福建省泉州市2019-2020学年度上学期教学质量跟踪监测考试七年级数学试题参考答案及评分意见

泉州市2019-2020学年度上学期教学质量跟踪监测考试
七年级数学试题参考答案及评分意见
一、选择题（每小题4分，共40分）
1．B 2．B 3．C 4．A 5．C 6．C 7．D 8．D
9．C
10．A
二、填空题（每小题4分，共24分）
11．＞
12．24
13．-3 14．58 15．2
5(1)8
πa
-
16．-2 三、解答题（共89分）
17. 解：原式=()114311264⎛⎫
−+⨯−−−⨯ ⎪⎝⎭ ······································································ 4分
4323=−−−+ ·
······················································································ 7分 6=− ···································································································· 8分 18.解：原式22222243264x x xy y x xy y =−+−+−+ ·························································· 3分
233xy y =−+ ··························································································· 5分
当1
3
x =，2y =-时
原式()()2
132323=−⨯⨯+⨯-- ··········································································· 6分
212=+ ································································ 7分
14= ···································································· 8分 19. 证明：∵//CE AB （已知）
∴A ACE ∠=∠（两直线平行，内错角相等） ·················· 2分
B DCE ∠=∠（两直线平行，同位角相等） ·
················· 4分 ∵B 、C 、D 三点共线
∴180ACE DCE ACB ∠+∠+∠=︒(平角的定义) ············· 6分
∴180A B ACB ∠+∠+∠=︒(等量代换) ··························· 8分
20. 解：(1) 如图线段BD 即为所求. ·········································· 1分
(2)如图直线CE 即为所求，点C 到直线AB 的距离为2. ······ 3分 (3)如图直线AF 即为所求. ············································ 5分 (4)CBD ∠的同位角:BAF ∠，BAC ∠，CED ∠. ················ 8分
21.(1)∵24AB =,且::3:2:1AC CD DB =，
∴286CD AB ==，1
46
DB AB ==， ·································· 1分
∴12CB CD DB =+=， ···················································································· 2分 ∵N 为CB 的中点，
∴1
62
CN CB ==， ·························································································· 3分
∴2ND CD CN =−= ························································································ 4分
E
B C D
A
E F D B C A
(2)证明：∵M 为AC 的中点，N 为CB 的中点，
∴12MC AC =
，1
2
CN CB =， ∴111
222
MN MC CN AC CB AB =+=+=， ··················································· 5分
∵::3:2:1AC CD DB =，
∴2163CD AB AB ==，1
6
DB AB =，
∴1
2
CB CD DB AB =+=， ······································································· 6分
∴1111
2224
CN CB AB AB ==⨯=，
∴111
3412DN CD CN AB AB AB =−=−=，
∴115
66)3122
CD DN AB AB AB +=⨯+=（）（， ················································ 7分
又15
5522
MN AB AB =⨯=，
∴56MN CD DN =+（）
············································································· 8分 22. 证明：(1) ∵AD ⊥BC ，
∴∠ADB =90°， ···················································· 1分 又∵∠ADG =35°
∴∠BDG =55° ······················································ 2分
又∵∠C =55° ∴∠C =∠BDG ······················································ 3分
∴DG ∥AC ·························································· 5分 (2)∵AD ⊥BC ，EF ⊥BC ∴AD ∥EF ··························································································· 6分 ∴∠FEC =∠DAC ················································································· 8分 由(1)可知，DG ∥AC ∴∠ADG =∠DAC ············································································································ 9分 ∴∠FEC =∠ADG ··········································································································· 10分
23.解：(1)当600x ≥时，实际付款：()6000.80.7600x ⨯+−=(0.760x +)元
答：当一次性购物x 元，600x ≥时，实际付款：(0.760x +)元 ····································· 4分 (2)①当300500a <≤时，则300800500a ≤−<，
购物实际付款：0.8800⨯=640(元) ·································································· 6分 ②当500600a <<时，则200800300a <−<，
购物实际付款：()0.80.9800a a +−=(0.1720a −+)元， ········································ 8分
G
E A D F
C B M N B
C D A
③当600800a ≤<时，则0800200a <−≤，
购物实际付款：()()6000.80.76000.9800a a ⨯+−+−=(0.2780a −+)元
故本次实际付款=6403005000.1+7205006000.2+780600800a a a a a <≤⎧⎪
−<<⎨⎪−≤<⎩， （），（），（） ···························································· 10分
24.解：(1)∵//BC AE ,
∴180ABC BAE ∠+∠=. ·············································································· 1分 ∵//DE CF ，
∴180CDE DCF ∠+∠=. ∵DCF ABC ∠=∠,
∴CDE BAE ∠=∠. ······································· 2分 ∵180BDC CDE ADE ∠+∠+∠=
180AED BAE ADE ∠+∠+∠=,
∴BDC AED ∠=∠, ····················································································· 3分 ∵ED 平分AEC ∠,
∴AED CED ∠=∠. ····················································································· 4分 ∴BDC CED ∠=∠. ····················································································· 5分 (2) 设,EDG DCG αβ∠=∠=
∵DG 平分ADE ∠,CG 平分DCF ∠,
∴2,2ADE DCF αβ∠=∠=. ·
···························· 6分 由(1)可知2180CDE DCF CDE β∠+∠=∠+=︒ ∴1802CDE β∠=︒−
······································ 7分 ∴()180G GDE CDE DCG ︒
∠=−∠+∠+∠
()1801802αβββα︒=−+︒−+=− ·········· 9分
又∵180AED ADE DAE ︒
∠=−∠−∠
∴()
()1802180218022AED CDE ααββα︒︒︒
∠=−−∠=−−−=−， ·
··················· 11分 ∴2ADE G ∠=∠. ·
··················································································· 12分 25.解：方法一：
(1)①∵3n = ∴1
3
a b c d a d −+−=−， ·
········································································ 1分 ∵a b c d <<<，
∴1
()3
b a d
c
d a −+−=−，
∴2
()3
c b
d a −=− ······················································································ 2分
∵6d a −=，
∴4c b −=.······························································································· 3分
b
a
G
E F A B C D
b
a
E
F
A
B
C
D
②∵b e c <<，49
b e a d −=−
∴4
()9
e b d a −=− ··················································································· 4分
∵2()3
c b
d a −=−
∴3
()2d a c b −=− ··················································································· 5分
∴432
()()923
e b c b c b −=⨯−=− ·································································· 6分
∴2233e b c b −=− ··················································································· 7分 ∴21
33
e c b =+. ······················································································· 8分
(2) ∵1
a b c d a d n
−+−=
−,a b e c d <<<< ∴()1122e b c c b =
−=−,()1122f a d d a =−=−，1
()()()b a d c a d n −+−=− ∴f e > ·
································································································· 9分 ∴()1
2
e f f e d a c b −=−=−−+ ································································ 10分
()()1
2b a d c =−+−⎡⎤⎣
⎦ ····································································· 11分 11
()2a d n
=⋅− 1
()2d a n
=
− ·
················································································ 12分 ∵1
10
e f a d −>−， ∴
11
()()210
d a d a n −>−，即
210d a d a n −−>， ∴210n <，
∴5n <， ································································································ 13分 ∵35n ≤<，且n 为正整数，
∴n 的最大值为4. ····················································································· 14分
方法二： (1)①把a ，b ，c ，d 四个数在数轴上分别用点A ,B ,C ,D 表示出来，如下图所示，
································ 1分
∵1
a b c d a d n
−+−=−， ∴AB +CD =
1
n
AD ···················································································· 2分 d
c
b
a
A B C D
又∵3n =，6d a −= ∴AD =6，AB +CD =2
∴BC =b c −=4c b −=. ············································································ 3分 ②e (b e c <<)用点E 表示数e 在数轴上表述出来，点E 在线段BC 上， ∵4
9
b e a d −=−， ∴BE =
4
9AD ······························································································ 4分 又∵BC =2
3AD ,
∴BE =
49AD =4392⨯BC =2
3BC ， ····································································· 5分 即2
3
b e b
c −=
− ·
······················································································ 6分 ∵b e c <<
∴22
33e b c b −=− ······················································································ 7分
∴21
33
e c b =+. ·························································································· 8分
(2) ∵1
a b c d a d n
−+−=
−,a b c d <<< ∴1
(1)()c b d a n
−=−−, ················································································ 9分
∵11,,22e b c f a d =−=−且110
e f a d −>−， ∴111
2210b c a d a d −−−>− ·
··································································· 10分 ∴1111
(1)2210
a d a d a d n ⨯−−−>−- ∴
11
210
a d a d n −>−，即210a d a d n −−> ······················································ 12分 ∵0a d −> ∴210n <
∴5n < ··································································································· 13分 ∵35n ≤<，且n 为正整数，
∴n 的最大值为4. ····················································································· 14分
e
d
c
b
a
A B C D E。