# Trapezoid With Perpendicular Diagonals

The diagonals of a trapezoid are perpendicular and have lengths 8 and 10. Find the length of the median of the trapezoid.

I have no idea about this problem except assuming that the trpezoid is a rhombus. This is a duplicate, but the solution on The diagonals of a trapezoid are perpendicular and have lengths 8 and 10. Find the length of the median of the trapezoid. is a little bit too advanced for me. Can anyone solve this using a simpler method?

Assume that the diagonals cut themselves in four segments with lengths $a,b,c,d$, in such a way that $b+d=10,a+c=8$ and the right triangle with legs $a,b$ is similar to the right triangle with legs $c,d$, so that $\frac{b}{a}=\frac{d}{c}$ or, equivalently, $ad=bc$. That leads to the equation: $$a(10-b)=b(8-a)$$ that is equivalent to $10a=8b$, hence $\frac{b}{a}=\frac{d}{c}=\frac{5}{4}$. So we have that all the trapezoids with perpendicular diagonals such that $a=x,\,b=\frac{5}{4}x,\,c=y,\,d=\frac{5}{4}y,\,x+y=8$ meet the given constraints. For such trapezoids, the bases have length $$\frac{x}{4}\sqrt{41},\qquad \frac{y}{4}\sqrt{41}$$ by the Pythagorean theorem, hence the "median" has length $$\frac{x+y}{8}\sqrt{41} = \color{red}{\sqrt{41}}.$$
In general, if a trapezoid has perpendicular diagonals with length $\ell_1,\ell_2$, the segment joining the midpoints of the diagonal sides has length $\frac{1}{4}\sqrt{\ell_1^2+\ell_2}$. Here it comes a proof (almost) without words: the red segments have the same length. 