Introduction to Pythagoras' Theorem

Welcome to one of the most famous and useful chapters in mathematics! Pythagoras' theorem is a fundamental tool used to find missing lengths in right-angled triangles. Whether you are calculating the distance across a field or finding the height of a building, this theorem is your "go-to" formula.

In this chapter, we will explore how to use this rule in simple 2D shapes and then move into the exciting world of 3D objects like cuboids and pyramids. Don't worry if it seems a bit "square" at first—once you see the pattern, it becomes much easier!

The Basics: Pythagoras in 2D

Pythagoras' theorem only works for right-angled triangles (triangles with a \(90^{\circ}\) angle). Before we start calculating, we need to know the names of the sides:

  • The Hypotenuse: This is the longest side of the triangle. It is always directly opposite the right angle. We usually call this side \(c\).
  • The Shorter Sides: These are the two sides that meet to form the right angle. We usually call these \(a\) and \(b\).

The Formula

The theorem states that in any right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides:

\(a^2 + b^2 = c^2\)

1. Finding the Hypotenuse (\(c\))

If you know the two shorter sides and want to find the longest side, follow these steps:

  1. Square both shorter sides (\(a^2\) and \(b^2\)).
  2. Add the results together.
  3. Take the square root to find the final length.

Example: A triangle has sides \(a = 3 \text{ cm}\) and \(b = 4 \text{ cm}\).
\(3^2 + 4^2 = 9 + 16 = 25\)
\(c = \sqrt{25} = 5 \text{ cm}\)

2. Finding a Shorter Side (\(a\) or \(b\))

If you already know the hypotenuse and one shorter side:

  1. Square both known sides.
  2. Subtract the smaller square from the larger square (\(c^2 - b^2\)).
  3. Take the square root to find the final length.

Example: The hypotenuse is \(13 \text{ m}\) and one side is \(5 \text{ m}\).
\(13^2 - 5^2 = 169 - 25 = 144\)
\(a = \sqrt{144} = 12 \text{ m}\)

Quick Review: Remember, if you are looking for the longest side, you add. If you are looking for a shorter side, you subtract.

Using Pythagoras in Other 2D Shapes

Sometimes a right-angled triangle isn't immediately obvious. You might need to "create" one:

  • Isosceles Triangles: You can drop a perpendicular line (an altitude) from the top vertex to the base. This splits the triangle into two identical right-angled triangles.
  • Rectangles and Squares: Drawing a diagonal creates two right-angled triangles.
  • Acute-angled Triangles: The syllabus mentions that you can use Pythagoras in any acute-angled triangle if an altitude is given or constructed. This height line creates the right angles you need to start calculating.

Common Mistake to Avoid: Always check your answer! The hypotenuse must always be the longest side. If you calculate a side and it’s longer than the hypotenuse, you probably added when you should have subtracted.

Surds and Exact Answers

In your exam, you might be asked for an exact answer. This means instead of writing a long decimal like \(4.2426...\), you should leave your answer as a surd.

Example: If \(c^2 = 18\), the exact length is \(c = \sqrt{18}\).
Following the rules from Topic 1 (Number), you can simplify this: \(\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}\).

Pythagoras in 3D

Pythagoras' theorem also works for finding distances in 3D shapes like cuboids or pyramids. Think of this as doing 2D Pythagoras twice!

The "Long Diagonal" of a Cuboid

If you want to find the distance from one bottom corner of a box to the opposite top corner, you are looking for the 3D diagonal. Let the dimensions of the box be length (\(x\)), width (\(y\)), and height (\(z\)).

The 3D formula is an extension of the 2D one:

\(d^2 = x^2 + y^2 + z^2\)

Step-by-Step for 3D:
  1. Identify the three dimensions: length, width, and height.
  2. Square all three numbers.
  3. Add them all together.
  4. Take the square root of the total.

Example: Find the diagonal of a room that is \(4 \text{ m}\) by \(3 \text{ m}\) by \(12 \text{ m}\).
\(d^2 = 4^2 + 3^2 + 12^2\)
\(d^2 = 16 + 9 + 144 = 169\)
\(d = \sqrt{169} = 13 \text{ m}\)

Pythagoras in Pyramids

In a pyramid, you often need to find the slant height or the vertical height. To do this, look for a right-angled triangle that connects the center of the base, the midpoint of an edge, and the top vertex (apex).

Did you know? 3D Pythagoras is used by architects and engineers every day to ensure that structures are stable and that beams are the correct length!

Key Takeaways for Success

  • Always draw a diagram: Even a quick sketch helps you identify where the right angle is.
  • Identify the Hypotenuse first: Label it \(c\) so you don't get mixed up in the formula.
  • Units matter: Ensure all your measurements are in the same SI units (e.g., all \(cm\) or all \(m\)) before you start calculating.
  • Show your working: In the Edexcel Spec B exam, you get marks for the process. Write down the squares and the addition/subtraction steps clearly.
  • Check the question: Does it ask for "3 significant figures" or an "exact value"? This tells you whether to use your calculator or leave it as a surd.

Note: For more advanced problems involving angles, remember to check the Trigonometry chapter, which covers the Sine and Cosine rules!