din C ducem perpendic, pe AB in E
EB = AB - CD = 2x -3 - x -2 = x -5
Δ CEB drept
BC² = CE² + EB² ; x² = 15² + ( x -5 ) ² ; x² =225 + x² -10x + 25
10x = 225 +25 ; 10x =250 ; x =25
AB=2·25 -3 = 47 ; BC=25 ; CD=27
linia mijlocie EF = ( AB +DC ) :2 = 37
perim ABEF = 47 + 25/2 + 37 + 15 /2 = 104
aria ABCD = ( AB + CD) ·AD /2 = ( 47 + 27 ) ·15 /2 = 555
