Wednesday, May 30, 2012

proving pythagoras theorem using algebraic

Algebraic proofs

Diagram of the two algebraic proofs
The theorem can be proved algebraically using four copies of a right triangle with sides a, b and c, arranged inside a square with side c as in the top half of the diagram.[16] The triangles are similar with area \tfrac12ab, while the small square has side ba and area (ba)2. The area of the large square is therefore
(b-a)^2+4\frac{ab}{2} = (b-a)^2+2ab = a^2+b^2. \,
But this is a square with side c and area c2, so
c^2 = a^2 + b^2. \,
A similar proof uses four copies of the same triangle arranged symmetrically around a square with side c, as shown in the lower part of the diagram.[17] This results in a larger square, with side a + b and area (a + b)2. The four triangles and the square side c must have the same area as the larger square,
(b+a)^2 = c^2 + 4\frac{ab}{2} = c^2+2ab,\,
giving
c^2 = (b+a)^2 - 2ab = a^2 + b^2.\,

No comments:

Post a Comment