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Proving the Pythagorean Theorem:

y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c . Proving the Pythagorean Theorem:.

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Proving the Pythagorean Theorem:

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  1. y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c Proving the Pythagorean Theorem: y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c a + b = c 2 2 2 y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c

  2. In any right triangle, the area of the square whose side is the hypotenuse(the side opposite the right angle) is equal to thesum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle). c 2 y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c a 2 b 2 y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c

  3. y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c Think of each of the expressions (the squares) in the equation as actual squares y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c

  4. y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c Watch as we now prove the smaller two squares are equivalent to the larger square. y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c

  5. y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c Watch as we now prove the smaller two squares are equivalent to the larger square. y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c

  6. y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c Watch as we now prove the smaller two squares are equivalent to the larger square. y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c

  7. y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c Watch as we now prove the smaller two squares are equivalent to the larger square. y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c

  8. As you can see there are now two equivalent squares. This visualization proves that; a + b = c 2 2 2 y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c y = mx + b V = lwh P = 2l + 2h P = 4s P = a + b + c P = ns A = bh A = lw a(b + c) = ab + ac (a x b) x c = a x (b x c) = a x b x c

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