Lesson The Pythagorean Theorem

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The Pythagorean Theorem



The Pythagorean Theorem establishes a remarkable relationship between the sides
of any right triangle. This theorem was proved by the Greek geometer Pythagoras
(Pythagoras of Samos), who lived from about 570 B.C. to about 475 B.C., and is
named after him. This Theorem is the basis and one of corner stones of the
Euclidean geometry. We give the Theorem and the proof close to as stated by
Euclid in his "Elements", famous books on Geometry.

The Pythagorean Theorem

The sum of the areas of the squares constructed on the legs of a right triangle  
is equal to the area of the square constructed on its hypotenuse
.

Figures 1 illustrates the Theorem, showing the right triangle ABC,
squares AGHC and CKMB constructed on its legs and the square ABFE
constructed on its hypotenuse. The Theorem states that the area of the square
ABFE is equal to the sum of the areas of squares AGHC and CKMB.

drawing%28+330%2C+345%2C+-3.0%2C+8.0%2C+-5.5%2C+6.0%2C+%0D%0A++++++++++++line%28+0%2C+0%2C+5%2C+++0+%29%2C+%0D%0A++++++++++++line%28+0%2C+0%2C+3.2%2C+2.4%29%2C%0D%0A++++++++++++line%28+5%2C+0%2C+3.2%2C+2.4%29%2C+%0D%0A%0D%0A++++++++++++locate+%28+0.1%2C+0%2C+++A%29%2C%0D%0A++++++++++++locate+%28+5.1%2C+0%2C+++B%29%2C%0D%0A++++++++++++locate+%28+3.4%2C+2.5%2C+C%29%2C%0D%0A%0D%0A++++++++++++line%28+0%2C++++0%2C+++-2.4%2C+3.2%29%2C+%0D%0A++++++++++++line%28+-2.4%2C+3.2%2C++0.8%2C+5.6%29%2C+%0D%0A++++++++++++line%28+0.8%2C++5.6%2C++3.2%2C+2.4%29%2C%0D%0A%0D%0A++++++++++++locate+%28-2.7%2C+3.2%2C+G%29%2C%0D%0A++++++++++++locate+%28+1.0%2C+5.9%2C+H%29%2C%0D%0A%0D%0A++++++++++++line%28+3.2%2C++2.4%2C++5.6%2C+4.2%29%2C+%0D%0A++++++++++++line%28+5.6%2C++4.2%2C++7.4%2C+1.8%29%2C+%0D%0A++++++++++++line%28+7.4%2C++1.8%2C++5.0%2C+0.0%29%2C%0D%0A%0D%0A++++++++++++locate+%28+5.8%2C+4.4%2C+K%29%2C%0D%0A++++++++++++locate+%28+7.4%2C+1.8%2C+M%29%2C%0D%0A%0D%0A++++++++++++line%28+0.0%2C++0.0%2C++0.0%2C+-5.0%29%2C+%0D%0A++++++++++++line%28+0.0%2C+-5.0%2C++5.0%2C+-5.0%29%2C+%0D%0A++++++++++++line%28+5.0%2C++0.0%2C++5.0%2C+-5.0%29%2C%0D%0A%0D%0A++++++++++++locate+%28+0.0%2C+-5.0%2C+E%29%2C%0D%0A++++++++++++locate+%28+5.0%2C+-5.0%2C+F%29%0D%0A%29
Figure 1. A right triangle and squares
        on its legs and its hypotenuse



The Proof of the Theorem

Let us draw the straight line CL perpendicularly to AB intersecting AB and EF at points
D and L respectively. Then the square ABFE is divided in two rectangles. Let us prove
that the rectangle ADLE has the same area as the square AGHC.

First, the triangle AEC has the area equal to half of the area of the rectangles ADLE,
because the triangle and the rectangle have the common base AE, and the altitude of the
triangle AEC to this base is congruent to the rectangle side AD.

Second, the triangle AGB has the area equal to half of the area of the square AGHC,
because the triangle and the square have the common base AG, and the altitude of the
triangle AGB to this base is congruent to the square side AC.

Third, triangles AEC and AGB are congruent having congruent sides AG and AC,
AC and AE (listed by pairs), as well as congruent angles GAB and CAE
concluded between these sides. Congruency of angles follows from the fact that each of
mentioned angles is equal to the right angle plus the common angle CAB.

drawing%28+330%2C+345%2C+-3.0%2C+8.0%2C+-5.5%2C+6.0%2C+%0D%0A++++++++++++line%28+0%2C+0%2C+5%2C+++0+%29%2C+%0D%0A++++++++++++line%28+0%2C+0%2C+3.2%2C+2.4%29%2C%0D%0A++++++++++++line%28+5%2C+0%2C+3.2%2C+2.4%29%2C+%0D%0A%0D%0A++++++++++++locate+%28+0.1%2C+0%2C+++A%29%2C%0D%0A++++++++++++locate+%28+5.1%2C+0%2C+++B%29%2C%0D%0A++++++++++++locate+%28+3.4%2C+2.5%2C+C%29%2C%0D%0A%0D%0A++++++++++++line%28+0%2C++++0%2C+++-2.4%2C+3.2%29%2C+%0D%0A++++++++++++line%28+-2.4%2C+3.2%2C++0.8%2C+5.6%29%2C+%0D%0A++++++++++++line%28+0.8%2C++5.6%2C++3.2%2C+2.4%29%2C%0D%0A%0D%0A++++++++++++locate+%28-2.7%2C+3.2%2C+G%29%2C%0D%0A++++++++++++locate+%28+1.0%2C+5.9%2C+H%29%2C%0D%0A%0D%0A++++++++++++line%28+3.2%2C++2.4%2C++5.6%2C+4.2%29%2C+%0D%0A++++++++++++line%28+5.6%2C++4.2%2C++7.4%2C+1.8%29%2C+%0D%0A++++++++++++line%28+7.4%2C++1.8%2C++5.0%2C+0.0%29%2C%0D%0A%0D%0A++++++++++++locate+%28+5.8%2C+4.4%2C+K%29%2C%0D%0A++++++++++++locate+%28+7.4%2C+1.8%2C+M%29%2C%0D%0A%0D%0A++++++++++++line%28+0.0%2C++0.0%2C++0.0%2C+-5.0%29%2C+%0D%0A++++++++++++line%28+0.0%2C+-5.0%2C++5.0%2C+-5.0%29%2C+%0D%0A++++++++++++line%28+5.0%2C++0.0%2C++5.0%2C+-5.0%29%2C%0D%0A%0D%0A++++++++++++locate+%28+0.0%2C+-5.0%2C+E%29%2C%0D%0A++++++++++++locate+%28+5.0%2C+-5.0%2C+F%29%2C%0D%0A%0D%0A+++++++blue%28line%28+3.2%2C++2.4%2C+3.2%2C+-5.0%29%29%2C%0D%0A++++++++red%28line%28-2.4%2C++3.2%2C+5.0%2C++0.0%29%29%2C%0D%0A++++++++red%28line%28+0.0%2C+-5.0%2C+3.2%2C++2.4%29%29%2C%0D%0A%0D%0A++++++++++++locate+%28+3.3%2C++0.0%2C+D%29%2C%0D%0A++++++++++++locate+%28+3.3%2C+-5.0%2C+L%29%0D%0A%29
Figure 2. The proof of the Theorem

Since triangles AEC and AGB are congruent, they have same areas.
Thus, this chain of arguments proves that the rectangle ADLE has the same area as the square BGHC.
Similarly, the rectangle DBFL has the same area as the square CKMB.
This implies that the area of the square ABFE is equal to the sum of the areas of squares AGHC and CKMB.
The Theorem is proved.

The Pythagorean Theorem has another, equivalent formulation:

If a and b denote the lengths of legs of the right triangle and c denotes the length of its hypotenuse, then a%5E2+%2B+b%5E2+=+c%5E2.

This formulation is equivalent to that given at the beginning of this lesson, because a%5E2, b%5E2 and c%5E2 are areas of respective squares mentioned in the original formulation.

The Pythagorean Theorem has many other proofs different from the given in this lesson.
Examples of different proofs of the Pythagorean Theorem are provided in the lesson More proofs of the Pythagorean Theorem.

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