Pythagoras Word Problems
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.