Trigonometry/The Pythagorean Theorem

In a right triangle
The square of the hypotenuse
is equal to
the sum of the squares of the other two sides.

Khan Academy

Video Links

The Khan Academy has video material relating to this topic which you may find easier to follow:

• The Pythagorean Theorem, gives the theorem.
• Pythagorean Theorem II, has worked through exercise on this topic.

The Theorem

A right-angle triangle

In a right triangle (right-angled triangle) the longest side is always furthest away from the right angle. It is called the hypotenuse. The length of the hypotenuse can be calculated from the lengths of the two other sides. In the diagram, c is the hypotenuse and we can calculate it from a and b.

This is a statement of the Pythagorean Theorem or Pythagoras' Theorem - as an equation relating the lengths of the sides a, b and c:

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

where c represents the length of the hypotenuse, and a and b represent the lengths of the other two sides.

This only works for right triangles!

For the right angle triangle with corners labelled by A,B,C as shown, the following says the same thing:

${\overline {BC}}^{2}+{\overline {AC}}^{2}={\overline {AB}}^{2}$

All we've done here is just use a different notation for the sides a, b and c.

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Geometrical Interpretation

The Pythagorean theorem: The sum of the areas of the two squares on the legs (a and b) equals the area of the square on the hypotenuse (c).

We can also show this equation with a diagram, as on the right, where each side of a right angle triangle has a square attached to it. This is the geometrical interpretation of the Pythagorean theorem, looking at the theorem as a theorem about areas. The areas of the smaller squares add up to the area of the larger square.

In terms of areas, the Pythagorean Theorem states:

In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).

The two ways of talking about the Pythagorean theorem amount to exactly the same thing, since the area of a square is just the length of one side squared.

History of the Theorem

Pythagoras, depicted on a 3rd-century coin

The Pythagorean theorem is named after the Greek mathematician Pythagoras, who by tradition is credited with its discovery and proof, although it is often argued that knowledge of the theorem predates him. There is much evidence that Babylonian mathematicians understood the formula.

Three Worked Examples

In a right triangle
The square of the hypotenuse
is equal to
the sum of the squares of the other two sides.

Example 1 An often used example for the Pythagorean Theorem is a right triangle in which $a=3$ and $b=4$ because the numbers come out particularly nicely: ${\begin{matrix}a^{2}+b^{2}&=&\\3^{2}+4^{2}&=&\\9+16&=&25\end{matrix}}$ So $c^{2}=25$ and $c={\sqrt {25}}$ and $c=5$

Example 2 Another often used example is a right triangle in which the two remaining angles are $45^{\circ }$ and both $a=1$ and $b=1$ . This time the answer doesn't come out as a whole number: ${\begin{matrix}a^{2}+b^{2}&=&\\1^{2}+1^{2}&=&\\1+1&=&2\end{matrix}}$ So $c^{2}=2$ and $c={\sqrt {2}}$ and we can stop here, or else give the answer as a decimal. $c\approx 1.414$ The $\approx$ symbol means 'approximately'.

Example 3 One more example is a right triangle in which the shorter sides are $p=11$ cm and $q=7$ cm. What's this about $p$ and $q$ ? It really doesn't matter what letter we use to name the lengths of the sides as long as we are consistent. The formula in the blue box still holds. In this case we could call the hypotenuse $r$ , and so we have $p^{2}+q^{2}=r^{2}$ . ${\begin{matrix}p^{2}+q^{2}&=&\\11^{2}+7^{2}&=&\\121+49&=&170\end{matrix}}$ So $r^{2}=170$ and $r={\sqrt {170}}$ cm and giving the answer as a decimal. $r\approx 13.04$ cm Does the answer make sense? It's worth checking the answer makes some kind of sense. If the short sides are 11 cm and 7 cm is it reasonable that the long side is about 13 cm? Yes, it is. If you'd got an answer of 20 cm (or 170 cm) it would be unreasonable. The two other sides won't stretch that far. Or if the answer were 11 cm or less than you'd know that you hadn't got the length of the hypotenuse, the longest side. We're OK here with this answer of 13.04 cm, which we probably should round down to 13 cm.

In 3D

Pythagoras' theorem in three dimensions relates the diagonal AD to the three sides.

Pythagoras' theorem can be applied to three dimensions as follows.

Consider the rectangular solid shown in the figure. The length of diagonal BD along the floor of the rectangular solid is found from Pythagoras' theorem as:

${\overline {BD}}^{2}={\overline {BC}}^{2}+{\overline {CD}}^{2}$

where these three sides form a right triangle. Using horizontal diagonal BD and the vertical edge AB, the length of diagonal AD then is found by a second application of Pythagoras' theorem as:

${\overline {AD}}^{2}={\overline {AB}}^{2}+{\overline {BD}}^{2}$

or, doing it all in one step:

${\overline {AD}}^{2}={\overline {AB}}^{2}+{\overline {BC}}^{2}+{\overline {CD}}^{2}$

This result is the three-dimensional expression for the diagonal ${\overline {AD}}$ in terms of the three mutually perpendicular sides.

This one-step formula may be viewed as a generalization of Pythagoras' theorem to higher dimensions. However, this result is really just the repeated application of the original Pythagoras' theorem.

The Pyramid of Khafre

An Example in 3 Dimensions The Khafre Pyramid; 274 cubits high; base 412 x 412 cubits Lines and labels for working out the length of the sloping edge AC This example is in three dimensions and again we have to use the Pythagorean Theorem twice. The great pyramid of Khafre is 274 cubits high. It has a square base and the square has sides of length 412 cubits. What is the length of the sloping diagonal edge from a corner up to the summit? To answer this question we work in two stages. See the diagram. The line ${\overline {AO}}$ is vertical and the line ${\overline {OB}}$ is horizontal, so there is a right angle triangle with sides ${\overline {AO}}$ and ${\overline {OB}}$ . This lets us calculate the length ${\overline {AB}}$ . We need to be careful. ${\overline {AO}}$ is 274 cubits, but ${\overline {OB}}$ is half the length of the side of the square, so ${\overline {OB}}$ is ${\tfrac {412}{2}}=206$ cubits. What is the length of the hypotenuse ${\overline {AB}}$ of the triangle $\triangle AOB$? Well, ${\begin{matrix}{\overline {AO}}^{2}+{\overline {OB}}^{2}&=&\\274^{2}+206^{2}&=&\\75,076+42,436&=&117,512\end{matrix}}$ So ${\overline {AB}}^{2}=117,512$ and ${\overline {AB}}={\sqrt {117,512}}$ cubits We've made progress, but we want the length ${\overline {AC}}$ , not the length ${\overline {AB}}$ . Now we have another right angle triangle, $\triangle ABC$ with the right angle at B. We want the length ${\overline {AC}}$ . We have just calculated ${\overline {AB}}$ and we know, because the base of the pyramid is a square, that ${\overline {BC}}$ is once again 206 cubits. This time ${\overline {AC}}$ is the hypotenuse (of $\triangle ABC$). So $\displaystyle {\begin{matrix}{\overline {AB}}^{2}+{\overline {BC}}^{2}&=&\\{\sqrt {117,512}}^{2}+206^{2}&=&\\117,512+42,436&=&159,948\end{matrix}}$ So ${\overline {AC}}^{2}=159,948$ and ${\overline {AC}}={\sqrt {159,948}}$ cubits ${\overline {AC}}\approx 400$ cubits

Other forms

As pointed out in the introduction, if c denotes the length of the hypotenuse and a and b denote the lengths of the other two sides, Pythagoras' theorem can be expressed as the Pythagorean equation:

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

or, solved for c:

$c={\sqrt {a^{2}+b^{2}}}$

That is what we have been doing in all the examples above. We are using the shorter sides to work out what the length of the hypotenuse is.

If the hypotenuse c is known, and the length of one of the legs must be found, the following equations can be used:

$b={\sqrt {c^{2}-a^{2}}}$

or

$a={\sqrt {c^{2}-b^{2}}}$

The Pythagorean equation provides a simple relation among the three sides of a right triangle so that if the lengths of any two sides are known, the length of the third side can be found.

Exercises

A cubical box is used to ship a printer. If one side is 90 cm, what is the longest distance between any two corners of the box?

An LCD television screen is measured 26 inches from corner to corner. The screen height is 13 inches. How wide is the screen? (To one-tenth of an inch)

What is the length of the longest line in this diagram? Pythagoras again and again In this problem you can assume that all the angles that look like right angles are.

Aliens from the fourth dimension are packing their holographic TV into a box ready for a trip to Earth. The box measures 1.5 Angstroms by 1.5 Angstroms by 1.5 Angstroms by 1.5 Angstroms (it is a 4 dimensional hypercube). What is the length of the longest diagonal?

If a room is 17 ft long, 14 ft wide, and 10 ft high, what is the length of the diagonal of (a) the floor, (b) an end wall, and (c) a side wall of the room?

Find the missing side (Pythagorean triples) In each of these examples you are told that All three sides of the triangle are whole numbers. The triangle is a right angle triangle You are given two of the sides. Find the third side. Which side is the hypotenuse? Sides you are told are 4,5, what is the third side? Sides you are told are 5,12, what is the third side? Sides you are told are 6,8, what is the third side? Sides you are told are 40,41, what is the third side?

Squaring small numbers What is $(-0.5)^{2}$ ? Of the numbers between -2 and +2, which ones end up closer to zero when you square them?