如图,将一矩形OABC放在直角坐标系中,O为坐标原点,点A在y轴正半轴上,点E是边AB上的

(1)设A(0, a), B(b, a), C(b, 0) 反比例函数y=k/x与AB(y = a)的交点为E(k/a, a) 反比例函数y=k/x与BC(x = b)的交点为F(b, k/b) AE = k/a, CF = k/b S1 + S2 = (1/2)*OA*AE + (1/2)OC*CF = (1/2)a*k/a + (1/2)b*k/b) = k/2 + k/2 =