2022學(xué)年第一學(xué)期初三數(shù)學(xué)練習(xí)卷（202303）（完卷時(shí)間分鐘，滿分150分）考生注意：125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．下列函數(shù)中，函數(shù)值y隨自變量x的值增大而減小的是（）xx22(A)y；(B)y；(C)y；(D)y．22xx2yx3（-BB2的坐標(biāo)是（▲）（A，1B-2，-（，D-，-）.3．在△中，點(diǎn)DE分別在邊AB、上，下列條件不能判定DE//的是（）ADAEBDCE（A）；（）；（C）；（D）．BDCEABAC4．如果點(diǎn)C是線段的中點(diǎn)，那么下列結(jié)論中正確的是（）（A）ACBC；B）；）0；D）2．5與x）13131313（A）sin；（）cos；C）tan；D）cot．6．如圖，以為斜邊作等腰直角三角形，再以點(diǎn)A為圓心，長為半徑作弧，交線段于點(diǎn)P，那么AP∶等于（▲）（A）∶2；（）∶3；（）2∶3；（D）∶3．圖1二、填空題（本大題共12題，每題4分，滿分48分）7．已知線段a4，b，如果線段c是a、b的比例中項(xiàng)，那么c的值是▲．1228．已知fxA，那么f1的值是▲．Bx1CD9．一次函數(shù)y3x1的圖像不經(jīng)過的象限是▲．EF10．如果兩個(gè)等邊三角形的邊長的比是1:4，那么它們的周長比是．如圖2，已知，它們依次交直線l、lCEDF．▲．圖212如果2，6，DF3，那么12．在△ABC中，如果AB=AC=7，BC=，那么osB的值是13中，是邊上的中線，GAD=6DG的長是▲．▲．▲．14．如圖，在△中，點(diǎn)DEF分別在邊ABAC、上，DE∥，EFAB，如果DEBC=∶5EFAB的值是▲．AADDEOBFCBC圖3圖4圖515．如圖4，在梯形ABCD中，ADBC，AC與相交于點(diǎn)O，如果AD=3：，那么S:S的值為16．已知一斜坡的坡度i1:3，高度為20米，那么這一斜坡的坡長▲.▲米．90，A、DECDE為▲．18sin的值叫做這個(gè)平4么這個(gè)平行四邊形的“變形系數(shù)”是▲．變形圖6三、解答題（本大題共7題，滿分78分）19101計(jì)算：4cos30sin60．2tan45cot3064已知拋物線yx2x3，將這條拋物線向左平移y23個(gè)單位，再向下平移2個(gè)單位．（1標(biāo)、對(duì)稱軸，并說明它的變化情況；（27所示的平面直角坐標(biāo)系內(nèi)畫出平移后的拋物1線．O1x圖721.（本題滿分分，每小題滿分5分）如圖，在△D在邊BC上，AB=BD=BC，E是的中點(diǎn)．12A（1）求證：BAE=C；（2a，b，BC用含向量a、b的式子表示向量．2210ED圖81在這個(gè)活動(dòng)中他們設(shè)計(jì)了以下兩種測量的方案：課題方案測量示意圖測量教學(xué)大樓的高度方案一方案二測得數(shù)據(jù)甲樓和乙樓之間的距離AC=20米，乙甲樓和乙樓之間的距離=20米，甲樓樓頂端D測得甲樓頂端B的仰角頂端B測得乙樓頂端D的俯角，測得甲樓底端A的俯角，測得乙樓底端C的俯角，40參考數(shù)據(jù)sin，，sin，，，，tan，，tan請你選擇其中一種方案，求甲樓和乙樓的高度（結(jié)果精確到1米）2312分，每小題滿分6分）已知：如圖9，在梯形中，ADBCE在邊對(duì)角線上，EAD∠BDC．（1）求證：；ADDE（2）如果點(diǎn)FDC上，且求證：EFBC．，DFEBC圖92412分，第小題滿分4分，第小題每小題滿分4分）bx的對(duì)稱軸為直線x=yax23如圖，在平面直角坐標(biāo)系xOy點(diǎn)為Ax軸相交于點(diǎn)、（30）（點(diǎn)BC的左邊），與y軸相交于點(diǎn)D．（1）求拋物線的表達(dá)式；y（DE．①DEACACDE的面積；②如果點(diǎn)E在直線DC上，點(diǎn)Q在平移后拋物線的對(duì)稱軸上，當(dāng)DQE=∠時(shí)，求點(diǎn)Q的坐標(biāo)．Ox圖102514分，第滿分4分，第（2小題滿分5分）E．E（1）求證：CE；（2）如果3DE6．①求CF的長；ADF②如果=10，求ABC值．BC圖2022學(xué)年度初三數(shù)學(xué)練習(xí)卷參考答案及評(píng)分說明（）6題，每題4分，滿分241．；2D；3；．B；5C；A.題，每題4分，滿分7.；8.；9.第四象限；10.1:4；325．；12.；；713．；3214．17．；15．53234516．2010；3；18．．7題33119．解：原式=4························································（82221313=53．·························································（223yx22x3x4．·············································（1220）解平移后新拋物線的表達(dá)式：yx22．····································（12該拋物線的開口方向向下，頂點(diǎn)坐標(biāo)為（2，x2，在直線x2的左側(cè)部分是上升的，右側(cè)部分是下降的．······························（4（2）在平面直角坐標(biāo)系中畫出正確的圖像．·······································（4（頂點(diǎn)、對(duì)稱軸、大致形狀正確）121）解AB=BD=BC，E是的中點(diǎn)．21，212∴．·······························································（112∵ABBD∴．·····················································（1∵∠ABC=EBA∴△ABC∽△EBA．···········································（2∴BAE=C．········································································（1(2)∵a，b，∴ba．············································（1∵2，∴b2a．·················································（2∴ababa．·······································（222．選擇方案一：過點(diǎn)D作DEAB，垂足為．··········································（1∵∠AED=EAC=∠ACD＝90°．∴四邊形ACDE是矩形．∴DEAC=20米，=．································（2在RtADE中，∠AED90°，，即400.84．20∴·······················································（3在RtBED中，∠BED90°，tan，即tan350.70．20∴·······························································（3∴·········································（1答：甲樓和乙樓的高度分別為17米．選擇方案二：延長CD，交水平線于H．··················································（1∵∠FBA=BAC=DCA＝90°．∴四邊形ACHB是矩形．∴BHAC=20米，=．································（2CHCH在Rt中，BHC90°，tan，即tan571.53．20∴······················································（3在Rt中，BHC90°，tan，即tan350.70．20∴······························································（3∴········································（1答：甲樓和乙樓的高度分別為17米．23.（）∵ADBC∴．················································（1∵∠EAD=BDC，∴△ADE∽△DBC．·················································（2∴．·················································································（2∴·······································································．（1（2∵△ADE∽△DBC，∴∵．∴，∴．·····························································（2，即．········································（2∴EFBC．·······················································································（224）由題意得，拋物線y對(duì)稱軸為直線x=，經(jīng)過點(diǎn)B(3，，代入得解得·············································（3∴拋物線的表達(dá)式是yx24x3．················································（1(2)①由題意得：原拋物線的頂點(diǎn)A的坐標(biāo)是（2-···························（1∴將拋物線向左或向右平移后，新拋物線的頂點(diǎn)E的坐標(biāo)為（x，1∵（2-1C（，0∴AC的表達(dá)式是：y=x-．∵DEACD的坐標(biāo)為（0，3，直線DE的表達(dá)式是：y=x+．∴點(diǎn)E的坐標(biāo)為（-，-1·······························································（1設(shè)直線DE與x軸交于點(diǎn)，則P的坐標(biāo)為（-0·····························（11∴SS636115．·························（12②將拋物線向左或向右平移后，新拋物線的頂點(diǎn)E的坐標(biāo)為（x，1∵D（，3C3，0∴DC的表達(dá)式是：yx3．∵點(diǎn)E在直線DC上，點(diǎn)E的坐標(biāo)為（4，1·································（1∴平移后新拋物線的對(duì)稱軸是直線x4．當(dāng)點(diǎn)Q在x軸上方的對(duì)稱軸上時(shí)，∵∠DQE=∠CDQ∴ED=EQ．∵（4，1D（03，∴42．······································（1∴點(diǎn)Q的坐標(biāo)為（4，421····························································（1當(dāng)點(diǎn)Q在x軸下方的對(duì)稱軸上時(shí)，同理42．∴點(diǎn)Q的坐標(biāo)為（4，421···························································（1綜上所述，當(dāng)DQE=時(shí)，點(diǎn)Q的坐標(biāo)為（，421）或（，42125）四邊形ABCD是平行四邊形，∴ADBC．∴∠ADBCBDDEC=∠BCF．∵∠DCE=ADB，DCE=∠CBD．∴△DCEFBC．··········································································（2DC∴．················································································（1CE∵ABCD∴，即CE···································（1．(2)①∵AD=3DE=6∴DE=2AE=4．DE∵ADBC，∴∵ADBC∴．DE1．314，即4·······························································（2．∴∵∠EDF=ECD，DEF=∠CED，DE∴△EDF∽△ECD∴．DE2，解得(負(fù)值舍去)···········································（21∴42∴4413．·····························································（1DF14②ADBC∴．5∵BD=10∴DF．2DFDE∵△EDF∽△ECD∴．DCDC．·····························································（2∵DE=2EC=4，∴5過點(diǎn)D作BC的延長線于點(diǎn)H．在Rt與Rt中，設(shè)CHx，得5xx)2222，213222，解得x．·································（14CH13在RtcosDCH．·····································