2015-2016七下期末试题2016.06.08

2015～2016学年度第二学期期末模拟测试题
七年级数学
本试题分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分．第Ⅰ卷共2页，满分为36分；第Ⅱ卷共4页，满分为84分．本试题共6页，满分为120分．考试时间为120分钟．答卷前，请考生务必将自己的姓名、准考证号、座号、考试科目涂写在答题卡上，并同时将考点、姓名、准考证号、座号填写在试卷规定的位置．考试结束后，将本试卷和答题卡一并交回．本考试不允许使用计算器．
第I卷（选择题共36分）
注意事项：第Ⅰ卷为选择题，每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑；如需改动，用橡皮擦干净后，再选涂其他答案标号．答案写在试卷上无效．
一、选择题（本大题共12个小题，每小题3分，共36分．在每小题给出的四个选项中，只有一项是符合题目要求的．）
1.以下是回收、绿色包装、节水、低碳四个标志，其中是轴对称图形的是
2. 二元一次方程组
3
24
x y
x
+=
⎧
⎨
=
⎩
的解是
A.
2
1
x
y
=
⎧
⎨
=-
⎩
B.
2
5
x
y
=
⎧
⎨
=
⎩
C.
2
5
x
y
=
⎧
⎨
=-
⎩
D.
2
1
x
y
=
⎧
⎨
=
⎩
3. 已知∠A＝60°，则∠A的补角是
A．160°B．120°C．60°D．30°
4. 在△ABC中，∠C=60°，∠B=70°，则∠A的度数是
A.70°
B. 55°
C. 50°
D. 40°
5. 如图，直线l1∥l2，若∠1=50°，则∠2的度数是
A．40°B．50°
C．90°D．130°
6.下列长度的三条线段，不能组成三角形的是
A.3，8，4
B.4，9，6
C.15，20，8
D.9，15，8
7.如图，已知∠1=∠2，则不一定能使△ABD≌△ACD的条件是
A. AB=AC
B. ∠B=∠C
C. BD=CD
D. ∠BDA=∠CDA
8.如图，AB∥CD，点E在BC上，且CD=CE，∠D=74︒，则∠B的度数为A．68︒B．32︒C．22︒D．16︒
9. 已知两数x、y之和是10，x比y的3倍大2，则下面所列方程组正确的是
A.
10
32
x y
y x
+=
⎧
⎨
=+
⎩
B.
10
32
x y
y x
+=
⎧
⎨
=-
⎩
C.
10
32
x y
x y
+=
⎧
⎨
=+
⎩
D.
10
32
x y
x y
+=
⎧
⎨
=-
⎩
10.如图，在△ABC中，AB=AC，∠ABC、∠ACB的平分线BD、CE相交于O点，且BD交AC于点D，CE 交AB于点E．某同学分析图形后得出以下结论：①△BCD≌△CBE；②△BAD≌△BCD；③△BDA≌△CEA；
④△BOE≌△COD；⑤△ACE≌△BCE；上述结论一定正确的是
A．①②③B．②③④C．①③⑤D．①③④
11.如图，在△ABC中，∠ABC和∠ACB的平分线交于点E，过点E作MN∥BC交AB于M，交AC于N，若BM+CN=9，则线段MN的长
A．6 B．7 C．8 D．9
12. 如图，AD是△ABC的角平分线，DF⊥AB，垂足为F，DE=DG，△ADG和△AED的面积分别为50和39，则△EDF的面积为
A. 11
B. 5.5
C. 7
D.
第Ⅱ卷（非选择题共84分）
注意事项：
所有答案必须用的黑色签字笔(不得使用铅笔和圆珠笔)写在答题卡各题目指定区域内（超出方框无效），不能写在试卷上，不能使用涂改液、修正带等.
不按以上要求做答，答案无效.
二、填空题(本大题共6个小题．每小题3分，共18分．把答案填在题中横线上．)
13. 如图，∠AOB = 90°，∠BOC = 30°，则∠AOC = 度.
14. 若x 、y 满足方程组37
35x y x y +=⎧⎨+=⎩
，则x －y 的值等于 .
15.如图所示，AB =DB ，∠ABD =∠CBE ，请你添加一个适当的条件__________________，使△ABC ≌△DBE ．（只
需添加一个即可）
16.如图，在直角△ABC 中，90BAC ∠=︒，CB =10，AC =6，DE 是AB 边的垂直平分线，垂足为D ，交BC 于点E ，连接AE ，则△ACE 的周长为 .
17.如图，在边长为a 的大正方形中剪去一个边长为b 的小正方形，再将图中的阴影拼成一个长方形，这个拼成的长方形的长为30，宽为20，则右图中Ⅱ部分的面积是 ．
18.如图，已知∠AOB =α，在射线OA 、OB 上分别取点A 1、B 1，使OA 1=OB 1，连结A 1B 1，在B 1A 1、B 1B 上分别取点A 2、B 2，使B 1B 2= B 1A 2，连结A 2B 2……按此规律继续下去，记∠A 2B 1B 2=α1，∠323A B B =α2……∠n+11A n n B B +=αn ，则αn = .
三、解答题(本大题共9个小题，共66分．解答应写出文字说明，证明过程或演算步骤．) 19(1) (本小题满分3分)解方程组25
4x y x y +=⎧⎨
-=⎩
19(2) (本小题满分4分)如图，∠B ＝30°，若AB ∥CD ，CB 平分∠ACD ，求∠A 的度数．
20．(本小题满分5分)
已知：如图，点A、F、C、D在同一直线上，点B和点E分别在直线AD的两侧，且AB＝DE，∠A＝∠D，AF＝DC．
求证：BC∥EF．
21．(本小题满分6分)
已知：如图，在△ABC中，∠C=90°，∠B=15°，AB的垂直平分线交BC于D，交AB•于E，DB=10.
求∠ADC的度数和边AC的长.
22．(本小题满分7分)
为了改善全市中、小学办学条件，计划集中采购一批电子白板和投影机，已知购买2块电子白板比购买3台投影机多4000元，购买4块电子白板和3台投影机共需44000元.问购买一块电子白板和一台投影机各需要多少元？
23．(本小题满分7分)
如图为一机器零件，∠A＝36°的时候是合格的，小明测得∠BDC＝98°，∠C＝38°，∠B＝23°.
请问该机器零件是否合格并说明你的理由.
24．(本小题满分8分)
如图，AB∥CD，直线MN分别交AB、CD于点E、F，EG平分∠AEF．EG⊥FG于点G，∠BEM=50°.
求∠CFG的度数．
25．(本小题满分8分)
如图，在10×10的正方形网格中，每个小正方形的边长都为1，网格中有两个格点A、B和直线l.
(1)求作点A关于直线l的对称点A1；
(2)P为直线l上一点，连接BP，AP，求△ABP周长的最小值.
26．(本小题满分9分)
如图，∠ABC=90°，D、E分别在BC、AC上，AD⊥DE，且AD=DE. 点F是AE的中点，FD的延长线与AB的延长线相交于点M，连接MC.
(1)求证：∠FMC=∠FCM；
(2)AD与MC垂直吗？说明你的理由.
27．(本小题满分9分)
如图，△ABC、△ADC、△AMN均为等边三角形，AM＞AB，AM与DC交于点E，AN与BC交于点
F.
(1)求证：△ABF≌△ACE；
(2)猜测△AEF的形状，并证明你的结论；
(3)请直接指出当F点在BC何处时，AC⊥EF.
参考答案与评分标准一、选择题
二、填空
13. 60°
14. －1
15. BC=BE(或∠D=∠BAC;或∠E=∠C)
16. 16
17. 100
18. (21)180
2
n
n
α
-⋅︒+
或90°+45°+……+
180
2n
︒
+
2n
α
三、解答题
19.解：(1) 解：①+②得3x=9，····························································· 1分∴x=3. ······························································································ 2分把x=3代入②得3－y=4
∴y=－1
∴方程组的解为
3
1
x
y
=
⎧
⎨
=-
⎩
. ······································································ 3分
（2）解：∵AB∥CD（已知）
∴∠B=∠BCD(两直线平行，内错角相等) ··············································· 1分∵∠B＝30°
∴∠BCD＝30°（等量代换）································································· 2分∵CB平分∠ACD（已知）
∴∠BCD＝∠ACB＝30°（角平分线定义） ··············································· 3分∴∠A ==180°－∠ACB－∠B=180°－30°－30°=120°（三角形内角和定理） ····· 4分
20. 证明：∵AF =DC ，（已知） ∴AF +FC =FC +DC ，（等式的性质） ························································ 1分 即AC =DF ，
又∵AB =DE ，∠A =∠D ，（已知） ∴△ACB ≌△DEF （SAS ） ···································································· 3分 ∴∠ACB =∠DFE ，(全等三角形的对应角相等) ········································ 4分 ∴BC ∥EF ．（内错角相等，两直线平行） ················································ 5分 21. 解：∵DE 为AB 的垂直平分线，DB =10 (已知) ∴AD=BD=10(线段垂直平分线定理) ······················································· 1分 ∴∠B =∠BAD=15°，(等边对等角) ························································· 2分 ∴∠ADC =15°+15°=30°(三角形外角定理) ················································· 4分 ∵∠C =90°(已知) ∴AC=
12AD =1
2
×10=5(直角三角形中30°角所对直角边等于斜边的一半) ······································································································ 6分 22. 解：设购买一块电子白板需x 元，设购买一台投影机需y 元， ················ 1分
234000
4344000x y x y -=⎧⎨
+=⎩
··········································································· 4分 ①+②得6x =48000， x =8000， ·························································································· 5分 把x =8000代入①得2×8000－3y =4000， 解得y =4000，
∴⎩⎨⎧x =8000，y =4000
················································································· 6分 答：购买一台电子白板需8000元，一台投影机需4000元. ·························· 7分 23.解：不合格 ··················································································· 1分 连接AD 并延长， ··············································································· 2分 ∴∠BDE ＝∠B ＋∠BAD （三角形外角定理） ∠CDE ＝∠C ＋∠CAD （三角形外角定理）············································· 4分 ∴∠BDE ＋∠CDE ＝∠B ＋∠BAD ＋∠C ＋∠CAD ，(等式的性质) 即∠BDC ＝∠B ＋∠C ＋∠BAC ， ···························································· 5分 ∵∠BDC ＝98°，∠C ＝38°，∠B ＝23° ∴∠BAC ＝98°－38°－23°＝37° ······························································ 6分 所以该机器零件不合格． ····································································· 7分
24. 解：∵AB ∥CD ，
∴∠AEF+∠CFE=180°，（两直线平行，同旁内角互补） ····························· 1分∵∠AEF=∠BEM=50°，（对顶角相等） ··················································· 2分∴∠CFE=130°， ················································································ 3分∵EG平分∠AEF，（已知）
∴∠GEF=1
2
∠AEF=25°(角平分线定义)，················································ 4分
∵EG⊥FG，（已知）
∴∠EGF=90°，（垂直定义）································································· 5分∴∠GFE=90°－∠GEF=65°，（直角三角形两锐角互余） ····························· 7分∴∠CFG=∠GFE=65°(等量代换)．························································· 8分25.（1）略 ························································································ 4分（2）连接B A1交于P，连接AP ···························································· 5分则AP=P A1························································································ 6分△ABP的周长的最小值为AB+AP+BP= AB+P A1+BP=4+B A1=4+6=10 ·········· 8分
26.解：（1）证明：∵△ADE是等腰直角三角形，F是AE的中点.
∴DF⊥AE，DF=AF=EF. ····································································· 1分又∵∠ABC=90°，∠DCF、∠AMF都与∠MAC互余，
∴∠DCF=∠AMF. ············································································· 2分又∵∠DFC=∠AFM=90°，
∴△DFC≌△AFM（ASA）. ································································· 3分∴CF=MF. ······················································································· 4分∴∠FMC=∠FCM. ············································································· 5分（2）AD⊥MC.
理由如下：
如图，延长AD交MC于点G.
由（1）知∠MFC=90°，FD=FE，FM=FC.
∴∠FDE=∠FMC=45°， ······································································ 6分∴DE//CM. ······················································································· 7分∴∠AGC=∠ADE=90°，······································································· 8分∴AG⊥MC，即AD⊥MC. ··································································· 9分
27.证明：(1)∵△ABC、△ADC均为等边三角形，(已知)
∴AB=AC,，∠B=∠BAC =∠DAC=∠ACD=60°(等边三角形的性质) ······································································································ 1分∴∠BAC－∠F AC=∠DAC－∠F AC，（等式的性质）··································· 2分即∠BAF=∠CAE
∴△ACE≌△ABF（AAS）···································································· 3分（2）△AEF为等边三角形 ··································································· 4分∵△ABC≌△ABC
∴AE=AF（全等三角形的对应边相等） ··················································· 5分∵△AMN为等边三角形，
∴∠MAN=60°(等边三角形的性质) ·························································· 6分∴△AEF为等边三角形(有一个角为60°的等腰三角形是等边三角形) ·············· 7分（3）当点F为BC中点AC⊥EF···························································· 9分。