Pythagorean Theorem Calculator
What is Pythagorean Theorem?
Pythagorean Theorem
The sum of the areas of the two squares on the legs (a and b) equals the area of the square on the hypotenuse (c).
In mathematics, the Pythagorean theorem, also known as Pythagoras' theorem, is a fundamental relation in Euclidean geometry among the three sides of a right triangle. It states that the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares on the other two sides. This theorem can be written as an equation relating the lengths of the sides a, b and c, often called the "Pythagorean equation":
a2+b2=c2
where c represents the length of the hypotenuse and a and b the lengths of the triangle's other two sides. The theorem, whose history is the subject of much debate, is named for the ancient Greek thinker Pythagoras.
The theorem has been given numerous proofs – possibly the most for any mathematical theorem. They are very diverse, including both geometric proofs and algebraic proofs, with some dating back thousands of years. The theorem can be generalized in various ways, including higher-dimensional spaces, to spaces that are not Euclidean, to objects that are not right triangles, and indeed, to objects that are not triangles at all, but n-dimensional solids. The Pythagorean theorem has attracted interest outside mathematics as a symbol of mathematical abstruseness, mystique, or intellectual power; popular references in literature, plays, musicals, songs, stamps and cartoons abound.
Rearrangement Proof
The two large squares shown in the figure each contain four identical triangles, and the only difference between the two large squares is that the triangles are arranged differently. Therefore, the white space within each of the two large squares must have equal area. Equating the area of the white space yields the Pythagorean theorem, Q.E.D.
Heath gives this proof in his commentary on Proposition I.47 in Euclid's Elements, and mentions the proposals of Bretschneider and Hankel that Pythagoras may have known this proof. Heath himself favors a different proposal for a Pythagorean proof, but acknowledges from the outset of his discussion "that the Greek literature which we possess belonging to the first five centuries after Pythagoras contains no statement specifying this or any other particular great geometric discovery to him." Recent scholarship has cast increasing doubt on any sort of role for Pythagoras as a creator of mathematics, although debate about this continues.
Other Proofs of the Theorem
This theorem may have more known proofs than any other (the law of quadratic reciprocity being another contender for that distinction); the book The Pythagorean Proposition contains 370 proofs.
Euclid's Proof
In outline, here is how the proof in Euclid's Elements proceeds. The large square is divided into a left and right rectangle. A triangle is constructed that has half the area of the left rectangle. Then another triangle is constructed that has half the area of the square on the left-most side. These two triangles are shown to be congruent, proving this square has the same area as the left rectangle. This argument is followed by a similar version for the right rectangle and the remaining square. Putting the two rectangles together to reform the square on the hypotenuse, its area is the same as the sum of the area of the other two squares. The details follow.
Let A, B, C be the vertices of a right triangle, with a right angle at A. Drop a perpendicular from A to the side opposite the hypotenuse in the square on the hypotenuse. That line divides the square on the hypotenuse into two rectangles, each having the same area as one of the two squares on the legs.
For the formal proof, we require four elementary lemmata:
- If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, then the triangles are congruent (side-angle-side).
- The area of a triangle is half the area of any parallelogram on the same base and having the same altitude.
- The area of a rectangle is equal to the product of two adjacent sides.
- The area of a square is equal to the product of two of its sides (follows from 3).
Next, each top square is related to a triangle congruent with another triangle related in turn to one of two rectangles making up the lower square.
The proof is as follows:
- Let ACB be a right-angled triangle with right angle CAB.
- On each of the sides BC, AB, and CA, squares are drawn, CBDE, BAGF, and ACIH, in that order. The construction of squares requires the immediately preceding theorems in Euclid, and depends upon the parallel postulate.
- From A, draw a line parallel to BD and CE. It will perpendicularly intersect BC and DE at K and L, respectively.
- Join CF and AD, to form the triangles BCF and BDA.
- Angles CAB and BAG are both right angles; therefore C, A, and G are collinear. Similarly for B, A, and H.
- Angles CBD and FBA are both right angles; therefore angle ABD equals angle FBC, since both are the sum of a right angle and angle ABC.
- Since AB is equal to FB and BD is equal to BC, triangle ABD must be congruent to triangle FBC.
- Since A-K-L is a straight line, parallel to BD, then rectangle BDLK has twice the area of triangle ABD because they share the base BD and have the same altitude BK, i.e., a line normal to their common base, connecting the parallel lines BD and AL. (lemma 2)
- Since C is collinear with A and G, square BAGF must be twice in area to triangle FBC.
- Therefore, rectangle BDLK must have the same area as square BAGF = AB2.
- Similarly, it can be shown that rectangle CKLE must have the same area as square ACIH = AC2.
- Adding these two results, AB2 + AC2 = BD × BK + KL × KC
- Since BD = KL, BD × BK + KL × KC = BD(BK + KC) = BD × BC
- Therefore, AB2 + AC2 = BC2, since CBDE is a square.
This proof, which appears in Euclid's Elements as that of Proposition 47 in Book 1, demonstrates that the area of the square on the hypotenuse is the sum of the areas of the other two squares. This is quite distinct from the proof by similarity of triangles, which is conjectured to be the proof that Pythagoras used.
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