Suppose in the isosceles trapezoid $ABCD$,$AB \parallel CD$,$I$ is the incenter and $AB=2x,BC=AD=h,CD=2y$.Let $G$ and $H$ be the projections of $I$ onto $AB$ and $CD$ and WLOG $AB<CD$.Easy to see that $G,I,H$ are collinear and by symmetry,$G,H$ are the midpoints of $AB,CD$.Since the area of $ABCD=28$,$x+y=\frac{28}{4}=7$.By Pitot's theorem on $ABCD$,we have $2(x+y)=2h$ or $x+y=7=h$.Let $X$ be the projection of $B$ onto $CD$.Considering the right angled $\bigtriangleup BXC$,we have $BX=4,CX=7-2x,BC=7$,by Pythagoras theorem,$x=\frac{7-\sqrt{33}}{2}$ and we are done.