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Proof For Pythagorean Theorem Jigar Bhatt 9769907927
Pythagorean Theorem ,[object Object],[object Object],[object Object],[object Object],A C B b a c
Proof  ,[object Object],[object Object],[object Object],A C B b a c
[object Object],[object Object],Proof  A C B b a c X 1 X 2
Proof  A C B b a c X 1 X 2 K 1 K 2 Now draw a line X 1 K 1  such that X 1 K 1  = b and X 1 K 1  is perpendicular  to CX 1 . Draw a line X 2 K 2  such that X 2 K 2  = a, and X 2 K 2  is perpendicular  to CX 2 .  b a b a
Proof  A C B b a c X 1 X 2 K 1 K 2 Extend a line X 1 K 1  to point X 3  such that K 1 X 3  = a and X 1 K 1. to CX 1 . Extend a line X 2 K 2  to point X 3 .   b a b a a X 3
Proof  A C B b a c X 1 X 2 K 1 K 2 Constructing this lines we have made a square whose side is  (a+b). If we observe carefully we have four right triangle at each  corner of square.  b a b a a X 3 b
Proof  A C B b a c X 1 X 2 K 1 K 2 We have four congruent right triangles. A square with side c, and  a square with side (a+b). Area of square with side (a+b) = 4 x Area of right triangle + Area  of square with side c  b a b a a X 3 c c c
Proof  Area of square with side (a+b) = 4 x Area of right triangle + Area  of square with side c  (a+b) 2  = 4 x (1/2) a x b + c 2 a2 + 2ab + b2 = 2ab + c 2   - 2ab    - 2ab a 2  + b 2  = c 2 Hence Proved.
Thank you

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Pythagorean Proof

  • 1. Proof For Pythagorean Theorem Jigar Bhatt 9769907927
  • 2.
  • 3.
  • 4.
  • 5. Proof A C B b a c X 1 X 2 K 1 K 2 Now draw a line X 1 K 1 such that X 1 K 1 = b and X 1 K 1 is perpendicular to CX 1 . Draw a line X 2 K 2 such that X 2 K 2 = a, and X 2 K 2 is perpendicular to CX 2 . b a b a
  • 6. Proof A C B b a c X 1 X 2 K 1 K 2 Extend a line X 1 K 1 to point X 3 such that K 1 X 3 = a and X 1 K 1. to CX 1 . Extend a line X 2 K 2 to point X 3 . b a b a a X 3
  • 7. Proof A C B b a c X 1 X 2 K 1 K 2 Constructing this lines we have made a square whose side is (a+b). If we observe carefully we have four right triangle at each corner of square. b a b a a X 3 b
  • 8. Proof A C B b a c X 1 X 2 K 1 K 2 We have four congruent right triangles. A square with side c, and a square with side (a+b). Area of square with side (a+b) = 4 x Area of right triangle + Area of square with side c b a b a a X 3 c c c
  • 9. Proof Area of square with side (a+b) = 4 x Area of right triangle + Area of square with side c (a+b) 2 = 4 x (1/2) a x b + c 2 a2 + 2ab + b2 = 2ab + c 2 - 2ab - 2ab a 2 + b 2 = c 2 Hence Proved.