**Problem**
The area of a trapezoid is 2 and the sum of its diagonals is 4. Find the height of the trapezoid.

*Solution*

Let $ABCD$ be the trapezoid, where $AB$ is parallel to $CD$. Notice that by increasing the length of $AB$ and decreasing $CD$ by the same amount, the conditions of the problem still hold (the diagonals $AD$ and $CB$ are being translated). Thus, without loss of generality, suppose that $ABCD$ is a paralellogram, i.e. $AB=CD=2d$. Therefore $CB$ and $AD$ meet at their midpoint $M$. Hence the triangle $CMD$ has $CD=2d$ and $CM+MD=2$, hence by locating $C$ at $(-d,0)$ and $D$ at $(d,0)$ we have that $M$ satisfies the ellipse

$x^2+\frac{y^2}{1-d^2}=1$

together with the condition $(2d)(2y)=2$. Therefore

$x^2=1-\frac{d^2y^2}{d^2(1-d^2)}=1-\frac{1/4}{d^2(1-d^2)}$,

and then $d^2(1-d^2)\geq 1/4$, but by AG-GM, $d^2(1-d^2)\leq 1/4$, thus $d=1/\sqrt{2}$, $x=0$, and $y=\frac{1}{\sqrt{2}}$ and the height of $ABCD$ is $\sqrt{2}$.