Pythagoras' Theorem: Proof, Triples and Applications
The Most Famous Theorem in Mathematics
Builders in ancient Egypt used a rope knotted into 12 equal sections to mark out perfect right angles: stretched into a triangle with sides of 3, 4 and 5 sections, it always makes a right angle. The rule behind this trick, named after the Greek mathematician Pythagoras, is used every day by carpenters, surveyors, engineers, game designers and navigators. This tutorial covers Pythagoras' theorem as tested in WAEC, NECO, BECE and JAMB mathematics: the statement and a proof, finding the hypotenuse and a shorter side, the converse for testing right angles, Pythagorean triples, applications to ladders, rectangles, isosceles triangles, bearings and coordinates, and the extension to three dimensions.
The Statement
In a right-angled triangle, the side opposite the right angle is the hypotenuse, and it is always the longest side. Pythagoras' theorem states:
The square on the hypotenuse equals the sum of the squares on the other two sides.
If the hypotenuse is c and the other two sides are a and b:
c² = a² + b²
The theorem is about areas: the area of the square drawn on the hypotenuse equals the total area of the squares drawn on the other two sides.
A Proof by Rearrangement
There are hundreds of proofs. Here is one of the simplest:
- Take four identical right-angled triangles with sides a, b and c, and arrange them inside a large square of side (a + b), so that their hypotenuses form a tilted square of side c in the middle.
- The large square's area is (a + b)² = a² + 2ab + b².
- It is also made up of the four triangles, each of area ½ab, plus the inner square: 4 × ½ab + c² = 2ab + c².
- Setting the two expressions equal: a² + 2ab + b² = 2ab + c², so a² + b² = c².
Finding the Hypotenuse
Add the squares of the two shorter sides, then take the square root.
Worked example 1
The two shorter sides of a right-angled triangle are 6 cm and 8 cm. Find the hypotenuse.
c² = 6² + 8² = 36 + 64 = 100, so c = √100 = 10 cm.
Worked example 2
Find the length of the diagonal of a rectangular field measuring 40 m by 25 m.
The diagonal is the hypotenuse of a right-angled triangle with sides 40 m and 25 m. d² = 1600 + 625 = 2225, so d = √2225 ≈ 47.2 m.
Finding a Shorter Side
Subtract the square of the known shorter side from the square of the hypotenuse: a² = c² − b².
Worked example 3: a ladder
A ladder 13 m long leans against a vertical wall, with its foot 5 m from the wall on level ground. How high up the wall does it reach?
The ladder is the hypotenuse. h² = 13² − 5² = 169 − 25 = 144, so h = 12 m.
Worked example 4: an isosceles triangle
An isosceles triangle has two equal sides of 10 cm and a base of 12 cm. Find its height and area.
The height from the apex meets the base at right angles at its midpoint, dividing the triangle into two right-angled triangles, each with hypotenuse 10 cm and base 6 cm.
h² = 10² − 6² = 100 − 36 = 64, so h = 8 cm. Area = ½ × 12 × 8 = 48 cm².
A common check is that the hypotenuse must be the longest side. If a "shorter side" comes out longer than the hypotenuse, you added when you should have subtracted.
The Converse: Testing for a Right Angle
The theorem also works in reverse. If the sides of a triangle satisfy c² = a² + b², where c is the longest side, then the triangle is right-angled, with the right angle opposite c.
- If c² is less than a² + b², all the angles are acute.
- If c² is greater than a² + b², the angle opposite c is obtuse.
Worked example 5
Do sides of 7 cm, 24 cm and 25 cm form a right-angled triangle? What about 6 cm, 9 cm and 11 cm?
7² + 24² = 49 + 576 = 625 = 25². Yes, it is right-angled.
6² + 9² = 36 + 81 = 117, but 11² = 121. Since 121 is greater than 117, the triangle is not right-angled. Its largest angle is obtuse.
This is the 3–4–5 rope trick: builders check that the corners of a foundation are square by measuring 3 m along one wall, 4 m along the other, and checking that the diagonal is exactly 5 m.
Pythagorean Triples
A Pythagorean triple is a set of three whole numbers that satisfy a² + b² = c². Recognising them saves time in examinations:
- 3, 4, 5 (and its multiples: 6, 8, 10; 9, 12, 15; 30, 40, 50 …)
- 5, 12, 13
- 8, 15, 17
- 7, 24, 25
- 20, 21, 29
Triples can be generated from any two whole numbers m and n with m greater than n: a = m² − n², b = 2mn, c = m² + n². For example, m = 3 and n = 2 give 5, 12 and 13.
Applications
Worked example 6: bearings and distance
A hunter walks 5 km due north, then 12 km due east. How far is the hunter from the starting point, and on what bearing?
The two legs are perpendicular, so the distance = √(5² + 12²) = √169 = 13 km.
The angle east of north is tan⁻¹(12/5) ≈ 67.4°, so the bearing is about 067°.
Worked example 7: distance between two points
Find the distance between the points A(1, 2) and B(7, 10).
The horizontal difference is 7 − 1 = 6, and the vertical difference is 10 − 2 = 8. AB = √(6² + 8²) = 10 units.
This is the distance formula of coordinate geometry: d = √((x₂ − x₁)² + (y₂ − y₁)²). It is simply Pythagoras' theorem in disguise.
Worked example 8: screen sizes
A television screen is advertised as "50-inch", which means its diagonal is 50 inches. If the ratio of its width to its height is 4 : 3, find its width and height.
Let the width be 4k and the height 3k. Then (4k)² + (3k)² = 25k² = 50², so 5k = 50 and k = 10. The width is 40 inches, and the height is 30 inches.
Pythagoras in Three Dimensions
For a cuboid of length l, width w and height h, the space diagonal (from one corner to the opposite corner through the inside) is:
d = √(l² + w² + h²)
This comes from using the theorem twice: first across the base, then up through the height.
Worked example 9
Find the length of the longest rod that fits inside a box measuring 3 m by 4 m by 12 m.
Base diagonal = √(3² + 4²) = 5 m. Space diagonal = √(5² + 12²) = √169 = 13 m. Equivalently, √(9 + 16 + 144) = 13 m.
Worked example 10: a pyramid
A pyramid has a square base of side 6 cm, and its vertex is directly above the centre of the base. Each slant edge is 5 cm long. Find the height of the pyramid.
The diagonal of the base is √(6² + 6²) = √72 ≈ 8.49 cm, so the distance from the centre of the base to a corner is half of this, √18 ≈ 4.24 cm.
Height² = 5² − (√18)² = 25 − 18 = 7, so the height = √7 ≈ 2.65 cm.
Common Mistakes with Pythagoras' Theorem
- Using the theorem on a triangle with no right angle. For other triangles, use the cosine rule.
- Adding when finding a shorter side. Subtract: a² = c² − b².
- Forgetting the square root at the end, giving c² instead of c.
- Taking the square root term by term: √(36 + 64) is 10, not 6 + 8 = 14.
- Using the wrong side as the hypotenuse. It is always opposite the right angle, and always the longest side.
Pythagoras' Theorem: Practice Questions
- Find the hypotenuse of a right-angled triangle with shorter sides 9 cm and 12 cm.
- A right-angled triangle has a hypotenuse of 17 cm and one side of 8 cm. Find the third side.
- Is a triangle with sides 10 cm, 24 cm and 26 cm right-angled?
- A square has a diagonal of 10 cm. Find the length of its side.
- A ship sails 9 km due west and then 12 km due south. How far is it from its starting point?
- Find the distance between (−2, 3) and (4, −5).
- Find the space diagonal of a cube of side 4 cm.
Answers: 1. √(81 + 144) = 15 cm. 2. √(289 − 64) = 15 cm. 3. Yes: 100 + 576 = 676 = 26². 4. 2s² = 100, so s = √50 ≈ 7.07 cm. 5. √(81 + 144) = 15 km. 6. √(6² + 8²) = 10 units. 7. √(16 + 16 + 16) = √48 ≈ 6.93 cm.
Conclusion
Pythagoras' theorem states that in a right-angled triangle, the square on the hypotenuse equals the sum of the squares on the other two sides: c² = a² + b². Add the squares to find the hypotenuse, and subtract to find a shorter side. The converse tests whether a triangle is right-angled, and Pythagorean triples such as 3–4–5 and 5–12–13 save time. The theorem gives the distance formula of coordinate geometry, solves problems with ladders, diagonals, bearings and screens, and extends to three dimensions as d = √(l² + w² + h²).