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参考答案:勾股定理 Pythagorean theorem
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The Pythagorean Theorem was one of the earliest theorems known to ancient civilizations. This famous theorem is named for the Greek mathematician and philosopher, Pythagoras. Pythagoras founded the Pythagorean School of Mathematics in Cortona, a Greek seaport in Southern Italy. He is credited with many contributions to mathematics although some of them may have actually been the work of his students.
The Pythagorean Theorem is Pythagoras' most famous mathematical contribution. According to legend, Pythagoras was so happy when he discovered the theorem that he offered a sacrifice of oxen. The later discovery that the square root of 2 is irrational and therefore, cannot be expressed as a ratio of two integers, greatly troubled Pythagoras and his followers. They were devout in their belief that any two lengths were integral multiples of some unit length. Many attempts were made to suppress the knowledge that the square root of 2 is irrational. It is even said that the man who divulged the secret was drowned at sea.
The Pythagorean Theorem is a statement about triangles containing a right angle. The Pythagorean Theorem states that:
"The area of the square built upon the hypotenuse of a right triangle is equal to the sum of the areas of the squares upon the remaining sides."
According to the Pythagorean Theorem, the sum of the areas of the two red squares, squares A and B, is equal to the area of the blue square, square C.
Thus, the Pythagorean Theorem stated algebraically is:
for a right triangle with sides of lengths a, b, and c, where c is the length of the hypotenuse.
Although Pythagoras is credited with the famous theorem, it is likely that the Babylonians knew the result for certain specific triangles at least a millennium earlier than Pythagoras. It is not known how the Greeks originally demonstrated the proof of the Pythagorean Theorem. If the methods of Book II of Euclid's Elements were used, it is likely that it was a dissection type of proof similar to the following:
"A large square of side a+b is divided into two smaller squares of sides a and b respectively, and two equal rectangles with sides a and b; each of these two rectangles can be split into two equal right triangles by drawing the diagonal c. The four triangles can be arranged within another square of side a+b as shown in the figures.
The area of the square can be shown in two different ways:
1. As the sum of the area of the two rectangles and the squares:
2. As the sum of the areas of a square and the four triangles:
Now, setting the two right hand side expressions in these equations equal, gives
Therefore, the square on c is equal to the sum of the squares on a and b. (Burton 1991)
There are many other proofs of the Pythagorean Theorem. One came from the contemporary Chinese civilization found in the oldest extant Chinese text containing formal mathematical theories, the Arithmetic Classic of the Gnoman and the Circular Paths of Heaven.
The proof of the Pythagorean Theorem that was inspired by a figure in this book was included in the book Vijaganita, (Root Calculations), by the Hindu mathematician Bhaskara. Bhaskara's only explanation of his proof was, simply, "Behold".
These proofs and the geometrical discovery surrounding the Pythagorean Theorem led to one of the earliest problems in the theory of numbers known as the Pythgorean problem.
The Pythagorean Problem:
Find all right triangles whose sides are of integral length, thus finding all solutions in the positive integers of the Pythagorean equation:
The three integers (x, y, z) that satisfy this equation is called a Pythagorean triple.
Some Pythagorean Triples:
x y z
3 4 5
5 12 13
7 24 25
9 40 41
11 60 61
The formula that will generate all Pythagorean triples first appeared in Book X of Euclid's Elements:
where n and m are positive integers of opposite parity and m>n.
In his book Arithmetica, Diophantus confirmed that he could get right triangles using this formula although he arrived at it under a different line of reasoning.
The Pythagorean Theorem can be introduced to students during the middle school years. This theorem becomes increasingly important during the high school years. It is not enough to merely state the algebraic formula for the Pythagorean Theorem. Students need to see the geometric connections as well. The teaching and learning of the Pythagorean Theorem can be enriched and enhanced through the use of dot paper, geoboards, paper folding, and computer technology, as well as many other instructional materials. Through the use of manipulatives and other educational resources, the Pythagorean Theorem can mean much more to students than just
and plugging numbers into the formula.
The following is a variety of proofs of the Pythagorean Theorem including one by Euclid. These proofs, along with manipulatives and technology, can greatly improve students' understanding of the Pythagorean Theorem.
The following is a summation of the proof by Euclid, one of the most famous mathematicians. This proof can be found in Book I of Euclid's Elements.
Proposition: In right-angled triangles the square on the hypotenuse is equal to the sum of the squares on the legs.
Figure 2
Euclid began with the Pythagorean configuration shown above in Figure 2. Then, he constructed a perpendicular line from C to the segment DJ on the square on the hypotenuse. The points H and G are the intersections of this perpendicular with the sides of the square on the hypotenuse. It lies along the altitude to the right triangle ABC. See Figure 3.
Figure 3
Next, Euclid showed that the area of rectangle HBDG is equal to the area of square on BC and that the are of the rectangle HAJG is equal to the area of the square on AC. He proved these equalities using the concept of similarity. Triangles ABC, AHC, and CHB are similar. The area of rectangle HAJG is (HA)(AJ) and since AJ = AB, the area is also (HA)(AB). The similarity of triangles ABC and AHC means
and therefore
or, as to be proved, the area of the rectangle HAJG is the same as the areaof the square on side AC. In the same way, triangles ABC and CHG are similar. So
and
Since the sum of the areas of the two rectangles is the area of the square on the hypotenuse, this completes the proof.
Euclid was anxious to place this result in his work as soon as possible. However, since his work on similarity was not to be until Books V and VI, it was necessary for him to come up with another way to prove the Pythagorean Theorem. Thus, he used the result that parallelograms are double the triangles with the same base and between the same parallels. Draw CJ and BE.
The area of the rectangle AHGJ is double the area of triangle JAC, and the area of square ACLE is double triangle BAE. The two triangles are congruent by SAS. The same result follows in a similar manner for the other rectangle and square. (Katz, 1993)
Click here for a GSP animation to illustrate this proof.
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The next three proofs are more easily seen proofs of the Pythagorean Theorem and would be ideal for high school mathematics students. In fact, these are proofs that students could be able to construct themselves at some point.
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The first proof begins with a rectangle divided up into three triangles, each of which contains a right angle. This proof can be seen through the use of computer technology, or with something as simple as a 3x5 index card cut up into right triangles.
Figure 4
Figure 5
It can be seen that triangles 2 (in green) and 1 (in red), will completely overlap triangle 3 (in blue). Now, we can give a proof of the Pythagorean Theorem using these same triangles.
Proof:
I. Compare triangles 1 and 3.
Figure 6
Angles E and D, respectively, are the right angles in these triangles. By comparing their similarities, we have
and from Figure 6, BC = AD. So,
By cross-multiplication, we get :
II. Compare triangles 2 and 3:
Figure 7
By comparing the similarities of triangles 2 and 3 we get:
From Figure 4, AB = CD. By substitution,
Cross-multiplication gives:
Finally, by adding equations 1 and 2, we get:
From triangle 3,
AC = AE + EC
so
Figure 8
We have proved the Pythagorean Theorem.
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The next proof is another proof of the Pythagorean Theorem that begins with a rectangle. It begins by constructing rectangle CADE with BA = DA. Next, we construct the angle bisector of <BAD and let it intersect ED at point F. Thus, <BAF is congruent to <DAF, AF = AF, and BA = DA. So, by SAS, triangle BAF = triangle DAF. Since <ADF is a right angle, <ABF is also a right angle.
Figure 9
Next, since m<EBF + m<ABC + m<ABF = 180 degrees and m<ABF = 90 degrees, <EBF and <ABC are complementary. Thus, m<EBF + m<ABC = 90 degrees. We also know that
m<BAC + m<ABC + m<ACB = 180 degrees. Since m<ACB = 90 degrees, m<BAC + m<ABC = 90 degrees. Therefore, m<EBF + m<ABC = m<BAC + m<ABC and m<BAC = m<EBF.
By the AA similarity theorem, triangle EBF is similar to triangle CAB.
Now, let k be the similarity ratio between triangles EBF and CAB.
.
Figure 10
Thus, triangle EBF has sides with lengths ka, kb, and kc. Since FB = FD, FD = kc. Also, since the opposite sides of a rectangle are congruent, b = ka + kc and c = a + kb. By solving for k, we have
and
Thus,
By cross-multiplication,
Therefore,
and we have completed the proof.
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The next proof of the Pythagorean Theorem that will be presented is one that begins with a right triangle. In the next figure, triangle ABC is a right triangle. Its right angle is angle C.
Figure 11
Next, draw CD perpendicular to AB as shown in the next figure.
Figure 12
Triangle 1
Compare triangles 1 and 3:
Triangle 1 (green) is the right triangle that we began with prior to constructing CD. Triangle 3 (red) is one of the two triangles formed by the construction of CD.
Figure 13
Triangle 1. Triangle 3.
By comparing these two triangles, we can see that
Compare triangles 1 and 2:
Triangle 1 (green) is the same as above. Triangle 2 (blue) is the other triangle formed by constructing CD. Its right angle is angle D.
Figure 14
Triangle 1. Triangle 2.
By comparing these two triangles, we see that
By adding equations 3 and 4 we get:
From Figures 11 and 12, with CD, we have that (p + q) = c. By substitution, we get
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The next proof of the Pythagorean Theorem that will be presented is one in which a trapezoid will be used.
Figure 15
By the construction that was used to form this trapezoid, all 6 of the triangles contained in this trapezoid are right triangles. Thus,
Area of Trapezoid = The Sum of the areas of the 6 Triangles
And by using the respective formulas for area, we get:
We have completed the proof of the Pythagorean Theorem using the trapezoid.
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The next proof of the Pythagorean Theorem that I will present is one that can be taught and proved using puzzles. These puzzles can be constructed using the Pythagorean configuration and then, dissecting it into different shapes.
Before the proof is presented, it is important that the next figure is explored since it directly relates to the proof.
Figure 16
In this Pythagorean configuration, the square on the hypotenuse has been divided into 4 right triangles and 1 square, MNPQ, in the center. Since MN = AN - AM = a - b. Each side of square MNPQ has length of a - b. This gives the following:
Area of Square on the hypotenuse = Sum of the Areas of the 4 triangles and the Area of Square MNPQ
As mentioned above, this proof of the Pythagorean Theorem can be further explored and proved using puzzles that are made from the Pythagorean configuration. Students can make these puzzles and then use the pieces from squares on the legs of the right triangle to cover the square on the hypotenuse. This can be a great connection because it is a "hands-on" activity. Students can then use the puzzle to prove the Pythagorean Theorem on their own.
Figure 17
To create this puzzle, copy the square on BC twice, once placed below the square on AC and once to the right of the square on AC as shown in Figure 17.
Proof Using Figure 17:
Triangle CDE is congruent to triangle ACB by leg-leg.
In triangle ACB, m<ACB =90 and the sides have lengths a, b, c.
In triangle CDE, m<CDE =90 and the sides have lengths a,b, c.
Triangle EGH is congruent to triangle ACB by leg-leg. The m<EGH =90 and its sides have lengths a and c. Since EF=b-a=AI, EG=b. Thus the diagonals CE and EH are both equal to c.
(Note: Pieces 4 and 7, and pieces 5 and 6 are not separated.)
By calculating the area of each piece, it can be shown that
Area 1:
Area 2:
Area 3:
Area 6 (and Area 5):
Area 7 (and Area 4):
By adding all of these areas together we get the following result:
and
Thus, we have proved the Pythagorean Theorem for the puzzle.