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22一選題本大題共小題,每小題3分,共18分在小題所給出的四個選項中,只有一項是符合題目要求的,請將確選項的字母代號填涂在答題卡相應(yīng)位置上)1.下圖形是中心對稱圖形的是·············································································()ABCDx2.若式有意義,則x滿的條件是···························································()xA.x0BxC.x3.≥.下列函數(shù)中,是反比例函數(shù)的為()A.y2x

B.y

2x

C.

y

3x

D.2y23.在代數(shù)式,,,中分式有()2aA2個B3C個D5個5.下等式成立的是·································································································()231abaA.+=;.=;.=;.=-ababa+ba+bab-b-b-a+ba+b63.某反比例函數(shù)的圖象經(jīng)過點,此數(shù)的圖象也經(jīng)過點()A,-3)B-3,)C,3)D,6菱具有而一般平行四邊形不具有的性質(zhì)是

()A對邊相等對角相等C.對角線相平分;D.對角線互相垂如,在平行四邊形中下結(jié)論中錯誤的是()A∠1=∠B.∠=∠BCDC=CDD.AC⊥BD下四個命題:①一組對邊平行且一組對角相等的四邊形是平行四邊形;②對角線互相垂直且相等的四邊形是正方形順次連結(jié)矩形四邊中點得到的四邊形是菱形等三角形既是軸對稱圖形又是中心對稱圖形.其中真命題共有()A.1個B2個C.3個D.個10.圖,在四邊形中∠A90°=,=7,點、分為線段BC、上的動點,點、分為DM、的中點,則長度的最大值為···················()A.

B..D2二填題本大題共8小題,每小題3分共24分.不需寫解答過程,請將答案直接寫在答題卡相應(yīng)位置上)x11.若式的為0則x=▲.1-.分式、的簡公分母是▲.3x13.形中,對角線=,=,菱形的面積是▲.14如圖,矩形ABCD的角線AC、交于點O,∠AOD,=4,則長▲.ADA

E

O

D

A

P

D

A

DB

O

FBC

B

B

P

E

F第14題圖

第15題圖

第17題圖

第題圖15圖eq\o\ac(□,)的角線相于點點F別是線段的點+=22,的周長是16cm,則的長為▲cm.16.知x0,分式的是▲.x圖形的長為別是邊的上點MC=MB=NC是對角上上點,則PM的最小值是▲..如圖所示,點為正方的對角線上一點,過點P作BC,垂足分別為點E、,連接EF.下列結(jié)論中是腰直角三角形;AP;AD=;PFE=BAP其中正確的結(jié)論是▲)三解題本題共小題,共76分,請在答卡指定區(qū)域內(nèi)作答,解答時應(yīng)寫出文字說明、推理過程或演算步驟)19)算.(1-+

+b

;

(2

a.a(chǎn)a20

)圖,點A、是的角線EF所在直線上的兩點且AE=CF.求證:四邊形平行四邊形.D

FEAB第20題圖xx1216)先化簡:),后在-,,2四數(shù)中找xx一個你認(rèn)為合適的代求值.226)方程:

x

31

.236觀等11=-,……45

111111=1-②=-=-④′22′3344′5試用含字母的式表示出你發(fā)現(xiàn)的規(guī)律,并證明該等成立;1111+++=)227本10圖27本10圖1圖248)圖,一次函數(shù)y=+b與比例函數(shù)=的圖象交于點A(16B(3n)兩點.求反比例函數(shù)和一次函數(shù)的表達(dá)式;點是一次函數(shù)=kx+圖位第一象限內(nèi)的一點,過點作MNx軸,垂足為點N過點作BDy軸,垂足點D,eq\o\ac(△,)MON的積小eq\o\ac(△,)BOD的面積,直接寫出點M的坐標(biāo)x取值范圍.第題258)eq\o\ac(□,)中過點D作DE⊥于,點在CD上,=BE,連接,.求證:四邊形DEBF是矩形;若平分∠,=,=,求的面積.

A

DFBE第題圖2610)圖1,方ABCD中點O是對角ACAO上(不與A、重)的一個動點,過點P作⊥PB且交于點.求證:=;過點E作于F,圖2.正方形ABCD的長為2,在點P運的過程,的度是否發(fā)生變化?若不變,請直接寫出這個不變的值;若變化,請說明理由.ADAPP

DB

O

EOF

E(把一張矩形片ABCD如圖方式折疊使點落邊AD(為點)點落在點分與邊、BC交于試在圖中連接,求證:四邊形是形;若AB,=,求線能取到的整數(shù)值.A

B

ADA

B

F

C

B

C

B

C第27題圖

備用圖

備用圖121212122810)面直角坐標(biāo)系xOy中已知函數(shù)=(>0)與y﹣(<)的圖象如圖所示,點AB是數(shù)=(x>)圖象上的兩點點P是=(x<)的圖象上的一點且∥x軸點Q是x軸一點設(shè)點B的坐標(biāo)分別為mm求的面積;若是等腰直角三角形,求點Q的標(biāo);若是以AB為的等腰三角形,求的.參答及分準(zhǔn)一選擇(330A

C

C

A

C

A

D

D

B

D二填題22011.;.

y

;;.;.2.5;.17.;.②④.三解題19)原=

(a-ba)+a+b

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

a+ba+b

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

a(a-×(+1)a-

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

aa+

.·············································································································320.明:連接DB交EF于O.∵四邊形是行四邊形,∴=OB,OE=.·······································································································2∵=,∴+=+,=.4∴四邊形是平行四邊形.·······················································································21)(x-1)x-21.:原式=···············································································x-+x=+.·············································································································4取x=.····························································································································5∴原式2+=.············································································································6(222.;驗不能漏。123)=-(為正整·······················································nn+1)eq\o\ac(□,S)eq\o\ac(□,S)證明:∵

1+1n-=-nn+(+1)

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

111=-.n(n++(2

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

y

6x

,y()x3

;25)四邊形是行四邊形,∥AB,即DF∥EB.又∵DF=,四邊形DEBF是行四邊形.2∵⊥,∴∠=.四邊形DEBF是形.()四邊形DEBF是矩形,∴==,BDDF.

4∵⊥,=

+DE

+

=.∵∥,∠=.∵平∠DAB,∠DAF=∠∴∠=∠.∴==.···········································································································∴=.∴ABAEBE=+=.∴=·BF=×=.·············································································826)圖,連接.∵四邊形是方形,∴,∠=DCA,∠BCD=.又∵PC=,△BCP≌DCP.∴=,PBCPDC.3∵⊥,BPE=.∴在四邊形BCEP中∠∠=-∠-∠=.又∵∠+∠,∴∠=PED∴∠=∠.∴=.·················································································································6∴=.·················································································································7(2P3BCCD11()

PE

的長度不發(fā)生變化,=.······································································10(OBPEFeq\o\ac(△,≌)82)連接BB′.由折疊知點B、關(guān)EF稱.是段BB的垂直平分線.=BEBFF.································································································2四形是矩形ADBC.B=.由折疊得B=.B=B.B′=BF················································································································=BE′F=.四形BFB是菱形.

4()圖1,點E與重時,四邊形′是正方形,此時BF最.5∵四邊形ABFB′正方形,∴==,最為.··············································································6如圖2,點與D重時BF最大.·····························································7設(shè)BF,=-x,==.在eq\o\ac(△,Rt)中,由勾股定理得+2=.∴-x)+=x2,得,即=10.···················································9∴≤≤.∴線段長能取到的整數(shù)值為8,,.·························································12eq\o\ac(△,S)1212eq\o\ac(△

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