Pythagoras Theorem in Two and Three Dimensions

The Fundamental Theorem

Pythagoras' theorem is one of the most elegant and essential rules in mathematics. It provides a simple algebraic method to calculate the length of a side in a right angled triangle when the other two sides are known. The theorem states that for any right angled triangle, the square of the longest side, which we call the hypotenuse, is equal to the sum of the squares of the two shorter sides. If we label the shorter sides $a$ and $b$ and the hypotenuse $c$, we write this as:

$a^2 + b^2 = c^2$

It is important to remember that this rule applies only to triangles containing a 90 degree angle. As we noted in our section on trigonometry, the cosine rule acts as a generalisation of Pythagoras for all triangles, where a correction term is subtracted: $a^2 = b^2 + c^2 - 2bc \cos A$. In a right angled triangle, the angle $A$ is 90 degrees, so $\cos 90 = 0$ and the correction term vanishes, leaving us with our standard theorem.

Pythagoras in Two Dimensions

A common application of this theorem is finding the distance between two points on the $xy$ plane. If we have two points $(x_1, y_1)$ and $(x_2, y_2)$, we can form a right angled triangle by drawing a horizontal line and a vertical line through the points. The horizontal distance is $\Delta x = x_2 - x_1$ and the vertical distance is $\Delta y = y_2 - y_1$. The distance $d$ between the points is the hypotenuse, giving:

$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$

This lead us directly to the coordinate geometry of the circle. Imagine drawing a circle of radius $r$ on the $xy$ plane with its centre at the origin. Every point $(x, y)$ on that circle is exactly a distance $r$ from the origin. Using Pythagoras' theorem, we find that all these points satisfy the equation:

$x^2 + y^2 = r^2$

As an aside, recall that all points inside the circle satisfy $x^2 + y^2 < r^2$, and all points outside the circle satisfy $x^2 + y^2 > r^2$. If the circle is translated so its centre is at $(a, b)$, we apply our graph shifting logic to get $(x - a)^2 + (y - b)^2 = r^2$, which is still just Pythagoras' theorem in disguise.

Extending to Three Dimensions

The TMUA and ESAT expect you to use this theorem in three dimensions too. This usually involves finding the distance between two opposite corners of a cuboid, often called the space diagonal. Consider a cuboid with dimensions $x$, $y$, and $z$. To find the diagonal $D$ that stretches from the bottom left front corner to the top right back corner, we proceed in two steps.

First, we find the diagonal $d_{base}$ of the base of the cuboid. This base is a rectangle with sides $x$ and $y$. Applying Pythagoras in 2D gives us $d_{base}^2 = x^2 + y^2$.

Second, we notice that this base diagonal, the vertical height $z$, and the space diagonal $D$ form another right angled triangle. The space diagonal $D$ is the hypotenuse of this new triangle. Therefore, $D^2 = d_{base}^2 + z^2$. Substituting our expression for the base diagonal, we get the general 3D formula for the distance from the origin to $(x, y, z)$:

$D^2 = x^2 + y^2 + z^2$

This elegant result shows that we can find distances in any number of dimensions by simply adding the square of the new dimension's length to our sum.

Worked Example: 3D Cuboid

Example

A cuboid has a base of 3cm by 4cm and a height of 12cm. Find the length of the internal space diagonal.

1. Calculate the diagonal of the base: $d_{base}^2 = 3^2 + 4^2 = 9 + 16 = 25$. So $d_{base} = 5$cm.

2. Use this to find the space diagonal $D$: $D^2 = d_{base}^2 + 12^2 = 25 + 144 = 169$.

3. Take the square root: $D = \sqrt{169} = 13$cm.

Notice that we could have done this in one calculation: $D = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$cm.

Finding Hidden Right Angles

In TMUA questions, you may not always be given a cuboid. You might be asked about a pyramid or a cone. In these cases, you must look for the vertical height which is perpendicular to the base. For example, in a square based pyramid with a vertex directly above the centre of the base, you can find the slant height by forming a triangle with the vertical height and half the diagonal of the square base.