## 11.2 Pythagoras’ Theorem

Knowing the coordinates of a point, it must be possible to work out its distance from the origin, but how? This is equivalent to the question: how can we work out the length of the hypotenuse of a right-angled triangle, given the lengths of the catheti (the two other sides)? If you were to carefully draw this triangle of three units on one side and four units on the other, you could measure the hypotenuse with a ruler and it would be about five units. But is it exactly five?

There are many ways to work out a length mathematically. If it’s made up of two smaller lengths, you can add them together; if it’s the width of a rectangle and you know the area and height, you can divide them; etc. There are many ideas we could explore, but one of the simplest ideas that proves to be helpful in this case is this: if we know the area of a square, we can work out its side length by taking the square root. It’s not immediately obvious to use this idea to solve this problem, but many different cultures all across the world eventually came across it after much trial and error.

As with most difficult problems, it can help to start with a simpler problem before working out the general solution, so let’s consider an isosceles right-angled triangle first.

##### Question 11.2.1 We can work out the side length of a square if we know its area. Use this fact to find the hypotenuse of a right-angled triangle with each cathetus one unit long. Show hint

This length cannot be expressed as a fraction (see Question 9.6.7), i.e. it is an irrational number (Section 8.1: Sets). This was of great concern to the cult of Pythagoras, a mysterious group of ancient Greek mathematicians led by Pythagoras of Samos, because they couldn’t accept that a length could be anything other than a ratio.

The theorem which we are about to discover, describing the relationship between the lengths of the sides of a right-angled triangle, is named after Pythagoras, although it was already known in the First Babylonian Dynasty, ancient India, and China (where it is known as the Shang Gao Theorem).

##### Question 11.2.2 By adapting your solution to the previous question, find the length $c$ of the hypotenuse of a right-angled triangle with catheti $a$ and $b$. Show answer

Remember:
Pythagoras’ theorem states that if $c$ is the hypotenuse of a right-angled triangle, and the other two sides have lengths $a$ and $b$, then $c^2=a^2+b^2.$

##### Question 11.2.3 Now let’s return to the question that started this section: what is the distance between the point $(4,3)$ and the origin? Show answer

##### Question 11.2.4 If a rectangle has a height of $5\,\mathrm{cm}$, and the diagonal is $13\,\mathrm{cm}$ long, what is its width? Show hint

You may have noticed that the past two questions involved numbers that worked out very neatly: right-angled triangles with sides $3,4,5$ and $5,12,13$, as opposed to the isosceles right-angled triangle with sides $1,1,\sqrt2$ where the hypotenuse was not a whole number. A set of three whole numbers that work in this way to make a right-angled triangle is called a Pythagorean triple. There are infinitely many Pythagorean triples; one way of finding them is presented in Section 11.2.1: Pythagorean Triples below.

##### Question 11.2.5 Find the distance between $(-2,1)$ and $(1,3)$. Show hint

##### Question 11.2.6 In general, what is the distance, $d$, between the points $(x_1,y_1)$ and $(x_2,y_2)$? Show hint

Remember:
The distance between two points is the hypotenuse of a right-angled triangle with catheti $\Delta x$ and $\Delta y$, the differences in the $x$ and $y$ coordinates.

You may skip ahead to Section 11.3: Plotting Graphs now, if you wish. The rest of this section covers some interesting applications of Pythagoras’ theorem.

##### Question 11.2.7 Suppose there’s a rhombus of side length $5\,\mathrm{cm}$, which is $9\,\mathrm{cm}$ wide (measured parallel to one of the sides). What is the area of the rhombus? Show hint

##### Question 11.2.8 This question is adapted from The Nine Chapters on the Mathematical Art. A vine wraps around a tree $6$ times, reaching a height of $5\,\mathrm{m}$. The circumference of the tree trunk is $2\,\mathrm{m}$. How long is the vine? Show answer

##### Question 11.2.9 You’re standing on the shore watching a ship sail away. When it reaches the horizon it starts disappearing due to the curvature of the Earth. Assuming your eyes are $1.5\,\mathrm{m}$ above sea-level and the Earth is a sphere of radius $6,\!400\,\mathrm{km}$, how far away is the horizon (along the straight line between your eyes and the base of the ship)? If you were instead on a cliff, $50\,\mathrm{m}$ above sea-level, how far away would the horizon be then? If you don’t have a calculator, estimate the distance. Show hint

##### Question 11.2.10 Where I live there are bike racks that look like circles of metal embedded in the ground. If the rack has a height of $h$ and a base length of $b$, find the radius, $r$, of the circle. Show hint

##### Question 11.2.11 If the length of a cube’s edge is $a$, what’s the distance between opposite corners of the cube? Show hint

##### Question 11.2.12 If you have two concentric circles as shown below, and the length of the straight line is $4\,\mathrm{cm}$, find the area of the shaded region. Show hint

##### Question 11.2.13  Show that the area of a triangle with side lengths $a$, $b$, and $c$ is: $A = \sqrt{s(s-a)(s-b)(s-c)}$ where $s$ is the semiperimeter (half the perimeter): $s = \frac{a + b + c}{2}.$ This is called Heron’s formula, after the ancient mathematician Heron (or Hero) of Alexandria. Show hint

##### Question 11.2.14 Imagine you’re standing $5\,\mathrm{km}$ north of a river whose bank runs east-west. You want to walk back to your camp, which is $15\,\mathrm{km}$ to the east, and $3\,\mathrm{km}$ north of the river, but on the way you need to get some water from the river. We’d like to know the shortest distance you have to walk to get to the camp, touching the river on the way. This question either requires some very sneaky thinking, or calculus to answer rigorously (we’ll do this in Question 15.3.26), but I’d like you to use your intuition and make an educated guess at what the shortest path will be like. Show hint

### 11.2.1 Pythagorean Triples

If we want to find three whole numbers $a$, $b$, and $c$ that satisfy the equation $c^2 = a^2 + b^2$ (so that they could form the sides of a right-angled triangle) we could use trial-and-error, trying various combinations of numbers for $a$ and $b$ to see if $a^2 + b^2$ is a perfect square. But hopefully we can work out a better way of finding these Pythagorean triples.

One idea that might occur to us is to factorise $b^2$ like so: \begin{align*} b^2 &= c^2 - a^2\\ &= (c-a)(c+a). \end{align*} Instead of picking values for $a$ and $b$ (which may result in $c$ being irrational), we could pick integer values for $b$ and $c+a$, then we can solve for $a$ and $c$.

It’s not a problem if $a$ and $c$ turn out to be fractions, because we can scale up $a$, $b$, and $c$ by the common denominator to make them all whole, thus generating a Pythagorean triple.

##### Question 11.2.15 1. Let $b=n$ and $c+a=m$, where $m$ and $n$ are integers. Given that $b^2 = (c-a)(c+a)$ find $a$ and $c$ in terms of $m$ and $n$. Show answer

2. Multiply $a$, $b$, and $c$ by the common denominator to make a Pythagorean triple (in fact, a formula for generating all Pythagorean triples). Show answer

3. Use your formulae to find a few Pythagorean triples, for example using the values of $m$ and $n$ in the table.

 $n$ $m$ $1$ $2$ $1$ $3$ $1$ $4$ $2$ $3$ $2$ $5$ $2$ $7$ $3$ $4$

In order for $a$ to be positive, $m$ must be greater than $n$. I’ve avoided using values of $m$ and $n$ with common factors, for example $n=2$ and $m=4$, because that would just give a multiple of the $n=1$, $m=2$ triangle. Show answer

This formula that we just derived is called Euclid’s formula after the ancient Greek mathematician. There is another derivation using complex numbers in Question 16.3.6.

When $a$, $b$, and $c$ have no common factors, it’s called a ‘primitive’ Pythagorean triple. We can generate all the other non-primitive triples just by scaling up a primitive one.

##### Question 11.2.16 If $m$ and $n$ are both odd, does Euclid’s formula generate a primitive Pythagorean triple? Show answer

Remember:
Euclid’s formula can generate any primitive Pythagorean triple: \begin{align*} a &= m^2 - n^2\\ b &= 2mn\\ c &= m^2 + n^2 \end{align*}for integers $m$ and $n$.