一道中考压轴题,帮忙解答,急如图①,在平面直角坐标系中,有一个矩形ABCD,边AB在x轴正半轴上,边CD在第一象限,AB

(1) A(m, 0), B(m+1, 0), C(m+1, 4), D(m, 4)
EH = HC = m+1, BH = BC - HC = 4 - (m+1) = 3 - m, H(m+1, 3 - m)
E(0, 3 - m)
(2)
EF与HB平行, 只需EF = BH即可
FG = GD = m, AG = AD - GD = 4 - m, G(m, 4 - m), F(0, 4-m)
EF = 4-m - (3-m) = 1
BH = 3- m
EF = BH, 3 - m = 1, m = 2
(3)
PQ的方程:y = 2 - x (i)
首先,m >3时,折叠后EF不可能在y轴上. 0 < m ≤ 3时, 须分三种情况考虑:
(a) 0 < m ≤ 1
此时E的纵坐标3-m > 2, P在E下方, PQ与AD, BC相交(交点R, S)
分别取x = m, x = m + 1, 由(1)可得:
R(m, 2-m), S(m+1, 1 - m)
阴影部分为梯形ABSR, S = (1/2)(AR + BS)*AB
= (1/2)(2-m + 1-m)*1 = (3 - 2m)/2
(b) 1 < m ≤ 2
此时E的纵坐标3-m < 2, F的纵坐标4 - m > 2, P在EF上, Q在AB上, PQ交EG于U
取y = 3-m, 由(1)可得:
U(m-1, 3 - m)
由(a): R(m, 2-m)
阴影部分为两个三角形EPU, ABR
S = (1/2)EU*EP + (1/2)AQ*AR
= (1/2)(m-1)(2 - 3 + m) + (1/2)(2-m)(2-m)
= (1/2)[(m-1)² + (m - 2)²]
(c) 2 < m ≤ 3
此时A在Q右方, PQ与EG, FH相交(交点U, V)
分别取y = 3 - m, y= 4-m, 由(1)可得:
U(m-1, 3-m), V(m-2, 4 - m)
阴影部分为梯形EUVF, S = (1/2)(EU + FV)*EF
= (1/2)(m-1 + m - 2)(40m - 3 + m)
= (2m -3)/2
(4)
AE斜率 = (0 - 3 + m)/(m- 0) = (m-3)/m
GH斜率 = ((4-m-3+m)/(m-m-1) = -1
(m-3)/m = -1
m = 3/2