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.
Is this problem the same as http://perplexus.info/show.php?pid=9293&cid=53702.
I didn't see it in the queue or I would have mentioned it.
Posted by broll
on 2017-06-17 07:03:51