2025北京平谷區(qū)初三（上）期末數(shù)學(xué)試題及答案

2025北京平谷初三（上）期末數(shù)學(xué)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(gè)選項(xiàng)，其中只有一個(gè)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(zhǎng)是929423（A）（B）（）（D）4y=2x23.在平面直角坐標(biāo)系中，將拋物線平移4個(gè)單位長(zhǎng)度后所得到的拋物線的表達(dá)式為先向左平移3個(gè)單位長(zhǎng)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(diǎn)A、B、C為⊙O上三點(diǎn)，∠ACB=30°，AB=3AB的長(zhǎng)是3A.B.C.D.425.如圖，已知正方形ABCD的邊長(zhǎng)為6，點(diǎn)E是邊上一點(diǎn)，CE=2，以CE為一邊作正方形CEMN，連接AM交CD于點(diǎn)H，則DH的長(zhǎng)為4323A．B．1C．3D．66.點(diǎn)A1，yB3，y）是反比例函數(shù)y=?圖象上的兩點(diǎn)，那么yy1212x小關(guān)系是（）A．y1＞2B．y1y2C．y1y2D．不能確定7.加工爆米花時(shí)，爆開(kāi)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(shí)間（分鐘）滿足的函數(shù)關(guān)系式為：p=at+bt+c(a0)2，如圖記錄了三次相同條件下實(shí)驗(yàn)的數(shù)據(jù)，根據(jù)上述函數(shù)模型和實(shí)驗(yàn)數(shù)據(jù)，可以得到最佳加工時(shí)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(diǎn)(B、點(diǎn)C)，連接AD，以AD為邊作等邊△AED.給出如下三個(gè)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(dá)點(diǎn)C處時(shí)，發(fā)現(xiàn)自己在地面上的影子CE長(zhǎng)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(diǎn)，如果∠AOC+∠EOD=180o，OD=5DE=6，那么AC的長(zhǎng)為_(kāi)________________.13.若拋物線y=kx2+2x?1的頂點(diǎn)在x軸上，則k的值為．14．如圖，點(diǎn)A、B在雙曲線5上，過(guò)點(diǎn)A作AC⊥x軸于點(diǎn)CB作⊥y軸于點(diǎn)D，連接y=xOA、OB，設(shè)△OBD的面積為SOAC的面積為S則S1S21，2，15.中國(guó)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(zhǎng)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(gè)展館，各展館參觀所需要的時(shí)間如下表，其中展館BE設(shè)有特定時(shí)間段的專業(yè)講解，若明明準(zhǔn)備：00進(jìn)科技館，1200離開(kāi)（各展館之間轉(zhuǎn)換時(shí)間忽略不計(jì)）.(1)若不考慮專業(yè)講解的情況下，明明最多可以參觀完___個(gè)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(fèi)時(shí)間的原則，請(qǐng)給出最合理的參觀順序_________.展館AB9：30-1100每半小時(shí)一場(chǎng)，共3場(chǎng)CDE1000-1200每1場(chǎng)，共2場(chǎng)F專業(yè)講解無(wú)無(wú)無(wú)無(wú)參觀所需時(shí)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(yīng)寫(xiě)出文字說(shuō)明、演算步驟或證明過(guò)程.1117．計(jì)算：2sin+3+2?1?18．18．如圖，四邊形ABCD是平行四邊形，CE⊥AD于點(diǎn)E，點(diǎn)E恰為AD中點(diǎn)，CF⊥AB于點(diǎn)F，當(dāng)BF=2，AD時(shí)，求AB的長(zhǎng).19.已知：如圖，△ABC中，AB=AC，AB>BC.1求作：線段BD，使得點(diǎn)D在線段AC上，且∠CBD=∠BAC.2作法：①以點(diǎn)A為圓心，AB長(zhǎng)為半徑畫(huà)圓；②以點(diǎn)C為圓心，BC長(zhǎng)為半徑畫(huà)弧，交⊙A于點(diǎn)P（不與點(diǎn)B③連接BP交AC于點(diǎn)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(diǎn)C在⊙A上.∵點(diǎn)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(duì)應(yīng)值如下表：xy……-1803102304……-1m（1）直接寫(xiě)出m的值，m=________；（2）求此二次函數(shù)的表達(dá)式.21.如圖，某槳輪船的輪子被水面截得的弦AB，過(guò)O作半徑OC⊥弦AB于點(diǎn)D,若輪子的半徑為5米，弦AB長(zhǎng)為8米，依題意補(bǔ)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(fēng)鈴，微風(fēng)吹過(guò)，玲聲悠揚(yáng)，是金海湖景區(qū)的主要景觀之一.某校組織九年級(jí)學(xué)生到金海湖景區(qū)參加社會(huì)實(shí)踐活動(dòng)，數(shù)學(xué)小組的同學(xué)最初的目標(biāo)是測(cè)量湖光塔的高度，但是他們通過(guò)網(wǎng)絡(luò)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(wù)確定為測(cè)量湖光塔所在的中心島小山的高度，數(shù)學(xué)小組設(shè)計(jì)的方案如圖所示，他們?cè)邳c(diǎn)C處用測(cè)角儀測(cè)得塔頂A的仰角為45°，此時(shí)，由于樹(shù)木的遮擋，看不清塔底，他們延水平方向向后走64米在點(diǎn)D處用測(cè)角儀測(cè)得塔底B的仰角為26.5°．請(qǐng)根據(jù)他們網(wǎng)上查到的數(shù)據(jù)和測(cè)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(biāo)系中，直線y1k與雙曲線=y的交點(diǎn)是．(a,x（1）求a和k的值；6（2）當(dāng)x3時(shí)，對(duì)于x的每個(gè)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(xiě)出m的取值范圍．24.如圖，已知△ABC中，AB=BCD是BC邊上一點(diǎn)，連接AD，以AD為直徑畫(huà)⊙O，與AB邊交于點(diǎn)E，與AC邊交于點(diǎn)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(zhǎng).525.某客運(yùn)站為了了解早高峰時(shí)間段運(yùn)營(yíng)情況，有效的緩解該時(shí)段乘客的等待時(shí)間，對(duì)早上6:00-8:00時(shí)間段內(nèi)，客運(yùn)站累計(jì)候車(chē)人數(shù)和累計(jì)承載人數(shù)進(jìn)行統(tǒng)計(jì)，為了便于記錄，將早上6:00開(kāi)始每分鐘記作一個(gè)單位時(shí)間，記為時(shí)間（0≤x12y1,累計(jì)承載人數(shù)記為y2.。下面是他們的調(diào)查過(guò)程，請(qǐng)補(bǔ)充完整：（1）他們調(diào)取了客運(yùn)站該時(shí)段內(nèi)累計(jì)候車(chē)人數(shù)y與累計(jì)承載人數(shù)y隨x的變化而變化的有關(guān)數(shù)據(jù):12時(shí)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(jì)候車(chē)人數(shù)y1（萬(wàn)人）累計(jì)承載人數(shù)y2（萬(wàn)人）m（1）補(bǔ)全表格，m的值為_(kāi)_______；（2）通過(guò)分析數(shù)據(jù)，發(fā)現(xiàn)可以用函數(shù)刻畫(huà)y與xy與x的關(guān)系，在給出的平面直角坐標(biāo)系中，補(bǔ)全表中1，2各對(duì)對(duì)應(yīng)值為坐標(biāo)的點(diǎn)，畫(huà)出這兩個(gè)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(wèn)題：①大約分時(shí)，客運(yùn)站滯留人數(shù)最多；②客運(yùn)站將在滯留乘客人數(shù)達(dá)到0.5萬(wàn)人及以上的時(shí)間段增派車(chē)次緩解供需壓力，公司約在分時(shí)間段增派車(chē)次更合理.點(diǎn)點(diǎn)分至點(diǎn)26．在平面直角坐標(biāo)系xOy中，已知二次函數(shù)y=x2?2+m2?4．（1）當(dāng)m=1時(shí)，求拋物線與x軸的交點(diǎn)坐標(biāo)；（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(diǎn)，yxyGxxy<y,求m的取值范圍.122121227.Rt△ABC中，∠ACB=°∠B=D是AB邊中點(diǎn)，點(diǎn)E是BC邊上一點(diǎn)（不與點(diǎn)B、點(diǎn)C重DE，將線段DE繞點(diǎn)D逆時(shí)針旋轉(zhuǎn)2α，得到線段DF，連接EF、（1）如圖1，若α=30°，點(diǎn)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(guān)系，從圖2、圖3中任選一種情況進(jìn)行證明.28.我們給出如下定義：在平面內(nèi)，已知點(diǎn)MG，點(diǎn)M到圖形G上所有點(diǎn)的距離的最小值稱作點(diǎn)M到圖形G的距離．（1）平面直角坐標(biāo)系下，已知點(diǎn)P(0,3),以O(shè)為圓心，1為半徑畫(huà)圓，則點(diǎn)P到⊙O的距離為_(kāi)_________；（2）平面直角坐標(biāo)系下，已知點(diǎn)P(0,3),在平面內(nèi)有一個(gè)矩形ABCD，A（-2,1B2,1D（-2，-1）.①當(dāng)矩形繞著點(diǎn)O旋轉(zhuǎn)時(shí)，點(diǎn)P到矩形的距離d的取值范圍為_(kāi)_________．②若MABCD上一點(diǎn)，連接OM，以O(shè)M為直徑畫(huà)圓，記作圓G，則點(diǎn)P到圓G的距離d的取值范圍為_(kāi)_________．參考答案一、選擇題（本題共16分，每小題2分）題號(hào)答案12345678DADACCBB二、填空題（本題共16分，每小題2題號(hào)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(yīng)寫(xiě)出文字說(shuō)明、演算步驟或證明過(guò)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(diǎn)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(diǎn)坐標(biāo)為（2，-1），··································2∴設(shè)拋物線的解析式為y=（x-2）1a）2··················································································································3∵拋物線過(guò)點(diǎn)（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(diǎn)為（0，3），······································2∴設(shè)拋物線的解析式為y=+bx+3（a）2··················································································································3∵拋物線過(guò)點(diǎn)（-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(bǔ)全圖形1解：∵半徑OC⊥弦AB于點(diǎn)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(shè)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(guò)點(diǎn)(a,x∴a=2······································································································1∵直線y1過(guò)點(diǎn)=+(2,∴k=1······································································································22（2）······························································································63（一個(gè)界值1分，符號(hào)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(huà)出函數(shù)圖象.....................................................................3(3)①7點(diǎn)20分...........................................................................4②6點(diǎn)45分至7點(diǎn)45分...................................................................................................626.解：（1）當(dāng)m=1時(shí)，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(gè)交點(diǎn)（3，0）和（-10）.............................................................................................................................................2(2)由y=x2?2+m2?4=(x?m)?42∴拋物線的頂點(diǎn)為（,-4）,拋物線的對(duì)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(gè)交點(diǎn)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(shí)，m?2?1解得，m1............................................................................................................................................4（m-2，0）（m+2，0））（，）0x=m-2-12m+12此時(shí)，有m+2?1.3解得，?3m?2.................................................................................63m1∴或?