Pythagoras' Theorem
Pythagoras' theorem provides a fundamental relationship between the side lengths of any right-angled triangle. For the ESAT, you must be able to apply this theorem rapidly in both 2D and 3D contexts, often in multi-step problems involving surds.
Part of the ESAT Mathematics 1 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- The theorem only works for right-angled triangles. The side opposite the right angle is called the hypotenuse and is always the longest side, 'c'.
- The formula can be rearranged to find a shorter side: a2 = c2 - b2. Be careful to subtract, not add.
- The distance between two coordinates (x1, y1) and (x2, y2) can be found by forming a right-angled triangle and using the formula d2 = (change in x)2 + (change in y)2.
- For 3D shapes like cuboids, the theorem can be extended to find the space diagonal: d2 = x2 + y2 + z2, where x, y, and z are the dimensions.
- Complex problems often require you to identify or construct a right-angled triangle within a larger shape (e.g., a pyramid or trapezium) and apply the theorem twice.
Diagram
Formulae
a2 + b2 = c2 To find a missing side length in a 2D right-angled triangle. 'a' and 'b' are the shorter sides adjacent to the right angle, and 'c' is the hypotenuse.
d2 = x2 + y2 + z2 To find the length of the longest diagonal ('space diagonal') passing through the interior of a cuboid with side lengths x, y, and z.
Definitions
- Hypotenuse
- The longest side of a right-angled triangle, which is always the side opposite the right angle.
- Pythagorean Triple
- A set of three positive integers (a, b, c) that satisfy the theorem a2 + b2 = c2. Common examples are (3, 4, 5) and (5, 12, 13).
Worked example
A square-based pyramid has a base with side length 10 cm and a vertical height of 12 cm. What is the length of one of its slant edges (the distance from a top corner to a bottom corner)?
- 1
First, find the length of the diagonal of the square base.
Using Pythagoras on the base:
dbase2 = 102 + 102 = 100 + 100 = 200 - 2
The vertical height meets the base at its centre.
The distance from the centre to a corner is half the length of the base diagonal.
Halfdbase2 = (√(200)/2)2 = 200 / 4 = 50.
- 3
Now, form a new right-angled triangle inside the pyramid.
The sides are the vertical height (12 cm), the half-diagonal (length2 = 50), and the slant edge is the hypotenuse (s).
- 4
Apply Pythagoras' theorem again:
s2 = height2 + (halfdbase)2 - 5 s2 = 122 + 50 = 144 + 50 = 194
- 6
The length of the slant edge is √(194) cm.
Answer: √(194) cm
Common mistakes
- ×Incorrectly identifying the hypotenuse. When finding a shorter side, students often add the squares (a2 + c2) instead of subtracting (c2 - a2).
- ×Forgetting the final step of taking the square root. The formula gives you the square of the length (c2), not the length itself (c).
- ×Arithmetic mistakes when squaring numbers or simplifying surds are frequent under time pressure and without a calculator.
- ×In 3D problems, using the wrong length to form the right-angled triangle. For example, confusing a pyramid's slant height with its slant edge.
No-calculator tips
- ✓Memorise common Pythagorean Triples and their multiples: (3,4,5), (6,8,10); (5,12,13); (8,15,17). This can provide an answer instantly.
- ✓Be fluent with square numbers up to 202 = 400. This saves critical time in calculations.
- ✓When simplifying surds like √(72), find the largest square factor. 72 = 36 × 2, so √(72) = √(36)*√(2) = 6*√(2).