Example 3
If a right triangle has a hypotenuse length of 9 feet and a leg length of 5 feet, what is the length of the third side? We use the Pythagorean theorem with the hypotenuse, 9, by itself on one side, and the other two lengths, 5 and x, on the other.
5^{2} + x^{2} = 9^{2}
x = √81 – 25 = √56 = 2√14 ≈ 7.48
The third side is about 7.48 feet long.
Pythagorean Word Problems
Many word problems involve finding a length of a right triangle. Identify whether each given length is a leg of the triangle or the hypotenuse. Then solve for the third length with the Pythagorean theorem.
Example 1
A diagonal board is needed to brace a rectangular wall. The wall is 8 feet tall and 10 feet wide. How long is the diagonal?
Having a rectangle means that we have a right triangle, and that the Pythagorean theorem can be applied. Because we are looking for the diagonal, the 10foot and 8foot lengths must be the legs, as shown in Figure 3.19.
The diagonal D must satisfy the Pythagorean theorem:
D^{2} = 10^{2} + 8^{2}
D^{2} = 100 + 64 = 164
D = √164 = 2√41 ≈ 12.81 feet
Example 2
A 100foot rope is attached to the top of a 60foot tall pole. How far from the base of the pole will the rope reach?
We assume that the pole makes a right angle with the ground, and thus, we have the right triangle depicted in Figure 3.20. Here, the hypotenuse is 100 feet.
The sides must satisfy the Pythagorean theorem:
x^{2} + 60^{2} = 100^{2}
x = √6,400 = 80 feet
Pythagorean Triples
Since the Pythagorean theorem was discovered, people have been especially fascinated by right triangles with whole number sides. The most famous one is the 345 right triangle, but there are many others, such as 51213 and 6810. A Pythagorean triple is a set of three whole numbers abc with a^{2} + b^{2} = c^{2}. Usually, the numbers are put in increasing order.
Example
Is 485573 a Pythagorean triple?
We calculate 48^{2} + 55^{2} = 2,304 + 3,025 = 5,329. It just happens that 73^{2} = 5,329. Thus, the numbers 48, 55, and 73 form a Pythagorean triple.
Generating Pythagorean Triples
The ancient Greeks found a system for generating Pythagorean triples. First, take any two whole numbers r and s, where r > s. We will get a Pythagorean triple abc if we set:
a = 2rs
b = r^{2} – s^{2}
c = r^{2} + s^{2}
This is because
a^{2} + b^{2}
= (2rs)^{2} + (r^{2} – s^{2})^{2}
= 4r^{2}s^{2} + r^{4} – 2r^{2}s^{2} + s^{4}
= r^{4} + 2r^{2}s^{2} + s^{4}
= (r^{2} + s^{2})^{2} = c^{2}
Thus, a^{2} + b^{2} = c^{2}.
Example
Find the Pythagorean triple generated by r = 7 and s = 2.
a = 2rs = 2(7)(2) = 28
b = r^{2} – s^{2} = 7^{2} – 2^{2} = 45
c = r^{2} + s^{2} = 7^{2} + 2^{2} = 53
Thus, r = 7 and s = 2 generate the 284553 Pythagorean triple.
If we plug in different values of r and s, we will generate many different Pythagorean triples, whole numbers a, b, and c with a^{2} + b^{2} = c^{2}. Around 1640, a French mathematician named Pierre de Fermat suggested that this could happen only with the exponent 2. He said that no positive whole numbers a, b, and c could possibly make a^{3} + b^{3} = c^{3} or a^{4} + b^{4} = c^{4} or a^{n} + b^{n} = c^{n} for any n > 2. Fermat claimed to have found a short and clever proof of this, but then died without writing it down.
For hundreds of years, mathematicians tried to prove this result, called "Fermat's Last Theorem." Only in 1995 was it finally proven, by a British mathematician named Andrew Wiles. Because his proof runs to more than 100 pages, it is clearly not the simple proof that Fermat spoke about. This assumes, of course, that Fermat had actually found a correct and simple proof.
Practice problems for this study guide can be found at:
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