Proof the Pythagorean theory

Answer 1

area squared of slanted face triangle (the "hpypotenuse") of right tetrahedon points (0,0,0), (a,0,0),(0,b,0),(0,0,c)

equals the three right triangle areas squared and summed

(hypotenuse triangle area)²=(1/4)(a²b²+a²c²+b²c²)= sum of areas² of right triangles (the legs)

is this the specific Pythagorean relation you are referring to? There are higher dimension forms and of course the distance form.

for this form, see base of slanted triangle is √(a²+b²)

and height of slanted triangle is √((a²b²/(a²+b²))+c²)

(this takes a bit of algebra and coordinate geometry)

so the area² of the slanted triangle is [(1/2)(√(a²+b²))( √((a²b²/(a²+b²))+c²))]² (the "hypotenuse")

= sum of area²=(1/4)((ab)²+(ac)²+(bc)²) of right triangles (the "three legs") since each right triangle has area equal to (1/2) the rectangle area of orthogonal sides.

for a four dimension right simplex of five points (0,0,0,0),(a,0,0,0),(0,b,0,0),(0,0,c,0),(0,0,0,d)

the 3d tetrahedron "face that is not a right tetrahedron (four points (a,0,0,0),(0,b,0,0),(0,0,c,0),(0,0,0,d)

,the "hypotenuse" face) has a volume, according to pythagorean theorem generalized, the volume squared is equal to the sum of the volumes squared of the four "leg":faces

that is volume² of (a,0,0,0),(0,b,0,0),(0,0,c,0),(0,0,0,d) tetrahedron in a right 4d simplex

=(1/36)((abc)²+(abd)²+(acd)²+(bcd)²)

since for a right tetrahedron the volume equals (1/6) volume of cube with sides equal three orthogonal side of tetrahedron.

Answer 2

There's about 100 different proofs of the Pythagorean Theorem.

Former President James A. Garfield provided one of them, when he was in the House of Representatives. See the New England Journal of Education, April 1, 1876 issue. Mathematical Association of America also published his proof in a more recent article, along with a modernized version of JAG's proof. It's geometrical. constructing a trapezoid with the hypotenuse as one side. then the opposite side going through the point of the right angle. then construct 2 parallel lines for the other two sides. you create two triangles of equal area. then caluclate the areas of those two plus the original right triangle. set that sum = the trapazoid area, simplify and c^2 = a^2 + b^2

Try googling Pythagorean Theorem Proofs.

Garfield was perhaps America's most intelligent president. He was ambidextrous and could write with both hands simultaneously, one hand in Latin, the other in Greek. Math is Greek to many, but Garfiled knew math like the back of his hand. He was also possibly the youngest Civil War general. Now no one has heard of him, just the comic strip cat keeps his name alive.

another method is try a specific example, a right triangle with sides 3, 4 and 5. 5^2 = 3^2 + 4^2

construct squares on each side of the triangle, largest is 5 by 5, other 2 squares are 3 by 3 and 4 by 4

the squares have areas 25, 9 and 16. 9+16 = 25