@ Proof of the Height of a Trapezoid Given Only the Lengths of it's Sides

In the top part of the diagram, the trapezoid has bases b and d, legs a and c, and height h.
In the bottom part of the diagram the right triangle defined by leg c, height h, and another side with length less than d has been translated to the left so that the translated triangle shares it’s height h with the right triangle defined by leg a, height h, and another side which has length less than d.

A triangle with sides of lengths a,c, and d − b has been formed.
The height of the triangle is identical to the height of the original trapezoid.
Applying the standard formula for the area of the triangle, the trapezoid area is:

$A=h\frac{d-b}{2}$

Using Heron’s formula to calculate the area:

$A=\frac{\sqrt{(d-b+a+c)(-d+b+a+c)(d-b-a+c)(d-b+a-c)}}{4}$

Both formulas express the same area, so:

$h=\frac{\sqrt{(d-b+a+c)(-d+b+a+c)(d-b-a+c)(d-b+a-c)}}{2(d-b)}$

Without_loss_of_generality it is assumed that $d > b$ , thus the absolute value in the denominator does not appear in the calculation of height.

References

http://www.murderousmaths.co.uk/books/trap.htm

Notes

John Blackburne, of Durham County, United Kingdom is convinced that the content of this post is not notable and does not belong on Wikipedia. If he’s wrong feel free to let him know.