(word完整版)楊浦區(qū)2018學(xué)年度第一學(xué)期期末考試初三數(shù)學(xué)試卷

楊浦區(qū)2018學(xué)年度第一學(xué)期期末質(zhì)量調(diào)研初三數(shù)學(xué)試卷2019.1（測試時(shí)間：100分鐘，滿分：150分）考生注意：1．本試卷含三個(gè)大題，共25題．答題時(shí)，考生務(wù)必按答題要求在答題紙規(guī)定的位置上作答，在草稿紙、本試卷上答題一律無效．2．除第一、二大題外，其余各題如無特別說明，都必須在答題紙的相應(yīng)位置上寫出證明或計(jì)算的主要步驟．一、選擇題：（本大題共6題，每題4分，滿分24分）1．下列四組線段中，成比例的是（A）1,1,2,3；（B）1,2,3,4；（C）2,2,3,3；（D）2,3,4,5．2．如果a:b3:2acb:c，且b是、的比例中項(xiàng)，那么等于（A）4:3；（B）3:4；（C）2:3；（D）3:2.13．如果△ABC中，∠C=90°，sinA，那么下列等式不正確的是22（B）cotA3；（C）sinB3；（D）tanB3．（A）cosA；224．下列關(guān)于向量的運(yùn)算中，正確的是（A）abba；（C）a(a)0；2(ab)2a2b（B）；0aa（D）．5．如果二次函數(shù)中函數(shù)值y與自變量x之間的部分對應(yīng)值如下表所示：112xy......031623......232144那么這個(gè)二次函數(shù)的圖像的對稱軸是直線（A）x0；（B）x1；（C）x3；（D）x1．246．如果以a、b、c為三邊的三角形和以4、5、6為三邊的三角形相，似那么a與b的比值不可能為2345（A）；（B）；（C）；（D）.3456二、填空題：（本大題共12題，每題4分，滿分48分）xxy35x7．如果，那么▲．y第1頁共8頁8．等邊三角形的中位線與高之比為▲．9．如果兩個(gè)相似三角形的面積比為4:9，較小三角形的周長為4，那么這兩個(gè)三角形的周長和為▲．10．在△ABC中，AB=3，AC=5，BC=6，點(diǎn)D、E分別在邊AB、AC上，且AD=1，如果△ABC∽△ADE，那么AE=▲．11．在△ABC中，AB=AC=5，BC=8，如果點(diǎn)G為重心，那么∠GCB的余切值為▲．12．如果開口向下的拋物線yax25x4a2(a0)過原點(diǎn)，那么a的值是▲．13．如果拋物線y或“>”）.2xbxc的對稱軸在y軸的左側(cè)，那么b▲0（填入“<”214．已知點(diǎn)A（x,y）、B（x,y）在拋物線yx2xm上，如果0xx，那2112212么y1▲y（填入“<”或“>”）.215．如圖，AG//BC，如果AF:FB=3:5，BC:CD=3:2，那么AE:EC=▲．16．某單位門前原有四級臺階，其橫截面如圖所示，每級臺階高為18cm，寬為30cm，為方便殘障人士，擬將它改成斜坡，設(shè)臺階的起點(diǎn)為A點(diǎn)，斜坡的起點(diǎn)為C點(diǎn)，準(zhǔn)備設(shè)計(jì)斜坡BC的坡度i1:5，則AC的長度是▲cm．17．如果拋物線C1的頂點(diǎn)在拋物線C2上時(shí)，拋物線C2的頂點(diǎn)也在拋物線C1上，此時(shí)我2們稱拋物線C1與C2是“互為關(guān)聯(lián)”的拋物線．那么與拋物線y2x是“互為關(guān)聯(lián)”▲（只需寫出一個(gè)）．18．Rt△ABC中，∠C=90°，AC=3，BC=2，將此三角形繞點(diǎn)A旋轉(zhuǎn)，當(dāng)點(diǎn)B落在直線BC上的點(diǎn)D處時(shí)，點(diǎn)C落在點(diǎn)E處，此時(shí)點(diǎn)E到直線BC的距離為▲.且頂點(diǎn)不同的拋物線的表達(dá)式可以是AGABFE30cm18cmABCDCBC（第15題圖）（第16題圖）（第18題圖）三、解答題：（本大題共7題，滿分78分）19．（本題滿分10分，第（1）小題6分，第（2）小題4分）如圖，已知□ABCD的對角線交于點(diǎn)O，點(diǎn)E為邊AD的中點(diǎn)，CE交BD于點(diǎn)G.（1）求OG的值；AEDDGGab（2）如果設(shè)ABa，BCb，試用、表示GO.OCB（第19題圖）第2頁共8頁20．（本題滿分10分，每小題各5分）3yaxbxc(a0)的圖像過點(diǎn)(1,2)和和(0,)．(1,0)y已知二次函數(shù)22（1）求此二次函數(shù)的解析式；543（2）按照列表、描點(diǎn)、連線的步驟，在如圖所示的平面直角坐標(biāo)系內(nèi)畫出該函數(shù)的圖像（要求至少5點(diǎn)）．21-3-2-1-1O1234x-2-3（第20題圖）21．（本題滿分10分，第（1）小題6分，第（2）小題4分）AD是△ABC的中線，tanB=1，cosC=2，AC=2．如圖，52A求：（1）BC的長；（2）∠ADC的正弦值．BCD（第21題圖）22．（本題滿分某學(xué)生為測量一棵大樹AH及其樹葉部分AB的高度，端A的仰角為30°，放在G處測得大樹頂端A的仰角為60°，樹葉部分下端B的仰角45°，已知點(diǎn)F、G與大樹底部H共線，點(diǎn)F、G相距15米，高度為1.5米.求該樹的高度AH和樹葉部分的高度AB．10分）將測角儀放在F處測得大樹頂為A測角儀BDGEFCH（第22題圖）23．（本題滿分6分）已知：如圖，在△ABC中，點(diǎn)ADDE；12分，每小題各D在邊AB上，點(diǎn)E在線段CD上，且∠ACD=∠B=∠BAE.（1）求證：BCACAAEAB.AD2證：2（2）當(dāng)點(diǎn)E為CD中點(diǎn)時(shí)，求DCEEC（第23題圖）B第3頁共8頁24．（本題滿分12分，每小題各4分）yaxbxc(a0)與y軸交于點(diǎn)在平面直角坐標(biāo)系xOy中，拋物線2C（0,2），1它的頂點(diǎn)為D（1,m），且tanCOD.3（1）求（2）將此拋物線向上平移后與y軸交于點(diǎn)B，且OA=OB.若點(diǎn)AE平移所得，求點(diǎn)2）的條件下，點(diǎn)P是拋物線對稱軸上的一點(diǎn)（位于求P點(diǎn)的坐標(biāo)m的值及拋物線的表達(dá)式；x軸正半軸交于點(diǎn)E的坐標(biāo)；A，與是由原拋物線上的點(diǎn)（3）在（x軸上方），且∠APB=45°..y54321-3-2-1O1234-1x-2-3（第24題圖）25．（本題滿分14分，第（1）小題4分，第（2）、（3）小題各5分）已知：梯形ABCD中，AD//BC，AB⊥BC，AD=3，AB=6，DF⊥DC分別交射線AB、射線CB于點(diǎn)E、F.（1）當(dāng)點(diǎn)（2）當(dāng)點(diǎn)E在邊AB上時(shí)（如圖2），聯(lián)結(jié)CE，試問：∠DCE的大小是出∠DCE的正切值；若不確定，則設(shè)AE=x，∠DCE的正切值為y，請求出y關(guān)于的函數(shù)解析式，（3）當(dāng)△AEF的面積為E為邊AB的中點(diǎn)時(shí)（如圖1），求BC的長；否確定？若確定，請求x并寫出定義域；3時(shí)，求△DCE的面積.ADADEECCFBFB（圖1）（圖2）（第25題圖）第4頁共8頁楊浦區(qū)初三數(shù)學(xué)期末試卷參考答案及評分建議2019.1一、選擇題：（本大題共6題，每題1．C；2．D；3．A；4．B；5．D；6．B；二、填空題：（本大題共12題，每題4分，滿分48分）534分，滿分24分）51:37．；8．；9．10；10．或；11．4；12．-2；2352413．<；14．<；15．3:2；16．270；17．y2x2+4x；18．；13三、解答題：（本大題共7題，滿分78分）19．解：（1）∵□ABCD，∴BO=OD，AD//BC，AD=BC.····························（3分）∴EDDG.········································································（1分）BCGB∵點(diǎn)E為邊AD的中點(diǎn)，∴ED1AD1BC.∴DG1.·············（1分）22GB2OG1.··························································（1分）∵BO=OD，∴DG2ABaBCbBDBAADBABCab.················（分）1（2）∵，，∴OG1OG1.···············································（2分）BD6∵BO=OD，，∴DG2∴GO1DB1(ab).····························································（1分）6631）∵二次函數(shù)yax2bxc圖像過點(diǎn)(1,2)、(1,0)和(0,)，20．解：（2a1,abc2,21232xx∴abc0,（3分）∴1,∴二次函數(shù)解析式為y.（2分）b2332c.c.2131yx2x(x1)22.···············································（1分）（2）222xy……-1001230……33-222································································································（2分）圖略··························································································（2分）21．解：（1）作AH⊥BC于H．2，∴AHAC=2.···································（1分）2在Rt△ACH中，∵cosC=2∵AC=2，∴CH=1.··································································（1分）第5頁共8頁∴AH=1.··················································································（1分）在Rt△ABH中，∵tanB=1，∴=.······································（1分）AH15BH5∴BH=5.···················································································（1分）∴BC=BH+CH=6．·······································································（1分）（2）∵BD=CD，BC=6，∴CD=3.························································（1分）∵CH=1，∴DH=2.∴AD=5.····················································（1分）在Rt△ADH中，5．··························（1分,1分）5sinADH=AH15AD∴∠ADC的正弦值為5．522．解：由題意可知∠AEC=30°，∠ADC=60°，∠BDC=45°，F(xiàn)G=15.············（3分）tanADCAC得AC=3x.··············（分）1設(shè)CD=x米，則在Rt△ACD中，由DC又Rt△ACE中，由cotAECEC得EC=3x.··········································（1分）AC∴3x=15+x.······················································································（1分）∴x=7.5.························································································（1分）∴AC=7.53.∴AH=7.53+1.5.····························································（分）1∵在Rt△BCD中，∠BDC=45°，∴BC=DC=7.5.∴AB=AC﹣BC=7.5(31)．··（1分）答：AH的高度是(7.53+1.5)米，AB的高度是7.5(31)米.························（1分）23．證明：（1）∵∠ACD=∠B，∠BAC=∠CAD，∴△ADC∽△ACB.············（2分）∵∠ACD=∠BAE，∠ADE=∠CDA，∴△ADE∽△CDA.·············（2分）∴△ADE∽△BCA.·····························································（1分）ADDE.·····································································（1分）∴BCACAEDE（2）∵△ADE∽△BCA，∴ABACAE，即DEACAB.·····························（1分）AC.····························（1分）AEDE∵△ADE∽△CDA，∴ACADAE，即DEADAE2∴ABACAB.····························································（2分）DE2ACADAD∵點(diǎn)E為CD中點(diǎn)，∴DECE.··················································（1分）AE∴CE2AB.···········································································（1分）AD2第6頁共8頁24．解：（1）作DH⊥y軸，垂足為H，∵D（1,m）（m0），∴DH=m，HO=1.13OH1∵tanCOD∴，∴m=3.......................................................................（1分），DH3yaxbxc的頂點(diǎn)為∴拋物線D（1,3）.2yaxbxc與y軸交于點(diǎn)又∵拋物線C（0,2），2ì?abc3,ì???a1,???b?íb2,∴拋物線的表達(dá)式為yx2x2........（1分）∴í1,（2分）∴2??2ac2.????c2.?????（2）∵將此拋物線向上平移，yx2x2k(k0)∴設(shè)平移后的拋物線表達(dá)式為......................................（1分），2則它與y軸交點(diǎn)B（0,2+k）.∵平移后的拋物線與x軸正半軸交于點(diǎn)A，且OA=OB，∴A點(diǎn)的坐標(biāo)為（2+k,0）..（1分）0(2k)22(2k)2k.∴k∴2,k11.2k0k1.∵，∴yx2x2向上平移了∴A（3,0），拋物線1個(gè)單位........................................（1分）2∵點(diǎn)A由點(diǎn)1個(gè)單位所得，∴E（3,-1）................................................（1分）（3）由（2）得A（3,0），B（0,3），∴AB32.P是拋物線對稱軸上的一點(diǎn)（位于P（1,y），設(shè)對稱軸與AB的交點(diǎn)為M，與E向上平移了∵點(diǎn)x軸上方），且∠APB=45°,原頂點(diǎn)D（1,3）,∴設(shè)x軸的交點(diǎn)為H，則H（1,0）.y∵A（3,0），B（0,3），∴∠OAB=45°,∴∠AMH=45°.∴M（1,2）.∴BM2.P∵∠BMP=∠AMH,∴∠BMP=45°.B∵∠APB=45°,∴∠BMP=∠APB.∵∠B=∠B，∴△BMP∽△BPA..........................................（2分）MBPBABP∴BM.∴BP2BABM3226xOHABP1(y3)26∴.∴y315，y35（舍）.......................................（1分）22∴P(1,35).....................................................................................................................（1分）25．（本題滿分14分，第（1）小題4分，第（2）、（3）小題各5分）DEAEAD.∵E為AB中點(diǎn)，∴解：（1）∵AD//BC，∴AE=BE.∴AD=BF，DE=EF.EFEBBF∵AD=3，AB=6，∴BF=3，BE=3.∴BF=BE.第7頁共8頁∵AB⊥BC，∴∠∴DF=2EF=62.·····················································································（1分）∵DF⊥DC，∠F=45°，∴CF=12.·····························································（1分）F=45°且EF=32.·························································（1分）∴BC=CFBF1239.····································································（1分）12.····················································（1分）tanDCE（2）∠DCE的大小確定，作CH⊥AD交AD的延長線于點(diǎn)H，∴∠HCD+∠HDC=90°.∵DF⊥DC，∴∠ADE+∠HDC=90°.∴∠HCD=∠ADE.又∵AB⊥AD，∴∠A=∠CHD.∴△AED∽△HDC.········································（2分）DEAD.