27．（10分）图①是一张∠AOB＝45°的纸片折叠后的图形，P、Q分别是边OA、OB上的点，且OP＝2 cm．将
∠AOB沿PQ折叠，点O落在纸片所在平面内的C处． （1）①当PC∥QB时，OQ＝ ▲ cm；
②在OB上找一点Q，使PC⊥QB（尺规作图，保留作图痕迹）； （2）当折叠后重叠部分为等腰三角形时，求OQ的长． B Q C O P A ①
- 6 -
B O P A 备用图1 （第27题）
B O P A 备用图2
2018年中考第一次模拟调研 数学参考答案及评分标准
说明：本评分标准每题给出了一种或几种解法供参考，如果考生的解法与本解答不同，参照本评分标准的精神给分．
一、选择题（每小题2分，共计12分）
题号 答案 1 A 2 C 3 B 4 A 5 B 6 D 二、填空题（每小题2分，共计20分）
7．x≤1 8．a(a+1)(a-1) 9．-2 10．6.75×104 11．＝ 7π2+2 12．（n，m） 13． 14．108° 15．a 16. 18°
3 2三、解答题（本大题共10小题，共计88分） 17．（本题6分）
a2＋2a＋1a2－1
解：原式 ＝÷
aaa2＋2a＋1a ＝·2
aa－1
(a＋1) 2a
＝·
a(a＋1)(a－1)
a＋1
＝． ·························································································· 6分
a－118．（本题7分）
解：解不等式①，得x＜2． ············································································· 2分
解不等式②，得x≥ —1． ········································································· 4分 所以，不等式组的解集是－1≤x＜2． ························································ 5分 画图 -1 0 2 ································································ 7分 1 19．（本题7分）
解：（1）∵四边形ABCD是平行四边形
∴AE∥CF
∴∠DAC＝∠BCA ········································································ 1分 在△AOE和△COF中
??∠DAC=∠ACB
? AO=CO ?∠AOE=∠COF?
∴△AOE≌△COF（ASA） ··································································· 3分 ∴AE＝CF
∴四边形AFCE是平行四边形 ·························································· 5分
- 7 -
（2）② ····························································································· 7分 20．（本题8分）
解：设批发了西红柿x千克，豆角y千克
?x?y?40 由题意得：? ····················································· 3分
3.6x?4.6y?180??x?4 解得：? ……………………………………………6分
y?36? (5.4 — 3.6)× 4＋(7.5 — 4.6)× 36 ＝ 111.6（元） ········································ 7分
答：卖完这些西红柿和豆角能赚111.6元． ···················································· 8分 21．（本题8分） 解：（1）1 ································································································ 2分 4
①（100，110）；②（100，120）；③（100，125）；④（110，120）；
⑤（110，125）；⑥（120，125） ··························································· 6分 总重量超过232g的结果有2种，即（110，125），（120，125） ··················· 7分 因此，总重量超过232g的概率是
22．（本题8分）
解：（1）① ···························································································· 2分 （2）① 60°，30° ··················································································· 4分
② 225 ··················································································· 6分 （3）两所学校都可以选择只要理由正确皆可得分 ············································· 8分 选择河西中学，理由是平均分相同，河西中学极差和方差较小，河西中学成绩更稳定． 选择复兴中学，理由是平均分相同，复兴中学A，B类频率和高，复兴中学高分人数更多． 23．（本题8分）
解：过点A作AP⊥EF，垂足为P
∵AD⊥DE，∴∠ADE＝90° ∵AD∥EF，∴∠DEP＝90°
∵AP⊥EF，∴∠APE＝∠APC＝90°,∴∠ADE＝∠DEP＝∠APE＝90° ∴四边形ADEP为矩形
∴EP＝AD＝0.5m ····················································································· 2分 ∠APC＝90°，∠ACB＝45° ∴∠CAP＝45°＝∠ACB，∠BAP＝∠CAP—∠CAB＝45°—30°＝15°
∴AP＝CP ······························································································· 4分 在Rt△APB中
BPtan ∠BAP＝＝tan15°＝0.27 ·································································· 5分
AP ∴BP＝0.27AP＝0.27CP,∴BC＝CP—BP＝CP—0.27CP＝0.73CP＝1.2m
∴CP＝1.64m ··························································································· 7分 ∴CF＝EF—EP—CP＝3.5—0.5—1.64＝1.36≈1.4m
······································ 8分
1
·················································· 8分 3
（2）共有6种等可能出现的结果，分别为 ······················································ 3分
- 8 -
24．（本题9分）
解：（1）由条件可得k1＝—80 1分
设y1＝—80x+b1，过点（2.5，160），可得方程160＝—80×2.5+b1
解得b1＝360 ····················································································· 3分 ∴y1 ＝—80x+360 ················································································ 4分 （2）当y1 ＝0时，可得x＝4.5
轿车和货车同时到达，终点坐标为（4.5，360） 设y2 ＝k2 x+b2 ，过点（2.5，160）和（4.5，360） 解得k2 ＝100，b2 ＝—90 ∴y2 ＝100x—90 图像如下图
························································ 7分
与x轴交点坐标为(0.9，0) ···································································· 8分 说明轿车比货车晚出发0.9h ·································································· 9分
25．（本题8分）
解：（1）表达式为y＝(—3x+90)(x—20)
化简为y＝—3x2+150x—1800 ································································· 4分 （2）把表达式化为顶点式y＝—3(x—25)2 +75 ·················································· 6分
当x＝25时，y有最大值75
答：当售价为25元时，有最大利润75元 ················································· 8分
26．（本题9分） （1）证明：连结OD
∵∠ACB＝90°，∴∠OED+∠EGC＝90° ·································································· 1分 ∵⊙O，∴OD＝OE，∴∠ODE＝∠OED
∵AG＝AD，∴∠ADG＝∠AGD ················································································ 3分 ∵∠AGD＝∠EGC
∴∠OED+∠EGC＝∠ADG+∠ODE＝∠ADO＝90°
∴OD⊥AB ························································································ 4分 ∵OD为半径
∴AB是⊙O的切线 ············································································· 5分 （2）连接OF．∵EF∥AB，AC:BC＝4:3，∴CF:CE＝4:3．
- 9 -
又∵EF＝5，∴CF＝4，CE＝3．
设半径＝r，则OF＝r，CF＝4，CO＝r－3．
25
在Rt△OCF中，由勾股定理，可得r＝． ………………………………………7分
6∵EF∥AB，∴∠CEF＝∠B，∴△CEF∽△DBO，∴
CFCE
＝， DO DB
25
∴BD＝． ······················································································ 9分
8
27．（本题10分）
解：（1）① 2； ………………………………………………………………………………2分 ② 分点C、P在BQ同侧和异侧两种情况，画对一种就给全分；
B
QB
C O A P
QC
O P A
图2
图1
·················································································································· 5分 （2）当点C在∠AOB的内部或一边上时，则重叠部分即为△CPQ．
因为△CPQ是由△OPQ折叠得到，所以当△OPQ为等腰三角形时，重叠部分必为等腰三角形．
如图1、2、3三种情况：
O
Q P 图4
A
O
P 图5
A
C O
图1
P
A
O
图2
P
A
O 图3
P
A
QB
Q
BQ B当QO＝QP时，
当PQ＝PO时，
当OQ＝OP时， OQ＝OP＝2cm
B Q OQ＝2
OP＝2cm OQ＝2OP＝22cm 2
当点C在∠AOB的外部时，
B C
的上方时（如图4）当点C在射线OB， 当点C在射线 OA的下方时（如图5），
- 10 -
OQ＝6－2(cm) OQ＝6＋2(cm)
………………………………………………………………………………10分
- 11 -