解过点B做BH垂直x轴与H设B(x1,y1),则A(x2,y2)其中y1=y2则S梯形ABHC=1/2(x1-x2)y2则SΔBOH=1/2x1y1则SΔAOB=S梯形ABHC-SΔBOH=1/2(x1-x2+x2)y2-1/2x1y1=1/2x1y2--1/2x1y1=1/2x1y1-1/2x1y1=1/2×5-1/2×2=3/2