Isosceles trapezoid $ABCD$ has parallel sides $\overline{AD}$ and $\overline{BC}$ with $AD=22$ and $BC=10$. The diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. What is the area of the trapezoid?
Let the diagonals meet at $P$. The trapezoid is isosceles, so $\angle PAD=\angle PDA$ and $PA=PD$; since $\angle APD=90^\circ$, triangle $APD$ is an isosceles right triangle, and its altitude from $P$ to $AD$ is $\frac{AD}2=11$.
In the same way, triangle $BPC$ is an isosceles right triangle with altitude $\frac{BC}2=5$ from $P$ to $BC$.
So the height of the trapezoid is $11+5=16$ and its area is $\frac12(22+10)\cdot16=\boxed{256}$.
