0580 · 6.1Pythagoras’ theorem

The Pythagoras relationship

0:00Imagine crossing a rectangular playground by walking straight across instead of following two edges. The diagonal is shorter than the combined distance along those edges. Pythagoras’ theorem lets us calculate that diagonal using the two perpendicular lengths. By the end, you will understand the relationship and use it to find different missing sides. You need two familiar ideas before we begin. Squaring a number means multiplying it by itself, so 6² = 36.

0:26Taking a square root reverses this process, so √36 is six. You also need to recognise a right angle, which measures ninety degrees. Pythagoras’ theorem applies to right-angled triangles. The side opposite the right angle is called the hypotenuse, and it is always the longest side. In this diagram, the hypotenuse is labelled h. The other two sides, labelled p and q, meet at the right angle. Imagine building a square on each side of the triangle.

0:54Each square has an area equal to its side length squared. Pythagoras’ theorem states that the square of the hypotenuse equals the sum of the squares of the other two sides. In our labels, p² + q² = h². This is a relationship between areas, not simply between lengths. The two smaller squares together contain exactly as much area as the largest square. That explains why the side lengths must be squared before we add them.

1:19Adding the unsquared lengths measures a journey around two edges, not the straight diagonal. A rearrangement shows why this area relationship works. Place four identical right-angled triangles inside a larger square, leaving a square between their hypotenuses. The middle square has side length h, so its area is h². Its corners are right angles because the two acute angles of each triangle total ninety degrees. Rearrange the same four triangles inside the same outer square.

1:46This time, the uncovered space forms two squares with areas p² and q². Neither the outer area nor the total triangle area has changed. Therefore, the uncovered areas must be equal, giving the theorem.

Finding missing triangle sides

1:59For our first example, calculate the hypotenuse when the shorter sides measure six centimetres and eight centimetres. First, identify the side opposite the right angle. That is the unknown side here. Since we want the largest square’s area, we add the two smaller square areas. Write h² = 6² + 8². Squaring the known lengths gives 36 + 64. Their sum makes h² equal to one hundred. Taking the positive square root gives h equal to ten centimetres.

2:27We choose the positive root because a side length cannot be negative. Our second example has a hypotenuse of thirteen centimetres and another side of five centimetres. Calculate the remaining side, labelled x. The largest square’s area is already known, so adding would make no sense. Instead, subtract the known smaller square’s area to find the missing one. Start with x² + 5² = 13². Evaluating the squares gives x² + 25 = 169.

2:56Subtract twenty-five from both sides, leaving x² equal to 169 − 25. The subtraction gives one hundred and forty-four. Its positive square root gives x equal to twelve centimetres. Both examples can be completed without a calculator. On Paper two, practise squaring whole numbers and recognising square roots such as one hundred and forty-four. Paper four allows a calculator, but the reasoning stays the same. Decide whether to add or subtract before entering any numbers.

Exam applications and common mistakes

3:24An examination question might show a rectangular floor, seven metres long and four metres wide. Calculate its diagonal, giving your answer to three significant figures. The diagonal divides the rectangle into two right-angled triangles. Use the length and width as the perpendicular sides, with the diagonal as the hypotenuse. Write d² = 7² + 4². Evaluating those squares gives 49 + 16. Adding gives d² equal to sixty-five. The positive square root shows eight point zero six two two five seven on the calculator, followed by more digits.

3:58Rounding only the final answer gives eight point zero six metres. A Paper four question might describe a ladder six point five metres long leaning against a vertical wall. Its foot stands one point eight metres from the wall on horizontal ground. Calculate how high up the wall the ladder reaches. The ladder is the hypotenuse, so this time we subtract. Write y² + 1.8² = 6.5². The squares are three point two four and forty-two point two five.

4:27Subtracting gives y² equal to thirty-nine point zero one. The positive square root is six point two four five eight on the calculator. To three significant figures, the ladder reaches six point two five metres up the wall. Keep the full calculator value until the final answer, rather than rounding during the calculation. Unless instructed otherwise, give answers that are not exact to three significant figures. Include the length unit, not square metres.

4:51The squared quantities represent areas, but the final square root returns a length. Common mistakes include choosing the hypotenuse by its position on the page instead of finding the side opposite the right angle. Another mistake is forgetting the final square root, which leaves a squared length rather than a length. Adding when a shorter side is missing also gives an impossible result. A shorter side must be less than the hypotenuse.

5:14Always establish that the triangle has a right angle before using this theorem. Identify the hypotenuse, then write the relationship between the 3² lengths. Add to find the hypotenuse, or subtract to find a shorter side. Finally, take the positive square root and check that your answer makes geometrical sense.