Three vertices of parallelogram $ABCD$ lie on rectangle $EFGH$. $BC=AF=10$, $CG=8$, and $AB=28$. Find $FG$.
BG = sqrt [ BC^2 -CG^2 ] = sqrt [ 10^2 - 8^2 ] = sqrt [100 - 64 ] = sqrt [ 36] = 6
FB =sqrt [ AB^2 - AF^2 ] =sqrt [28^2 - 10^2 ] = sqrt [ 784 -100 ] =sqrt [684] = 6sqrt (19)
FG = FB + BG = 6 + 6sqrt (19) = 6 ( 1 + sqrt (19) ) ≈ 32.15