Slice off the corner of a right rectangular prism so that the result is a tetrahedron with three right triangular faces mutually perpendicular to each other. The fourth face is a triangle formed by the hypotenuses.
Prove: The sum of the squares of the areas of the three right triangles is equal to the square of the area of the fourth.
let the vertices have coordinates
(0,0,0)
(x,0,0)
(0,y,0)
(0,0,z)
Then the right triangle faces have areas:
xy/2=P
xz/2=Q
yz/2=R
for the 4th face has side lengths a,b,c given by
a^2=x^2+y^2
b^2=x^2+z^2
c^2=y^2+z^2
let s=(a+b+c)/2
then from herons formula we get that the area A of the 4th face is given by
A^2=s*(s-a)*(s-b)*(s-c)
substitute, expand, and simplify to get
A^2=x^2y^2/4+x^2z^2/4+y^2z^2/4
A^2=(xy/2)^2+(xz/2)^2+(yz/2)^2
A^2=P^2+Q^2+R^2
Thus the square of the area of the 4th face is equal to the sum of the squares of the areas of the other 3 sides.
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Posted by Daniel
on 2017-06-16 11:56:04 |
solution
