Download I. Introduction

The Pythagorean Theorem
I. Introduction
There are an uncountable number of topics that students are expected to cover
each year in school. For example, they are expected to learn about right triangles, similar
triangles, and polygons. We expect them to learn about angles, lines, and graphs. One of
the topics that almost every high school geometry student learns about is the Pythagorean
Theorem.
When asked what the Pythagorean Theorem is, students will often state that
a2+b2=c2 where a, b, and c are sides of a right triangle. However, students often don't
know why this is true. Most have never proved it. In the paper that follows, there is
history, several proofs, and ways for high school students to use the proof in real life
situations.
To understand the following information, a general knowledge of geometry and
algebra should be sufficient. The proofs are not difficult and with some thought it is clear
that the Pythagorean theorem works and is important.
II.
The Pythagorean Theorem
To begin, the Pythagorean theorem states that the square on the hypotenuse of a
right triangle has an area equal to the combined areas of the squares on the other two
sides. (Gardner 152) One will find the converse of this statement to also be true.
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The Pythagorean theorem was a mathematical fact that the Babylonians knew and
used. However 1000 years later, between the years of 580-500 BC, Pythagoras of Samos
was the first to prove the theorem. It is possible that someone proved the theorem before
Pythagoras, but no proof has been found. Because of this, Pythagoras is given credit for
the first proof. (MacTutor History of Mathematics Archive)
Before a proof was ever given, besides the Babylonians it was thought that
"Egyptian temple builders used ropes in laying foundations, suggested that perhaps the
obtained accurate right angles by using marked ropes that could be stretched around
stakes to form a 3,4,5 right triangle." There is no documented evidence of this but
Cantor, a historian of mathematics agreed this could be true. (Gardner 155)
Following the first proof, many proofs followed. Proofs have been found by
Euclid, Socrates, and even President Garfield. The Pythagorean Proposition, by Elisha S.
Loomis gives 367 different proofs of this theorem. With the Pythagorean theorem being
such a popular topic, it is no wonder high school students study the theorem. (Gardner
153,155)
Before the proofs, it is important for students to see the theorem as it is worded.
"The square on the hypotenuse of a right triangle has an area equal to the combined areas
of the squares on the other two sides" can be drawn to get a better idea about what this
theorem is stating.
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I+II=III where I, II, and III represent areas of the squares.
This definition gives rise to the definition that is used in today's classroom textbooks. "In
a right triangle, the sum of the squares of the lengths of the legs is equal to the square of
the length of the hypotenuse." (Bellman 414)
a2`+b2=c2
With a statement of the Pythagorean theorem, it is only logical to show that it is
true. As mentioned earlier, this has been done hundreds of ways. It seems logical to
begin with a proof that very well may predate Pythagoras. This proof is appealing
because of its simplicity. For this reason, one such as this is suitable to show in a
classroom.
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Proof 1. (Gardner 154)
Begin with the figure on the left below. Arrange four duplicated right triangles in
this pattern. Now by rearranging the four triangles inside the largest square, the figure on
the right can be reached. The two white squares are each squares on the legs of the
triangle. Since their area is that of the large square minus the four triangles, we know it
must equal the area of the white square on the left. Therefore, the square on the
hypotenuse (left picture) has an area equal to the area of the sum of the squares on the
legs (right picture).
Proof 2. James A. Garfield's Proof. (Gardner 155)
Garfield stated "we think it something on which the members of both houses can
unite without distinction of the party." This referred to his proof which uses the diagram
below.
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The basic right triangle is shaded. On the hypotenuse the right isosceles triangle CBE is
drawn. Extend AC, then from a point E a perpendicular is drawn to the extended line,
meeting at D. The shaded triangle is congruent with triangle DCE, therefore AB=DC and
AC=DE.
From here, the area of trapezoid ABED is the product of the base and half the sum
of its sides, x and y.
( x  y )( x  y ) x 2 y 2


2
2
2
The area of the trapezoid is also the sum of the areas of the three triangles. These
areas are z2/2, xy/2 and xy/2. So, another way of writing the area is z2 /2 + 2(xy)/2. By
setting these two expressions equal to one another Garfield got x2+y2 = z2 . This was his
proof of the Pythagorean theorem.
III.
The Law of Cosines
The Pythagorean theorem is basically a generalization of the Law of Cosines.
The Law of Cosines is used in any triangle when given three sides (SSS) or two sides and
their included angle (SAS). Using the same triangle as earlier, the Law of Cosines states
the following:
a 2b  b 2  c 2  2bcCosA
b 2  a 2  c 2  2acCosB
c 2  b 2  a 2  2abCosC
b2  c2  a2
CosA 
2bc
2
a  c2  b2
CosB 
2ac
2
a  b2  c2
CosC 
2ab
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Those on the left are used to find a side length and those on the right are used to find the
size of an angle. The proof for the Law of Cosines involves the triangle below that has
three acute angles.
B has the coordinates (c,0)
C has the coordinates (x,y) where x=bcosA and y=bsinA
Using the distance formula, the side a has length of
a  ( x  c) 2  ( y  0) 2
From this point the equation can be altered using substitution.
a 2  (b cos A  c) 2  (b sin A) 2
a 2  b 2 cos 2 A  2bc cos A  c 2  b 2 sin 2 A
a 2  b(sin 2 A  cos 2 A)  c 2  2ab cos A
a 2  b 2  c 2  2ab cos A
This is one form of the Law of Cosines. The others can be found using the same method.
Now, in this situation what happens if A is 90The cosine of 90is 0. This leaves the
equation a2+b2=c2, which is the Pythagorean theorem.
(Larson and Hostetler 522-524)
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IV. Exercises
1. Make a conjecture about what other figures can be placed on the edges of a right
triangle such that the sum of the areas of the figures on the legs equals the area of the
figure on the hypotenuse? In deciding what figures work, use either Geometer's
Sketchpad or Peanut Geometry to make your drawings.
2. What is the smallest number of matches needed to form simultaneously, on a plane,
two different (non-congruent) Pythagorean triangles? The matches represent units of
length and must not be broken or split in any way. (Gardner 160)
3. The following idea is taken from Algebra, A Challenging World, chapter 9. In the
textbook, it asks students to think about the construction of a box kite and the lengths of
the diagonal struts. However, students would enjoy actually building a kite and by
constructing the kite, they work with measurements and the Pythagorean theorem.
The assignment: Design a box kite in the shape of a rectangular solid. This will require
researching the kite-making process to determine what materials would best work for the
kite. The opening at each end of the kite is a square. In making your kite a) determine the
length of the diagonal struts using the Pythagorean theorem. This assignment is does not
necessarily require the kites to fly, but instead allow students to be creative while
learning math. b) If the square on the end of the solid has sides of length two then what
is the length of the diagonal c) If the kite has squares with a side length of s, what is the
length of the diagonal?
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4) In the figure below, determine the length of side x.
C
A
B
Angle A is 35 degrees. The sides adjacent to a are of length 3 and 5. What is the length
of BC.
IV. Solutions
1. Any figure can be drawn on the three sides as long as the three figures are similar.
While using Geometer’s Sketchpad, as similar figures are drawn on the sides of a
triangle, the sum of the two figures on the legs have the same area as the area on the
hypotneuse. Geometer’s Sketchpad allows many figures to be drawn without the hassle
of computing the area of each figure by hand. (Gardner 158)
2. (Gardner 162)
3. a) The length of the diagonal depends on the side length, so answers will vary.
b) If the side is of length 2 then the diagonal will be 2 times the square root of 2.
c) If the side length is s then the diagonal will be s times the square root of 2.
4. Using the law of cosines you get the formula a2 =32+52 –2(3)(5)cos35. Solve for an
and one will get 3.07.
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V. Bibliography
Martin Gardner, “The Pythagorean Theorem.” Sixth Book of Mathematical Games from
Scientific American. Charles Scribner’s Sons: New York, 1971. Chapter 16.
MacTutor History of Mathematics Archive. University of St. Andrews: Scotland.
http://www.bulletproof.org/math.
Bellman, Allan., et al. Algebra, Tools for a Changing World. Massachusetts and New
Jersey: Prentice Hall, 1998.
Larson, Roland and Hostetler, Robert. Precalculus. Boston and NewYork: Houghton
Mifflin Co., 1997. 522-524.