2014 AMC 10B Problem 21

Let the feet of the perpendiculars from $D$ and $C$ to $AB$ be $E$ and $F$, respectively. As drawn, take $AD=10$ and $BC=14$; interchanging the two legs only reflects the trapezoid. Let $AE=x$ and let the altitude be $h$. Since $EF=CD=21$ and $AB=33$, we have $FB=12-x$. The two right triangles give $$10^2=x^2+h^2$$ and $$14^2=(12-x)^2+h^2.$$ Subtracting yields $96=144-24x$, so $x=2$ and $h^2=96$. The shorter diagonal is $AC$, whose horizontal displacement is $AE+EF=2+21=23$. Therefore $$AC=\sqrt{23^2+96}=\sqrt{625}=25.$$ Thus, the correct answer is B .