The Pythagorean theorem, perhaps the most famous theorem of geom-etry and indeed one of the most famous formulas in mathematics, de-serves a section of its own. In a right triangle, the side opposite the right angle is called the hypotenuse. The other two sides are called legs.
The right triangle below has legs BC and AC and hypotenuse AB, with respective lengths a, b, and c.
C B
b A
a c
Pythagorean theorem: For a right triangle with leg lengths a and b and hypotenuse length c,
a2+b2 =c2
There are reportedly over three hundred proofs of the Pythagorean theorem, but we’ll present the simplest ones here. Feel free to skip some of the proofs. My goal is that at least one of the proofs makes you smile or say, “That’s pretty cool!”
Proof 1: In the picture on the next page, we have assembled four right triangles to create a giant square.
Question: What is the area of the giant square?
Answer 1: Each side of the square has length a+b, so the area is (a+b)2 =a2+2ab+b2.
Answer 2: On the other hand, the giant square consists of four trian-gles, each with area ab/2, along with a tilted square in the middle with area c2. (Why is the middle object a square? We know that all four sides are equal and we can use symmetry to see that all four angles are equal:
if we rotated the figure 90 degrees, it would be identical, and so each of the middle square’s angles must be the same. Since the sum of the angles in a quadrilateral is 360 degrees, we know that each angle must be 90 degrees.) Therefore the area is 4(ab)/2+c2=2ab+c2.
Equating answers 1 and 2 gives us
a2+2ab+b2 =2ab+c2
b
a b
a b a
b
a c
c c
c
Compute the area of the big square in two different ways.
When you compare your two answers, the Pythagorean theorem pops out.
Subtracting 2ab from both sides gives us a2+b2 =c2
as desired.
Proof 2:Using the same picture as above, we rearrange the triangles as in the figure below. In the first picture the area not occupied by trian-gles is c2. In the new picture, the area not occupied by triangles is seen to be a2+b2. Thus c2 =a2+b2, as desired.
b
a b
a b a
b
a c
c c
c
b
a
a
b b b
a a
Compare the area of the white space in this figure and the previous one:a2+b2=c2 Proof 3: This time, let’s rearrange the four triangles to form a more compact square on the opposite page with area c2. (One reason this object is a square is that each corner is composed of∠A and∠B, which
sum to 90◦.) As before, the four triangles contribute an area of 4(ab/2) = 2ab. The tilted square in the middle has area(a−b)2 = a2−2ab+b2. Hence the combined area equals 2ab+ (a2−2ab+b2) = a2+b2, as
desired.
c
c c
c c
c a { b
a { b b a
a
b
The area of this figure is bothc2anda2+b2
Proof 4:Here’s a similar proof, and by that I mean, let’s exploit what we know about similar triangles. In right triangle ABC, draw the line segment CD perpendicular to the hypotenuse, as shown below. Notice
C B
A D
Both of the two smaller triangles are similar to the large one
that the triangle ADC contains both a right angle and ∠A, so its third angle must be congruent to∠B. Likewise, triangle CDB has a right an-gle and∠B, so its third angle must be congruent to∠A. Consequently, all three triangles are similar:
ACB ∼ ADC ∼ CDB
Note that the order of the letters matters. We have∠ACB = ∠ADC =
∠CDB = 90◦ are all right angles; likewise,∠A = ∠BAC = ∠CAD =
∠BCD, and∠B= ∠CBA = ∠DCA = ∠DBC. Comparing side lengths from the first two triangles gives us
AC/AB =AD/AC ⇒ AC2 =AD× AB
Comparing side lengths from the first and third triangles gives us CB/BA=DB/BC⇒ BC2 =DB× AB
Adding these equations, we have
AC2+BC2=AB×(AD+DB)
And since AD+DB= AB=c, we have our desired conclusion:
b2+a2 =c2
The next proof is purely geometrical. It uses no algebra, but it does require some visualization skills.
Proof 5: This time we start with two squares, with areas a2and b2, placed side by side, as below. Their combined area is a2+b2. We can dissect this object into two right triangles (with side lengths a and b and hypotenuse length c) and a third strange-looking shape. Note that the angle at the bottom of the strange shape must be 90◦since it is sur-rounded by∠A and∠B. Imagine a hinge placed in the upper left corner of the big square and the upper right corner of the smaller square.
a b b
a
a
a c
b c b
b b
a a
These two squares with areaa2+b2can be transformed into . . .
Now imagine that the bottom left triangle is “swiveled” 90◦ in a counter-clockwise direction so it rests on the outside of the top of the large square. Then swivel the other triangle clockwise 90◦ so that the right angles match up and it sits comfortably in the corner made by the
a a
c
b b
a a
b c b
c c
c
c
c b
a
b
a b
. . . a square of areac2!
two squares, as shown in the figure above. The end result is a tilted square of area c2. Thus a2+b2=c2, as promised. We can apply the Pythagorean theorem to explain question 4 about the football field at the beginning of the chapter with a rope of length 361 feet connecting two goalposts that are 360 feet apart.
h 180.5
180
By the Pythagorean theorem,h2+1802=180.52
The distance from either goalpost to the middle of the field is 180 feet. After the rope is raised to its highest point, h, we create a right triangle, as shown below, with leg length 180 and hypotenuse 180.5.
Thus by the Pythagorean theorem, and a few lines of algebra, we get h2+1802 =180.52
h2+32,400=32,580.25 h2 =180.25 h=√180.25≈13.43 feet
Hence the rope would be high enough that most trucks could drive under it!