The sides of a trapezoid are 5, 8, 11, and 13, and its diagonals are also integer numbers; what are they?

(In reply to re: Solution (Proof by Hand) by David Shin)

You're right David. I did leave out the possibility that the there may only be 1 obtuse angle. I guess whenever I picture a trapezoid, it's always with double obtuse angle. Never seem to picture the trapezoid that looks more like a parallelogram.

Here goes. It turns out that for this second case that the equations come out identical for x1. The set up is the same as before. Except it looks more like a parallelogram.

This time drop the perpendicular from the top left down and "raise" the perpendicular from the bottom right up. Let the left altitude break the segment d into x1 and d-x1 with x1 the leg of the right triangle. Similarly, x2 for the right side.

x1^2+h^2=a^2 (Pythagorean Theorem)
x2^2+h^2=c^2 (Pythagorean Theorem)
d-x1=b-x2 (trapezoid and how we split up the segments)

x1^2-x2^2=a^2-c^2.
x2=b-d+x1

As you can see this x2 works out to be the negative of the other one. But since we're dealing with x2^2, the answer for x1 comes out to be the same:

x1=(a^2+b^2-c^2+d^2-2bd)/(2d-2b)

The difference is that x2 is that x2=b-d+x1.

Again, checking the 24 possibilities, only four of them have positive x1 and x2's with x1<d and x2<b. However, for all of these x1>a which cannot be since a is the hypotenuse.

Therefore, there is no such trapezoid of the given lengths where one angle of one side is acute and the other is obtuse. This trapezoid must have obtuse angles on the same side, which I addressed in the earlier post.

Posted by np_rt on 2005-01-18 00:53:18