Pythagoras Word Problems
Pythagoras Word Problems turns real situations into right-angled triangles in NSW Year 8 maths (NESA outcome MA4‑PYT‑C‑01). The first step is always a sketch: mark the right angle, label the lengths in one unit and spot the sloping length, which is the hypotenuse.
You will solve problems about ladders, kites, ramps, handrails, TV screens and tents, find the diagonal of a soccer field or a rhombus tile, and work out the distance between two points on a grid. Some problems need an area first, a second triangle or a side written as \(3x\). You will round only at the end and finish with a sentence that answers the question. It is part of Year 8 Pythagoras.
Theory
A word problem becomes a Pythagoras question the moment you draw the triangle and mark the right angle. The sloping length (a ladder, cable, brace, diagonal or straight-line distance) is the hypotenuse. This Year 8 maths page shows how to set up and solve worded problems and answer them in a sentence.
Draw and label. Sketch the situation and mark the right angle. A wall meets level ground at a right angle; so do north and east, and so do the sides of a rectangle or square. Put every length in the same unit.
Name the hypotenuse. The sloping length (a ladder, a rope, a cable, a brace, a diagonal, a straight-line distance) is opposite the right angle, so it is the hypotenuse.
Add or subtract. Hypotenuse wanted: \(c^2 = a^2 + b^2\). Shorter side wanted: \(a^2 = c^2 - b^2\). Take the square root last.
Round at the end, to the places asked for. If a later step uses the answer, keep the full value, or its square.
Another formula may come first. A perimeter can give a side of a square (\(44 \div 4 = 11\)). The area of a triangle, \(\dfrac{1}{2} \times \text{base} \times \text{height}\), can give a missing height. Then Pythagoras’ theorem gives the third side. In a rhombus or kite, the diagonals cross at right angles; in a rhombus they also cut each other in half.
On a grid, the distance across and the distance up or down are the two shorter sides, and the straight line joining the points is the hypotenuse.
Sides with a pronumeral? Square each one: \((2x)^2 = 4x^2\), so \((2x)^2 + x^2 = 5x^2\).
Finish with a sentence that answers the question, with its unit. If it asks whether something fits, compare your answer with the limit.
Setting up a word problem:
| In the words | In the triangle |
|---|---|
| ladder, rope, cable, zip line, brace, diagonal, straight-line distance | the hypotenuse |
| height up a wall, distance along the ground | the two shorter sides |
| north then east, across then up | the two shorter sides |
| the diagonal of a rectangle or square | \(d^2 = \text{length}^2 + \text{width}^2\) |
How to solve a Pythagoras word problem
- Draw a diagram and mark the right angle.
- Label the lengths in one unit and the unknown with a letter.
- Decide: hypotenuse (add) or shorter side (subtract).
- Work out and take the square root; round at the end.
- Answer in a sentence with the unit.
Each side is \(44 \div 4 = 11\) cm. The diagonal is the hypotenuse.
| \(d^2\) | \(=\) | \(11^2 + 11^2 = 242\) |
| \(d\) | \(=\) | \(\sqrt{242} = 15.5563\ldots\) |
The diagonal is about \(15.6\) cm.
(i) Find \(AB\).
(ii) Find \(AC\).
| \(\dfrac{1}{2} \times 3 \times AB\) | \(=\) | \(6\) |
| \(AB\) | \(=\) | \(4\) |
| \(AC^2\) | \(=\) | \(4^2 + 3^2 = 25\) |
(i) \(3 \times AB = 12\), so \(AB = 4\) cm
(ii) \(AC = \sqrt{25} = 5\) cm
| \(b^2\) | \(=\) | \(1.6^2 + 1.2^2\) |
| \(\) | \(=\) | \(2.56 + 1.44 = 4\) |
| \(b\) | \(=\) | \(\sqrt{4} = 2\) |
The brace is \(2\) m long.
The zip line is the hypotenuse, so the height is a shorter side: subtract.
| \(h^2\) | \(=\) | \(30^2 - 28^2 = 900 - 784 = 116\) |
| \(h\) | \(=\) | \(\sqrt{116} = 10.7703\ldots\) |
The platform is about \(10.77\) m high.
| \(d^2\) | \(=\) | \(3.6^2 + 1.8^2\) |
| \(\) | \(=\) | \(12.96 + 3.24 = 16.2\) |
| \(d\) | \(=\) | \(\sqrt{16.2} = 4.0249\ldots\) |
She is about \(4.02\) km from her start.
(i) the exact height
(ii) the area, correct to \(1\) decimal place.
Half the base is \(2.5\) m and the \(3.5\) m edge is the hypotenuse: subtract.
| \(h^2\) | \(=\) | \(3.5^2 - 2.5^2\) |
| \(\) | \(=\) | \(12.25 - 6.25 = 6\) |
| \(A\) | \(=\) | \(\dfrac{1}{2} \times 5 \times 2.4494\ldots\) |
| \(\) | \(=\) | \(6.1237\ldots\) |
(i) \(h = \sqrt{6}\) m
(ii) Carry the full value of \(\sqrt{6}\) and round last: about \(6.1\) m\(^2\).
(ii) A rhombus has diagonals \(24\) cm and \(10\) cm. Find its perimeter.
(iii) A rectangle is \(x\) m wide and \(2x\) m long, with a \(15\) m diagonal. Find its area.
| \(PQ^2\) | \(=\) | \(6^2 + 3^2 = 45\) |
| \(s^2\) | \(=\) | \(12^2 + 5^2 = 169\) |
| \((2x)^2 + x^2\) | \(=\) | \(15^2\) |
| \(5x^2\) | \(=\) | \(225\) |
(i) \(\sqrt{45}\) units
(ii) The diagonals cut each other in half at right angles: \(12\) and \(5\), side \(13\) cm, perimeter \(4 \times 13 = 52\) cm
(iii) \(x^2 = 45\), so the area is \(x \times 2x = 2x^2 = 90\) m\(^2\)
| \(h^2\) | \(=\) | \(100^2 - 87^2 = 2431\) |
| \(h\) | \(=\) | \(\sqrt{2431} = 49.30\ldots\) |
The TV is about \(49.31\) cm high, which is less than \(50\) cm, so it fits.
Common pitfalls
Frequently asked questions
How do I start a Pythagoras word problem?
Draw a diagram, mark the right angle and label the lengths.
Which side is the hypotenuse in a ladder problem?
The ladder: it is opposite the right angle between the wall and the ground.
How do you find the distance between two points on a grid?
Use the distance across and the distance up as the shorter sides: from \((1, 1)\) to \((4, 5)\) is \(\sqrt{3^2 + 4^2} = 5\).
How do you find the diagonal of a square?
Use \(d^2 = s^2 + s^2\). A square with sides \(11\) cm has \(d = \sqrt{242} \approx 15.6\) cm.
When should I round?
Only at the end, to the number of places the question asks for.