苏科版八年级数学下册第9章-第11章综合测试

22一选题本大题共小题，每小题3分，共18分在小题所给出的四个选项中，只有一项是符合题目要求的，请将确选项的字母代号填涂在答题卡相应位置上）1．下图形是中心对称图形的是·············································································（）ABCDx2．若式有意义，则x满的条件是···························································（）xA．x0BxC．x3．≥．下列函数中，是反比例函数的为（）A．y2x

B．y

2x

C．

y

3x

D．2y23．在代数式，，，中分式有（）2aA2个B3C个D5个5．下等式成立的是·································································································（）231abaA．+＝；．＝；．＝；．＝-ababa+ba+bab-b-b-a+ba+b63．某反比例函数的图象经过点，此数的图象也经过点（）A，-3）B-3，）C，3）D，6菱具有而一般平行四边形不具有的性质是

()A对边相等对角相等C．对角线相平分；D．对角线互相垂如，在平行四边形中下结论中错误的是（）A∠1=∠B．∠=∠BCDC=CDD．AC⊥BD下四个命题：①一组对边平行且一组对角相等的四边形是平行四边形；②对角线互相垂直且相等的四边形是正方形顺次连结矩形四边中点得到的四边形是菱形等三角形既是轴对称图形又是中心对称图形．其中真命题共有（）A．1个B2个C．3个D．个10．图，在四边形中∠A90°＝，＝7，点、分为线段BC、上的动点，点、分为DM、的中点，则长度的最大值为···················（）A．

B．．D2二填题本大题共8小题，每小题3分共24分.不需写解答过程，请将答案直接写在答题卡相应位置上）x11．若式的为0则x＝▲．1-．分式、的简公分母是▲．3x13．形中，对角线＝，＝，菱形的面积是▲．14如图，矩形ABCD的角线AC、交于点O，∠AOD，＝4，则长▲．ADA

E

O

D

A

P

D

A

DB

O

FBC

B

B

P

E

F第14题图

第15题图

第17题图

第题图15图eq\o\ac(□,)的角线相于点点F别是线段的点＋＝22，的周长是16cm，则的长为▲cm．16．知x0，分式的是▲．x图形的长为别是边的上点MC＝MB＝NC是对角上上点，则PM的最小值是▲．．如图所示，点为正方的对角线上一点，过点P作BC，垂足分别为点E、，连接EF．下列结论中是腰直角三角形；AP；AD＝；PFE＝BAP其中正确的结论是▲）三解题本题共小题，共76分，请在答卡指定区域内作答，解答时应写出文字说明、推理过程或演算步骤）19）算．（1-+

+b

；

（2

a．aa20

）图，点A、是的角线EF所在直线上的两点且AE＝CF．求证：四边形平行四边形．D

FEAB第20题图xx1216）先化简：)，后在－，，2四数中找xx一个你认为合适的代求值．226）方程：

x

31

．236观等11＝-，……45

111111＝1-②＝-＝-④´22´3344´5试用含字母的式表示出你发现的规律，并证明该等成立；1111+++＝）227本10图27本10图1图248）图，一次函数y=+b与比例函数=的图象交于点A（16B（3n）两点．求反比例函数和一次函数的表达式；点是一次函数=kx+图位第一象限内的一点，过点作MNx轴，垂足为点N过点作BDy轴，垂足点D，eq\o\ac(△,)MON的积小eq\o\ac(△,)BOD的面积，直接写出点M的坐标x取值范围．第题258）eq\o\ac(□,)中过点D作DE⊥于，点在CD上，＝BE，连接，．求证：四边形DEBF是矩形；若平分∠，＝，＝，求的面积．

A

DFBE第题图2610）图1，方ABCD中点O是对角ACAO上（不与A、重）的一个动点，过点P作⊥PB且交于点．求证：＝；过点E作于F，图2．正方形ABCD的长为2，在点P运的过程，的度是否发生变化？若不变，请直接写出这个不变的值；若变化，请说明理由．ADAPP

DB

O

EOF

E（把一张矩形片ABCD如图方式折叠使点落边AD（为点）点落在点分与边、BC交于试在图中连接，求证：四边形是形；若AB，＝，求线能取到的整数值．A

B

ADA

B

F

C

B

C

B

C第27题图

备用图

备用图121212122810）面直角坐标系xOy中已知函数=（＞0）与y﹣（＜）的图象如图所示，点AB是数=（x＞）图象上的两点点P是=（x＜）的图象上的一点且∥x轴点Q是x轴一点设点B的坐标分别为mm求的面积；若是等腰直角三角形，求点Q的标；若是以AB为的等腰三角形，求的．参答及分准一选择（330A

C

C

A

C

A

D

D

B

D二填题22011．；．

y

；；．；．2.5；．17．；．②④．三解题19）原＝

(a-ba)+a+b

··························································································2＝

a+ba+b

．·········································································································3（）式＝

a(a-×(+1)a-

··································································································2＝

aa+

．·············································································································320．明：连接DB交EF于O．∵四边形是行四边形，∴＝OB，OE＝．·······································································································2∵＝，∴＋＝＋，＝．4∴四边形是平行四边形．·······················································································21)(x-1)x-21．：原式＝···············································································x-+x＝+．·············································································································4取x＝．····························································································································5∴原式2＋＝．············································································································6（222．；验不能漏。123）＝-（为正整·······················································nn+1)eq\o\ac(□,S)eq\o\ac(□,S)证明：∵

1+1n-＝-nn+(+1)

n+-n＝＝．······················4n(+n(n+∴

111＝-．n(n++（2

．···························································································6（124）

y

6x

，y（）x3

；25）四边形是行四边形，∥AB，即DF∥EB．又∵DF＝，四边形DEBF是行四边形．2∵⊥，∴∠＝．四边形DEBF是形．（）四边形DEBF是矩形，∴＝＝，BDDF．

4∵⊥，＝

+DE

＝

+

＝．∵∥，∠＝．∵平∠DAB，∠DAF＝∠∴∠＝∠．∴＝＝．···········································································································∴＝．∴ABAEBE＝＋＝．∴＝·BF＝×＝．·············································································826）图，连接．∵四边形是方形，∴，∠＝DCA，∠BCD＝．又∵PC＝，△BCP≌DCP．∴＝，PBCPDC．3∵⊥，BPE＝．∴在四边形BCEP中∠∠＝－∠－∠＝．又∵∠＋∠，∴∠＝PED∴∠＝∠．∴＝．·················································································································6∴＝．·················································································································7（2P3BCCD11（）

PE

的长度不发生变化，＝．······································································10（OBPEFeq\o\ac(△,≌)82）连接BB′．由折叠知点B、关EF称．是段BB的垂直平分线．＝BEBFF．································································································2四形是矩形ADBC．B＝．由折叠得B＝．B＝B．B′＝BF················································································································＝BE′F＝．四形BFB是菱形．

4（）图1，点E与重时，四边形′是正方形，此时BF最．5∵四边形ABFB′正方形，∴＝＝，最为．··············································································6如图2，点与D重时BF最大．·····························································7设BF，＝-x，＝＝．在eq\o\ac(△,Rt)中，由勾股定理得＋2＝．∴-x)+＝x2，得，即＝10．···················································9∴≤≤．∴线段长能取到的整数值为8，，．·························································12eq\o\ac(△,S)1212eq\o\ac(△