2025北京平谷區(qū)初三（上）期末數(shù)學試題及答案

2025北京平谷初三（上）期末數(shù)學2025年1月一、選擇題（本題共16分，每小題2分）下面各題均有四個選項，其中只有一個是符合題意的.．．1．在RtABC中，∠＝90°，AC＝4，AB＝，則sinA43453435A．B．C．D．2．如圖，直線lll，直線ll被直線ll、l所截，截得的線段分別12345123為AB，BCDE，EF，若AB＝4，BC＝6，DE＝，則EF的長是929423（A）（B）（）（D）4y=2x23.在平面直角坐標系中，將拋物線平移4個單位長度后所得到的拋物線的表達式為先向左平移3個單位長度，再向下y=2(x?2+4y=2(x?2?4?4A．C．B．D．y=2(x2+4y=2(x24．如圖，點A、B、C為⊙O上三點，∠ACB=30°，AB=3AB的長是3A.B.C.D.425.如圖，已知正方形ABCD的邊長為6，點E是邊上一點，CE=2，以CE為一邊作正方形CEMN，連接AM交CD于點H，則DH的長為4323A．B．1C．3D．66.點A1，yB3，y）是反比例函數(shù)y=?圖象上的兩點，那么yy1212x小關系是（）A．y1＞2B．y1y2C．y1y2D．不能確定7.加工爆米花時，爆開且不糊的粒數(shù)的百分比稱為“可食用率”.在一定條件下，可食用率P與加工時間（分鐘）滿足的函數(shù)關系式為：p=at+bt+c(a0)2，如圖記錄了三次相同條件下實驗的數(shù)據(jù)，根據(jù)上述函數(shù)模型和實驗數(shù)據(jù)，可以得到最佳加工時間為A.3.5分鐘C.4分鐘B.3.75分鐘D.4.258.如圖，等邊△ABCD是BC邊上一點(B、點C)，連接AD，以AD為邊作等邊△AED.給出如下三個結(jié)論：34SS①BE=DC;②△DBE∽△；③1上述結(jié)論一定正確的是(A)①(B)①③(D)①②③(C)②③二、填空題（本題共16分，每小題2分）9．函數(shù)2x?4的自變量x的取值范圍是．mn232mm+n10．若=，則=．如圖，身高1.6米的小林從一盞路燈下B處向前走了8米到達點C處時，發(fā)現(xiàn)自己在地面上的影子CE長2米，則路燈的高AB為米．12．如圖，在⊙O中，AB是⊙O的直徑，C，D，E是⊙O上的點，如果∠AOC+∠EOD=180o，OD=5DE=6，那么AC的長為_________________.13.若拋物線y=kx2+2x?1的頂點在x軸上，則k的值為．14．如圖，點A、B在雙曲線5上，過點A作AC⊥x軸于點CB作⊥y軸于點D，連接y=xOA、OB，設△OBD的面積為SOAC的面積為S則S1S21，2，15.中國古代建筑中的斜脊結(jié)構(gòu)，既有利于排水，又有利于保溫，是古代工匠智慧的體現(xiàn).如圖，房屋的屋頂截面結(jié)構(gòu)為等腰三角形，若斜脊AB的坡度i為1：2，房子側(cè)寬BC為12米，AB的長為米.16.周末，明明要去科技館參觀，該科技館共有、B、CDE、F六個展館，各展館參觀所需要的時間如下表，其中展館BE設有特定時間段的專業(yè)講解，若明明準備：00進科技館，1200離開（各展館之間轉(zhuǎn)換時間忽略不計）.(1)若不考慮專業(yè)講解的情況下，明明最多可以參觀完___個展館；(2)若B、E展館必須參觀且正好趕上專業(yè)講解，本著不浪費時間的原則，請給出最合理的參觀順序_________.展館AB9：30-1100每半小時一場，共3場CDE1000-1200每1場，共2場F專業(yè)講解無無無無參觀所需時間（分603045156090鐘）三、解答題(68分，第171819、202122題，每小題5分；第23、24、2526題，每小題6分；第27、28題，每小題7)解答應寫出文字說明、演算步驟或證明過程.1117．計算：2sin+3+2?1?18．18．如圖，四邊形ABCD是平行四邊形，CE⊥AD于點E，點E恰為AD中點，CF⊥AB于點F，當BF=2，AD時，求AB的長.19.已知：如圖，△ABC中，AB=AC，AB>BC.1求作：線段BD，使得點D在線段AC上，且∠CBD=∠BAC.2作法：①以點A為圓心，AB長為半徑畫圓；②以點C為圓心，BC長為半徑畫弧，交⊙A于點P（不與點B③連接BP交AC于點D.線段BD就是所求作的線段.（1（2）完成下面的證明.證明：連接PC.∵AB=AC，∴點C在⊙A上.∵點P在⊙A上，1∴∠CPB=∠BAC（_______.2∵BC=PC，∴∠CBD=_______.1∴∠CBD=∠BAC.220．已知二次函數(shù)幾組x與y的對應值如下表：xy……-1803102304……-1m（1）直接寫出m的值，m=________；（2）求此二次函數(shù)的表達式.21.如圖，某槳輪船的輪子被水面截得的弦AB，過O作半徑OC⊥弦AB于點D,若輪子的半徑為5米，弦AB長為8米，依題意補全圖形，并求輪子的吃水深度CD為多少米.22.湖光塔坐落在平谷區(qū)金海湖中心島的山頂，七層八角形樓閣式建筑掛滿風鈴，微風吹過，玲聲悠揚，是金海湖景區(qū)的主要景觀之一.某校組織九年級學生到金海湖景區(qū)參加社會實踐活動，數(shù)學小組的同學最初的目標是測量湖光塔的高度，但是他們通過網(wǎng)絡搜索發(fā)現(xiàn)，網(wǎng)上可以查到湖光塔的塔高為30米，所以他們把任務確定為測量湖光塔所在的中心島小山的高度，數(shù)學小組設計的方案如圖所示，他們在點C處用測角儀測得塔頂A的仰角為45°，此時，由于樹木的遮擋，看不清塔底，他們延水平方向向后走64米在點D處用測角儀測得塔底B的仰角為26.5°．請根據(jù)他們網(wǎng)上查到的數(shù)據(jù)和測量數(shù)據(jù)求中心島小山BE的高度約為sin26.526.526.5）623.在平面直角坐標系中，直線y1k與雙曲線=y的交點是．(a,x（1）求a和k的值；6（2）當x3時，對于x的每個值，函數(shù)ym0)既大于函數(shù)=（y的值，又小于函數(shù)xy=+1的值，直接寫出m的取值范圍．24.如圖，已知△ABC中，AB=BCD是BC邊上一點，連接AD，以AD為直徑畫⊙O，與AB邊交于點E，與AC邊交于點F，EF=AF，連接DE.（1）求證：BC是⊙O的切線；3（2）若BC=10，cos=，求AC的長.525.某客運站為了了解早高峰時間段運營情況，有效的緩解該時段乘客的等待時間，對早上6:00-8:00時間段內(nèi)，客運站累計候車人數(shù)和累計承載人數(shù)進行統(tǒng)計，為了便于記錄，將早上6:00開始每分鐘記作一個單位時間，記為時間（0≤x12y1,累計承載人數(shù)記為y2.。下面是他們的調(diào)查過程，請補充完整：（1）他們調(diào)取了客運站該時段內(nèi)累計候車人數(shù)y與累計承載人數(shù)y隨x的變化而變化的有關數(shù)據(jù):12時間段0123456789100.51.11.62.22.93.64.25.15.76.06.36.56.60.51.01.52.02.53.03.54.04.55.05.56.512累計候車人數(shù)y1（萬人）累計承載人數(shù)y2（萬人）m（1）補全表格，m的值為________；（2）通過分析數(shù)據(jù)，發(fā)現(xiàn)可以用函數(shù)刻畫y與xy與x的關系，在給出的平面直角坐標系中，補全表中1，2各對對應值為坐標的點，畫出這兩個函數(shù)的圖象；(3)根據(jù)以上數(shù)據(jù)與函數(shù)圖象，解決下列問題：①大約分時，客運站滯留人數(shù)最多；②客運站將在滯留乘客人數(shù)達到0.5萬人及以上的時間段增派車次緩解供需壓力，公司約在分時間段增派車次更合理.點點分至點26．在平面直角坐標系xOy中，已知二次函數(shù)y=x2?2+m2?4．（1）當m=1時，求拋物線與x軸的交點坐標；（2）將拋物線在x軸上方的部分沿x軸翻折，其余部分保持不變，得到的新函數(shù)記為G，若點，yxyGxxy<y,求m的取值范圍.122121227.Rt△ABC中，∠ACB=°∠B=D是AB邊中點，點E是BC邊上一點（不與點B、點C重DE，將線段DE繞點D逆時針旋轉(zhuǎn)2α，得到線段DF，連接EF、（1）如圖1，若α=30°，點F剛好落在BC邊上，BE=1，則AF=，AC=；（2）判斷AF、BE和BC的數(shù)量關系，從圖2、圖3中任選一種情況進行證明.28.我們給出如下定義：在平面內(nèi)，已知點MG，點M到圖形G上所有點的距離的最小值稱作點M到圖形G的距離．（1）平面直角坐標系下，已知點P(0,3),以O為圓心，1為半徑畫圓，則點P到⊙O的距離為__________；（2）平面直角坐標系下，已知點P(0,3),在平面內(nèi)有一個矩形ABCD，A（-2,1B2,1D（-2，-1）.①當矩形繞著點O旋轉(zhuǎn)時，點P到矩形的距離d的取值范圍為__________．②若MABCD上一點，連接OM，以OM為直徑畫圓，記作圓G，則點P到圓G的距離d的取值范圍為__________．參考答案一、選擇題（本題共16分，每小題2分）題號答案12345678DADACCBB二、填空題（本題共16分，每小題2題號答案9101181281314=1516x21454;F-B-E35三、解答題(本題共6817-22題，每小題5分；第23-26題，每小題6分；第、28題，每小題7分)解答應寫出文字說明、演算步驟或證明過程.17.解：2=2+3+2?1?32··············································································42=2-2········································································································518.解：∵四邊形ABCD是平行四邊形∴∠D=B，AD=BCAB=DC·······································································1∵CE⊥ADCF⊥AB，∴∠CED=∠CFB=90°···················································································2∴CDE··················································································································.................................................3CD∴=····················································································4∵AD=BC=6，E是AD中點DE=3∴∴CD32=6AB=CD=9·····················································································5∴19.··················································································································2圓周角定理···································································································4∠CPB·········································································································520．解：（1）3·········································································································1（2）由表中數(shù)據(jù)可知拋物線的頂點坐標為（2，-1），··································2∴設拋物線的解析式為y=（x-2）1a）2··················································································································3∵拋物線過點（3，0），∴（3-2）?1=02··················································································································4解得a=1，∴拋物線的解析式為y=（x-2）?1.2··················································································································5法2：（2）由表中數(shù)據(jù)可知拋物線與軸交點為（0，3），······································2∴設拋物線的解析式為y=+bx+3（a）2··················································································································3∵拋物線過點（-1，8），（10）a?b+3=8a+b+3=0··················································································································4解得a=1，b=-4.∴拋物線的解析式為y=x?4x+3.2··················································································································521.依題意補全圖形1解：∵半徑OC⊥弦AB于點D，AB=8∴BD=4·····································································································2連接OB·······································································································3∵OB=5由勾股定理OD=3···························································································4∴CD=5-3=2··································································································5∴吃水深度為2米.22.解：由題意，∠AED=90°，∠ACE=45°，∠BDE=26.5°，AB=30，CD=64···················1設BE=xEC=x+30.····················································································226.50.50x∴0.50·····················································································4x+30+64解得x94·································································································5答：小山高度約為94米.623.(1)∵雙曲線y過點(a,x∴a=2······································································································1∵直線y1過點=+(2,∴k=1······································································································22（2）······························································································63（一個界值1分，符號完全正確滿分）24.1）證明∵AD為⊙O的直徑，∴∠AED=90°···························································1∵BA=BC∴∠BAC=BCA∵EF=AF∴∠BAC=FEA·························································2∴∠BCA=FEA∵∠DEF=DAC∴∠DAC+BCA=∠DEA+AEF=90°∴AD⊥BC∴BC為⊙O的切線····················································3（2）∵為⊙O的切線∴∠ADE+BDE=90°∴∠B+BDE=90°∴∠B=ADE3∵cosAFE=53∴B=5BDAB35==∴∴BD3510∴BD=6····································································4由勾股，AD=8∵BC=10∴DC=10-6=4····························································5由勾股=45······························································625.1）6................................1y8765432112xO123456789畫出函數(shù)圖象.....................................................................3(3)①7點20分...........................................................................4②6點45分至7點45分...................................................................................................626.解：（1）當m=1時，y=x?2x?32y=x?2x?32令y=0，得x=x=?1解得12∴拋物線與x軸的兩個交點（3，0）和（-10）.............................................................................................................................................2(2)由y=x2?2+m2?4=(x?m)?42∴拋物線的頂點為（,-4）,拋物線的對稱軸為x=.............................................................................................................................................3令y=，得(x?m)2?4=0解得x=m+，x=m-2，12∴拋物線與x軸的兩個交點為（,0）,（,0）.（3）由題意，圖象G如圖所示，分以下兩種情況：（m-2，0）（m+2，0）x=m-1-2此時，m?2?1解得，m1............................................................................................................................................4（m-2，0）（m+2，0））（，）0x=m-2-12m+12此時，有m+2?1.3解得，?3m?2.................................................................................63m1∴或?