \(h\) | \(=\) | \(\sqrt{30.25}\) |
| \(\) | \(=\) | \(\mathbf{5.5}\ \textbf{m}\) |
Check: \(5.5\) m is shorter than the \(7.3\) m ladder.
The right angle is at \(M\), the middle letter. The side that does not touch \(M\) is \(LN\), so \(LN = 8.5\) is the hypotenuse, and it is given. \(MN\) is a shorter side. Subtract:
| \(MN^2\) | \(=\) | \(LN^2 - LM^2\) |
| \(\) | \(=\) | \(8.5^2 - 3.4^2\) |
| \(\) | \(=\) | \(72.25 - 11.56\) |
| \(\) | \(=\) | \(60.69\) |
| \(MN\) | \(=\) | \(\sqrt{60.69}\) |
| \(\) | \(=\) | \(7.7903\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{7.8}\) |
With no diagram, a quick sketch with the right angle at \(M\) makes the decision easy.
(i) its perpendicular height
(ii) its area.
The height cuts the base in half at a right angle, making two right-angled triangles. Work with one of them: half the base is \(10 \div 2 = 5\) cm.
(i) In the half-triangle, the \(13\) cm side faces the right angle, so it is the hypotenuse. Subtract:
| \(h^2\) | \(=\) | \(13^2 - 5^2\) |
| \(\) | \(=\) | \(169 - 25\) |
| \(\) | \(=\) | \(144\) |
| \(h\) | \(=\) | \(\sqrt{144}\) |
| \(\) | \(=\) | \(\mathbf{12}\ \textbf{cm}\) |
(ii) Area of a triangle \(= \dfrac{1}{2} \times \text{base} \times \text{height}\), using the whole base:
| \(A\) | \(=\) | \(\dfrac{1}{2} \times 10 \times 12\) |
| \(\) | \(=\) | \(\mathbf{60}\ \textbf{cm}^\mathbf{2}\) |
Common pitfalls
Frequently asked questions
How do I know whether to add or subtract in Pythagoras' theorem?
Look at the side opposite the right angle. If it is the unknown, add the squares of the other two sides. If it is already known, subtract the square of the other known side from its square.
What if there is no diagram?
Sketch one. A statement such as \(\angle LMN = 90^\circ\) puts the right angle at \(M\), the middle letter. The side that does not touch \(M\), \(LN\), is the hypotenuse.
How do you find the height of an isosceles triangle?
Draw the height from the top to the middle of the base. It cuts the base in half at a right angle. Then subtract: \(h^2 = \text{equal side}^2 - \text{half base}^2\).
How do you find the distance from the start after walking east then north?
East and north are at right angles, so the straight-line distance is the hypotenuse. Add any legs in the same direction first, then add the squares and take the square root.
Can the answer to a Pythagoras question be bigger than the hypotenuse?
No. The hypotenuse is always the longest side. If a shorter side comes out longer than the hypotenuse, you added when you should have subtracted.