2015年苏科版八年级上元旦假期作业含答案

八年级数学月考复习卷

张 志 2015.1.3

一、选择题

1．2的算术平方根是················································································· （ ）

A

B．2 C

D．±2

2．2013年12月2日，“嫦娥三号”从西昌卫星发射中心发射升空，并于12月14日在月球上成功实施软着陆．月球距离地球平均为384401000米，用四舍五入法取近似值，精确到1000000米，并用科学计数法表示，其结果是 ················································ （ ）

A．3.84×107米 B．3.8×107米 C．3.84×108米 D．3.8×108米

3．在实数：3.21，π

−

A．1个 22中，无理数的个数有 ·························· （ ） 7B．2个 C．3个 D．4个

4．在平面直角坐标系中，点P（3，−5）在 ·········································· （ ）

A．第一象限 B．第二象限 C．第三象限 D．第四象限

5．如图是一个风筝设计图，其主体部分（四边形ABCD）关于BD所在的直线对称，AC与BD相交于点O，且AB≠AD，则下列判断不正确的是 ························· （ ）

A．△ABD≌△CBD B．△ABC是等边三角形

C．△AOB≌△COB D．△AOD≌△COD

第5题

第7题

6．一次函数y＝kxb，当k＜0，b＜0时，它的图象大致为············· （ ）

ABCD

7．一次函数y2x1的图象不经过 （ ）

（A）第一象限 （B）第二象限 （C）第三象限 (D)第四象限

8．如图1，直线ykxb与x轴交于点(2，0)，则y＜0时，x的取值范围是( )

（A）x>2 （B）x0 (D)x

D 图1 图2

9．如图2，∠AOP=∠BOP=15°，PC∥OA交OB于C，PD⊥OA于D，若PC=3，则PD等于

1

( )

（A）3 （B）2 （C）1.5 (D)1

10．如图，正方形网格中，已有两个小正方形被涂黑，再将图中其余小正方形涂黑一个，使整个

被涂黑的图案构成一个轴对称图形，那么涂法共有 ························ （ ）

A．3种 B．4种 C．5种 D．6种

二、填空题（本大题共10小题，每小题2分，共20分）

11．点P（2，3）到x轴的距离是＿＿＿＿＿．

12．比较大小：

7．（填“＞”、“＝”、“＜”）

13．已知等腰三角形的一个外角是80°，则它顶角的度数为＿＿＿＿＿．

14．若直角三角形的两条直角边的长分别是6和8，则斜边上的中线长为＿＿＿＿＿． 15．如图，在△ABC中，∠C＝90°，AD平分∠CAB，BC＝10cm，BD＝6cm，那么D点到直线

AB的距离是＿＿＿＿＿cm．

BBCDEGC 第15题 第16题

第18题

16．在平面直角坐标系中，一青蛙从点A（−1，0）处向左跳2个单位长度，再向下跳2个单位长

度到点A′处，则点A′的坐标为＿＿＿＿＿．

17．写出同时具备下列两个条件的一次函数关系式＿＿＿＿＿．（写出一个即可）

（1）y随x的增大而减小；（2）图像经过点（1，−2）．

18．如图，在△ABC中，AB的垂直平分线分别交AB、BC于点D、E，AC的垂直平分线分别交

AC、BC于点F、G，若∠BAC＝100°，则∠EAG＝＿＿＿＿＿°．

19．如图，已知直线y＝axb，则关于x的方程ax1＝b的解x＝＿＿＿＿＿．

三、解答题

20．（1）求x的值：4x29＝0； （2

）计算：(1)0

21．如图，一木杆在离地某处断裂，木杆顶部落在离木杆底部8米处，已知木杆原长16米，求木

杆断裂处离地面多少米？

2

22．（在平面直角坐标系中，已知A（−1，5）、B（4，2）、C（−1，0）三点．

（1）点A关于原点O的对称点A′的坐标为＿＿＿＿＿，点B关于x轴的对称点B′的坐标为＿

＿＿＿＿，点C关于y轴的对称点C′的坐标为＿＿＿＿＿；

（2）求以（1）中的点A′、B′、C′为顶点的△A′B′C′的面积．

23．如图，四边形ABCD是梯形，AD∥BC，∠A＝90°，BD＝CB，CE⊥BD，

垂足为E． （1）求证：△ABD≌△ECB； （2）若∠DBC＝50°，求∠DCE的度数．

CB

24．如图，点E是∠AOB的平分线上一点，EC⊥OA，ED⊥OB，垂足分别是C、D．

求证：（1）∠EDC＝∠ECD；

B（2）OC＝OD；

（3）OE是线段CD的垂直平分线．

AO C

2)及点B(0，4)． 25．已知一次函数ykxb的图象经过点A(3，

⑴求此一次函数的解析式，并画出图象；

⑵求此函数图象与两坐标轴所围成的三角形的面积．

3

4

八年级数学参考答案及评分标准

（阅卷前请认真校对，以防答案有误！）

一、选择题（每小题3分，共24分）

二、填空题（每小题2分，共20分）

9．3． 10．＜．

13．4． 14．（−3，−2）．

16．20． 17．4．

三、解答题（共76分） 11．100°． 12．5． 15．答案不唯一，如y＝x1等． 18．①②④⑤．

19．（1）4x2＝9， ················································································································ 1分

9x2＝， ·············································································································· 2分 4

3········································································································ 4分 x＝±． ·2

（2）原式＝1＋2＋2 ······································································································· 3分

＝5． ·······································

········

································································ 4分

3说明：第（1）题答案写成x＝扣1分； 2

第（2）题(1)01分．

20．作出线段垂直平分线，·································································································· 3分

作出角平分线． ············································································································· 6分

21．设木杆断裂处离地面x米，由题意得 ··········································································· 1分

x282＝(16x)2． ······································································································ 3分 解得x＝6 ······················································································································· 5分 答：木杆断裂处离地面6米． ······················································································ 6分

22．（1）（1，−5）；（4，−2）；（1，0）． ·············································································· 3分

151（2）S△A′B′C′＝5(41)＝． ····················································· 6分 22

23．（1）∵AD∥BC，

∴∠ADB＝∠EBC．

∵CE⊥BD，

5

∴∠BEC＝90°．

∵∠A＝90°，

∴∠A＝∠BEC． ·············································································································· 1分 在△ABD和△ECB中，

ABEC······································································································· 2分 ADBEBC， ·

BDCB

∴△ABD≌△ECB（AAS）． ··························································································· 3分

（2）∵BD＝CB，∠DBC＝50°，

11∴∠BDC＝(180DBC)＝(18050)＝65°． ··················································· 4分 22

∠65°∴在Rt△CDE中，∠DCE＝90°BDC＝90°＝25°． ······································· 6分

24．（1）∵点E是∠AOB的平分线上一点，EC⊥OA，ED⊥OB，

∴ED＝EC． ··················································································································· 3分 ∴∠EDC＝∠ECD． ········································································································ 4分

（2）∵EC⊥OA，ED⊥OB，

∴∠EDO＝∠ECO＝90°． ······························································································· 5分 由（1）知∠EDC＝∠ECD，

∴∠EDO∠EDC＝∠ECO∠ECD，即∠ODC＝∠OCD． ······································· 6分 ∴OC＝OD． ·················································································································· 7分

（3）∵OC＝OD，∠EOC＝∠EOD，

∴OE⊥CD，OE平分CD，即OE是线段CD的垂直平分线． ································ 10分

25．（1）AB

． ········································································· 3分

（2）AB＝5(1)＝6． ································································································ 6分

（3）△ABC是直角三角形． ·························································································· 7分 理由：∵AB

AC

BC

5，

∴AB

2＋AC2

＝22＝25，BC2＝52＝25．

∴AB2＋AC2＝BC2． ······································································································· 9分 ∴△ABC是直角三角形． ····························································································· 10分

26．（1）3600，20． ············································································································· 2分

（2）当50≤x≤80时，设y与x的函数关系式为y＝kxb，根据题意得 ··············· 3分 当x＝50时，y＝1950；当x＝80时，y＝3600．···················································· 4分 50kb1950∴． 80kb3600

k55解得．············································································································ 6分 b800

∴y与x的函数关系式为y＝55x800． ···································································· 7分

（3）缆车到山顶的路线长为3600÷2＝1800（m）． ···················································· 8分 缆车到达终点所需时间为1800÷180＝10（min）． ····················································· 9分 爸爸到达缆车终点时，小华行走的时间为10＋50＝60（min）． ······························ 10分 把x＝60代入y＝55x800，得y＝55×60800＝2500． ···································· 11分 ∴当爸爸到达缆车终点时，小华离缆车终点的路程是36002500＝1100（m） ···· 12分

27．（1）∵△ABC和△ADE都是等边三角形，

6

∴AB＝AC＝BC，AD＝AE，∠BAC＝∠DAE＝60°．

∴∠BAC∠CAD＝∠DAE∠CAD，即∠BAD＝∠CAE． ············································ 1分 在△ABD和△ACE中，

ABACBADCAE，

ADAE

∴△ABD≌△ACE（SAS）． ···························································································· 3分 ∴BD＝CE． ··················································································································· 4分 ∵BC＝BD＋CD，AC＝BC，

∴AC＝CE＋CD． ·········································································································· 5分

（2）AC＝CE＋CD不成立，

AC、CE、CD之间存在的数量关系是：AC＝CECD． ············································ 6分 理由：∵AB＝AC＝BC，AD＝AE，∠BAC＝∠DAE＝60°．

∴∠BAC＋∠CAD＝∠DAE＋∠CAD，即∠BAD＝∠CAE． ············································ 7分 在△ABD和△ACE中，

ABACBADCAE，

ADAE

∴△ABD≌△ACE（SAS）． ···························································································· 8分 ∴BD＝CE． ··················································································································· 9分 ∴CECD＝BDCD＝BC＝AC，即AC＝CECD． ··············································· 10分

（3）补全图形（如图）． ······························································································ 11分 AC、CE、CD之间存在的数量关系是：AC＝CDCE． ·········································· 12分

DE

说明：解答题中，考生若使用其它解法，请参考评分标准酌情给分． 7

八年级数学月考复习卷

张 志 2015.1.3

一、选择题

1．2的算术平方根是················································································· （ ）

A

B．2 C

D．±2

2．2013年12月2日，“嫦娥三号”从西昌卫星发射中心发射升空，并于12月14日在月球上成功实施软着陆．月球距离地球平均为384401000米，用四舍五入法取近似值，精确到1000000米，并用科学计数法表示，其结果是 ················································ （ ）

A．3.84×107米 B．3.8×107米 C．3.84×108米 D．3.8×108米

3．在实数：3.21，π

−

A．1个 22中，无理数的个数有 ·························· （ ） 7B．2个 C．3个 D．4个

4．在平面直角坐标系中，点P（3，−5）在 ·········································· （ ）

A．第一象限 B．第二象限 C．第三象限 D．第四象限

5．如图是一个风筝设计图，其主体部分（四边形ABCD）关于BD所在的直线对称，AC与BD相交于点O，且AB≠AD，则下列判断不正确的是 ························· （ ）

A．△ABD≌△CBD B．△ABC是等边三角形

C．△AOB≌△COB D．△AOD≌△COD

第5题

第7题

6．一次函数y＝kxb，当k＜0，b＜0时，它的图象大致为············· （ ）

ABCD

7．一次函数y2x1的图象不经过 （ ）

（A）第一象限 （B）第二象限 （C）第三象限 (D)第四象限

8．如图1，直线ykxb与x轴交于点(2，0)，则y＜0时，x的取值范围是( )

（A）x>2 （B）x0 (D)x

D 图1 图2

9．如图2，∠AOP=∠BOP=15°，PC∥OA交OB于C，PD⊥OA于D，若PC=3，则PD等于

1

( )

（A）3 （B）2 （C）1.5 (D)1

10．如图，正方形网格中，已有两个小正方形被涂黑，再将图中其余小正方形涂黑一个，使整个

被涂黑的图案构成一个轴对称图形，那么涂法共有 ························ （ ）

A．3种 B．4种 C．5种 D．6种

二、填空题（本大题共10小题，每小题2分，共20分）

11．点P（2，3）到x轴的距离是＿＿＿＿＿．

12．比较大小：

7．（填“＞”、“＝”、“＜”）

13．已知等腰三角形的一个外角是80°，则它顶角的度数为＿＿＿＿＿．

14．若直角三角形的两条直角边的长分别是6和8，则斜边上的中线长为＿＿＿＿＿． 15．如图，在△ABC中，∠C＝90°，AD平分∠CAB，BC＝10cm，BD＝6cm，那么D点到直线

AB的距离是＿＿＿＿＿cm．

BBCDEGC 第15题 第16题

第18题

16．在平面直角坐标系中，一青蛙从点A（−1，0）处向左跳2个单位长度，再向下跳2个单位长

度到点A′处，则点A′的坐标为＿＿＿＿＿．

17．写出同时具备下列两个条件的一次函数关系式＿＿＿＿＿．（写出一个即可）

（1）y随x的增大而减小；（2）图像经过点（1，−2）．

18．如图，在△ABC中，AB的垂直平分线分别交AB、BC于点D、E，AC的垂直平分线分别交

AC、BC于点F、G，若∠BAC＝100°，则∠EAG＝＿＿＿＿＿°．

19．如图，已知直线y＝axb，则关于x的方程ax1＝b的解x＝＿＿＿＿＿．

三、解答题

20．（1）求x的值：4x29＝0； （2

）计算：(1)0

21．如图，一木杆在离地某处断裂，木杆顶部落在离木杆底部8米处，已知木杆原长16米，求木

杆断裂处离地面多少米？

2

22．（在平面直角坐标系中，已知A（−1，5）、B（4，2）、C（−1，0）三点．

（1）点A关于原点O的对称点A′的坐标为＿＿＿＿＿，点B关于x轴的对称点B′的坐标为＿

＿＿＿＿，点C关于y轴的对称点C′的坐标为＿＿＿＿＿；

（2）求以（1）中的点A′、B′、C′为顶点的△A′B′C′的面积．

23．如图，四边形ABCD是梯形，AD∥BC，∠A＝90°，BD＝CB，CE⊥BD，

垂足为E． （1）求证：△ABD≌△ECB； （2）若∠DBC＝50°，求∠DCE的度数．

CB

24．如图，点E是∠AOB的平分线上一点，EC⊥OA，ED⊥OB，垂足分别是C、D．

求证：（1）∠EDC＝∠ECD；

B（2）OC＝OD；

（3）OE是线段CD的垂直平分线．

AO C

2)及点B(0，4)． 25．已知一次函数ykxb的图象经过点A(3，

⑴求此一次函数的解析式，并画出图象；

⑵求此函数图象与两坐标轴所围成的三角形的面积．

3

4

八年级数学参考答案及评分标准

（阅卷前请认真校对，以防答案有误！）

一、选择题（每小题3分，共24分）

二、填空题（每小题2分，共20分）

9．3． 10．＜．

13．4． 14．（−3，−2）．

16．20． 17．4．

三、解答题（共76分） 11．100°． 12．5． 15．答案不唯一，如y＝x1等． 18．①②④⑤．

19．（1）4x2＝9， ················································································································ 1分

9x2＝， ·············································································································· 2分 4

3········································································································ 4分 x＝±． ·2

（2）原式＝1＋2＋2 ······································································································· 3分

＝5． ·······································

········

································································ 4分

3说明：第（1）题答案写成x＝扣1分； 2

第（2）题(1)01分．

20．作出线段垂直平分线，·································································································· 3分

作出角平分线． ············································································································· 6分

21．设木杆断裂处离地面x米，由题意得 ··········································································· 1分

x282＝(16x)2． ······································································································ 3分 解得x＝6 ······················································································································· 5分 答：木杆断裂处离地面6米． ······················································································ 6分

22．（1）（1，−5）；（4，−2）；（1，0）． ·············································································· 3分

151（2）S△A′B′C′＝5(41)＝． ····················································· 6分 22

23．（1）∵AD∥BC，

∴∠ADB＝∠EBC．

∵CE⊥BD，

5

∴∠BEC＝90°．

∵∠A＝90°，

∴∠A＝∠BEC． ·············································································································· 1分 在△ABD和△ECB中，

ABEC······································································································· 2分 ADBEBC， ·

BDCB

∴△ABD≌△ECB（AAS）． ··························································································· 3分

（2）∵BD＝CB，∠DBC＝50°，

11∴∠BDC＝(180DBC)＝(18050)＝65°． ··················································· 4分 22

∠65°∴在Rt△CDE中，∠DCE＝90°BDC＝90°＝25°． ······································· 6分

24．（1）∵点E是∠AOB的平分线上一点，EC⊥OA，ED⊥OB，

∴ED＝EC． ··················································································································· 3分 ∴∠EDC＝∠ECD． ········································································································ 4分

（2）∵EC⊥OA，ED⊥OB，

∴∠EDO＝∠ECO＝90°． ······························································································· 5分 由（1）知∠EDC＝∠ECD，

∴∠EDO∠EDC＝∠ECO∠ECD，即∠ODC＝∠OCD． ······································· 6分 ∴OC＝OD． ·················································································································· 7分

（3）∵OC＝OD，∠EOC＝∠EOD，

∴OE⊥CD，OE平分CD，即OE是线段CD的垂直平分线． ································ 10分

25．（1）AB

． ········································································· 3分

（2）AB＝5(1)＝6． ································································································ 6分

（3）△ABC是直角三角形． ·························································································· 7分 理由：∵AB

AC

BC

5，

∴AB

2＋AC2

＝22＝25，BC2＝52＝25．

∴AB2＋AC2＝BC2． ······································································································· 9分 ∴△ABC是直角三角形． ····························································································· 10分

26．（1）3600，20． ············································································································· 2分

（2）当50≤x≤80时，设y与x的函数关系式为y＝kxb，根据题意得 ··············· 3分 当x＝50时，y＝1950；当x＝80时，y＝3600．···················································· 4分 50kb1950∴． 80kb3600

k55解得．············································································································ 6分 b800

∴y与x的函数关系式为y＝55x800． ···································································· 7分

（3）缆车到山顶的路线长为3600÷2＝1800（m）． ···················································· 8分 缆车到达终点所需时间为1800÷180＝10（min）． ····················································· 9分 爸爸到达缆车终点时，小华行走的时间为10＋50＝60（min）． ······························ 10分 把x＝60代入y＝55x800，得y＝55×60800＝2500． ···································· 11分 ∴当爸爸到达缆车终点时，小华离缆车终点的路程是36002500＝1100（m） ···· 12分

27．（1）∵△ABC和△ADE都是等边三角形，

6

∴AB＝AC＝BC，AD＝AE，∠BAC＝∠DAE＝60°．

∴∠BAC∠CAD＝∠DAE∠CAD，即∠BAD＝∠CAE． ············································ 1分 在△ABD和△ACE中，

ABACBADCAE，

ADAE

∴△ABD≌△ACE（SAS）． ···························································································· 3分 ∴BD＝CE． ··················································································································· 4分 ∵BC＝BD＋CD，AC＝BC，

∴AC＝CE＋CD． ·········································································································· 5分

（2）AC＝CE＋CD不成立，

AC、CE、CD之间存在的数量关系是：AC＝CECD． ············································ 6分 理由：∵AB＝AC＝BC，AD＝AE，∠BAC＝∠DAE＝60°．

∴∠BAC＋∠CAD＝∠DAE＋∠CAD，即∠BAD＝∠CAE． ············································ 7分 在△ABD和△ACE中，

ABACBADCAE，

ADAE

∴△ABD≌△ACE（SAS）． ···························································································· 8分 ∴BD＝CE． ··················································································································· 9分 ∴CECD＝BDCD＝BC＝AC，即AC＝CECD． ··············································· 10分

（3）补全图形（如图）． ······························································································ 11分 AC、CE、CD之间存在的数量关系是：AC＝CDCE． ·········································· 12分

DE

说明：解答题中，考生若使用其它解法，请参考评分标准酌情给分． 7