CD平行于AB 可设CD边所在直线为：y=x+k，（k为在y轴的截距） AB、CD边的距离 = a = |k-4|/根号2 ...(1) CD与抛物线y^2=x交点：C(x1,y1), D(x2,y2) ==> y^2 = x = y-k ===> y^2 -y +k = 0 ==> |CD|^2 = (x1-x2)^2 +(y1-y2)^2 = 2*(y1-y2)^2 = 2*[(y1+y2)^2 -4*y1*y2] ==> [|k-4|/根号2]^2 =a^2= |CD|^2 =2*[(y1+y2)^2 -4*y1*y2] =2*[(1)^2 -4*k] ==> k = -2 或 -6 因此，正方形的面积 = a^2 = [|k-4|/根号2]^2 = 18 or 50。
因在正方形内 AB∥CD 则设CD:x-y+b=0 所以两直线距离d=|b-4|/根号2 将CD代入得 y^2-y+b=0 设C(x1,y1) D(x2,y2) 则y1+y2=1 y1y2=b 代入CD得x1+x2=y1-b+y2-b=1-2b x1x2=(y1-b)(y2-b)=y1y2-b(y1+y2)+b^2=b^2 |CD|=根号下[(x1-x2)^2+(y1-y2)^2]=d 又因为 (x1-x2)^2=(x1+x2)^2-4x1x2 (y1-y2)^2=(y1+y2)^2-4y1y2 所以 |CD|=根号下(2-8b)=|b-4|/根号2 平方得 2-8b=(b^2-8b+16)/2 (2-8b>0) 得b=-2或-6 S=d^2=(b-4)^2/2=18或50
答：解: 设正方形边长为a. 在等腰Rt△ABC和ACD中,斜边AC中点为M. 则: AC=(√2)a CM=AM=(√2)×a/2 BM=DM=CM (...详情>>
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