2007學年第二學期第一次考試九年級數(shù)學試題歡送你參加檢測，祝你取得好成績！請先看清以下三點考前須知：1．檢測范圍：浙教版?數(shù)學?實驗教材九年級上、下兩冊的全部內(nèi)容2．全卷總分值為150分．試卷共7頁，有三大題，24小題．3．本試卷分卷Ⅰ、卷Ⅱ兩局部，請將卷Ⅰ的答案做在卷Ⅱ的相應位置上．卷Ⅰ共2頁、卷Ⅱ共6頁．溫馨提示：請仔細審題，細心答題，相信你一定會有出色的表現(xiàn)！卷Ⅰ一、選擇題〔此題有10小題，每題4分，共40分．請選出各題中一個符合題意的正確選項，不選、多項選擇、錯選，均不給分〕1．生活處處皆學問．如圖，眼鏡鏡片所在的兩圓的位置關系是【▲】A．外離B．外切C．內(nèi)含D．內(nèi)切正面圖〔甲〕〔甲〕C、A、D、Ｂ正面圖〔甲〕〔甲〕C、A、D、Ｂ、3．明明的學校有30個班，每班50名學生，學校要從中抽出1名學生參加社會實踐活動，那么明明被選中的概率是【▲】A．D．不確定4．Rt△ABC中，∠C=90°，AB=13，BC=5，那么tan∠A=【▲】A． B． C． D．5．DBA、ChDBA、ChpOHhpOHhpOHHhpO6．以下命題中，屬于真命題的是【▲】A．所有的等腰三角形都相似B．所有的直角三角形都相似C．所有的等邊三角形都相似D．所有的矩形都相似7．二次函數(shù)圖象的大致位置如圖，以下判斷錯誤的選項是【▲】A．B．C．D．8．如圖，A、B是兩座燈塔，在弓形內(nèi)有暗礁，游艇C在附近海面游弋，且∠AOB=80°，要使游艇C不駛入暗礁區(qū)，那么航行中應保持∠ACB【▲】A．小于40°B．大于40°C．小于80°D．大于80°〔第7題〕m〔第7題〕m(第8題)668(第9題)〕9．用一張扇形的紙片卷成一個如下圖的圓錐模型，要求圓錐的母線長為6cm，底面圓的直徑為8cm，那么這張扇形紙片的圓心角度數(shù)是【▲】A． B．C． D．10．如圖，在7×12的正方形網(wǎng)格中有一只可愛的小狐貍，觀察畫面中由黑色陰影組成的五個三角形，那么相似三角形有【▲】A．1對B．2對C．3對D．4對〔第1〔第10題圖〕數(shù)學卷=12\*ROMANII〔非選擇題，共110分〕考前須知：1.卷=12\*ROMANI〔選擇題，共2頁〕的答案填寫在下面的答案表中;2.卷Ⅱ共5頁，用蘭、黑色鋼筆或圓珠筆直接答在試卷上．選擇題答案表題號12345678910得分答案 二、填空題：〔本大題共6小題，每題5分，共30分.只要求填出最后結果〕11．假設x∶y=1∶2，那么=_____________．12．如圖，一寬為2cm的刻度尺在圓上移動，當刻度尺的一邊與圓相切時，另一邊與圓兩個交點處的讀數(shù)恰好為“2”和“8”(單位：cm)，那么該圓的半徑為cm．13．如圖，電燈P在橫桿AB的正上方，AB在燈光下的影子為CD，AB∥CD，AB=2m，CD=5m，點P到CD的距離是3m，那么P到AB的距離是m.．〔第14題〕〔第14題〕20468〔第12題圖〕〔第〔第13題〕14．拋物線y=x2－2x+0.5如下圖，利用圖象可得方程x2－2x+0.5=0的近似解為．〔第16題〕A301〔第15題〕15．如圖，是反比例函數(shù)在第一象限內(nèi)的圖象，且過點A〔3，1〕，l2與關于軸對稱，那么圖象的函數(shù)解析式為〔〔第16題〕A301〔第15題〕16．如圖，把直角三角形ABC的斜邊AB放在定直線l上，按順時針方向在l上轉動兩次，使它轉到ΔA"B"C"的位置．設BC＝1cm，AC=cm，那么頂點A運動到點A"的位置時，點A經(jīng)過的路線與直線l所圍成的圖形的面積是____cm2．三、解答題〔此題有8小題，第17～20題每題8分，第21題10分，第22、23題每題12分，第24題14分，共80分〕17.(此題總分值8分)計算:18.(此題總分值8分)如圖，甲為四等分數(shù)字轉盤，乙為三等分數(shù)字轉盤，同時自由轉動兩個轉盤，當轉盤停止轉動后〔假設指針在邊界處那么重轉〕，請用畫樹狀圖或列表格的方法，求兩個轉盤指針指向數(shù)字之和不超過4的概率.19.(此題總分值8分)某蓄水池的排水管每時排水8m3，6小時〔〔1〕如果增加排水管，使每時的排水量增加到Q〔m3〕,那么將滿池水排空所需的時間為t(h)，試寫出t與Ｑ之間的關系式；〔4分〕〔2〕如果準備在５h內(nèi)將滿池水排空，那么每時的排水量至少為多少？〔2分〕〔3〕排水管的最大排水量為每時12m3〔2分〕20.(此題總分值8分))二次函數(shù)圖像的對稱軸為直線x=2,函數(shù)的最小值為-4,且圖象經(jīng)過點(-1,5).〔1〕求此二次函數(shù)的解析式.〔5分〕〔2〕求二次函數(shù)圖象與x軸的交點坐標.〔5分〕21．(此題總分值10分)如圖，在△ABC中，AD⊥BC于D，BE⊥AC于E,AD交BE于F.(1)求證：△ADC∽△BEC；〔5分〕(2)假設S△ABC=9，S△DCE=1，求DC與AC的比值.〔5分〕22．(此題總分值12分)如圖，兩建筑物AB和CD的水平距離為30米，從A點測得D點的俯角為30°，測得C點的俯角為60°，那么建筑物CD的高為多少米?23．(此題總分值12分)如圖，以的邊為直徑的⊙交邊于點，其中∠CAB=90°，為的中點，且，．〔1〕求的值和的長；〔4分〕〔2〕連結，判斷與是否垂直？為什么？〔4分〕〔3〕判斷是否是⊙的切線？假設是，試求出切線的長，假設不是，請說明理由；〔4分〕24.(此題總分值14分〕：如圖，梯形ABCD中，AB∥CD，∠ABC=90°，AB=8，CD=6，在AB邊上取動點P，連結DP，作PQ⊥DP，使得PQ交射線BC于點E，設AP=x，BE=y.〔1〕當BC=4時，試寫出y關于x的函數(shù)關系式；〔4分〕〔2〕在滿足〔1〕的條件下，假設△APD是等腰三角形時，求BE的長；〔4分〕〔3〕在滿足〔1〕的條件下，點E能否與C點重合，假設存在，求出相應的AP的長，假設不存在，請說明理由；〔4分〕〔4〕當BC在什么范圍內(nèi)，存在點P，使得PQ經(jīng)過C〔直接寫出結果，不必寫出相應的解題過程〕.〔2分〕備用備用圖〔1〕備用備用圖〔2〕2007學年第二學期第一次考試九年級數(shù)學試題〔參考答案〕一、選擇題：〔每題4分，共計40分〕題號12345678910答案ABCADCDADB二、填空題：〔每題6分，共30分〕題號111213141516答案1.2x1≈0.3,x2≈1.7三、解答題〔此題有8小題，第17～20題每題8分，第21題10分，第22、23題每題12分，第24題14分，共80分〕17.(此題總分值8分)解原式=····························〔4分〕=6-······················································〔2分〕·························································〔2分〕18.(此題總分值8分)解：方法一、列表如下：····················〔4分〕*123412*3*4*523*4*5634*567所以兩次摸到不同顏色球的概率為：p=.··········〔4分〕方法二：畫樹狀圖〔略〕19.19.(此題總分值8分)解：〔1〕t=····························································〔4分〕〔2〕當t=5時，Q==7.2(m3)·······································〔1分〕答：如果準備在５h內(nèi)將滿池水排空，那么每時的排水量至少為7.2m3〔3〕當Q=12時，t==4(h)··························································〔1分〕答：排水管的最大排水量為每時12m3,那么最少需要時間4h可將滿池水全部排空····················〔120.(此題總分值8分))解：〔1〕根據(jù)題意，設二次函數(shù)的解析式是y=a(x-2)2-4·································〔1分〕把點〔-1，5〕代入上式得：5=a(-1-2)2-4·············································〔1分〕解得：a=1·································································································〔1分〕∴二次函數(shù)的解析式是y=(x-2)2-4···························································〔1分〕(2)當y=0時，得=a(x-2)2-4·········································································〔1分〕解得：x=4或x=0····················································································〔1分〕∴二次函數(shù)圖象與x軸的交點坐標為〔4，0〕、〔0，0〕··················〔1分〕21．(此題總分值10分)〔1〕證明：∵AD⊥BC，BE⊥AC，∴∠ADC=∠BEC=90°·······························〔2分〕又∵∠C=∠C···················································〔1分〕∴△ADC∽△BEC········································〔2分〕〔2〕解：由〔1〕可得·······························〔1分〕又∵∠C=∠C···················································〔1分〕∴△CDE∽△CAB········································〔1分〕∴········································〔1分〕∴··········································〔1分〕22．(此題總分值12分)解：過A點作AE⊥CD于E，那么AE=BC=30·········〔1分〕∵tan30°=························〔2分〕∴DE=30=10··················〔2分〕∴AD=2DE=20·····················〔2分〕又∵∠ACD=60°-30°=30°········································〔2分〕∴∠ACD=∠CAD=30°···············································〔2分〕∴CD=AD=20······················································〔1分〕23．(此題總分值12分)解：〔1〕∵∠CAB=90°，，∴BC=·······················································〔1分〕∴=··················································〔1分〕∵AB為⊙的直徑∴∠ADB=90°································〔1分〕∵=∴=∴AD==4.8··········〔1分〕〔2〕∵∠CDA=90°，為的中點∴點E在AD的中垂線上··················································〔1分〕∵AO=DO∴點O也在AD的中垂線上·············································〔1分〕∴OE為AD的中垂線∴AD⊥OE···········································································〔2分〕〔3〕在△EOA和△EOD中：AE=DE，AO=DO，EO=EO·······································〔1分〕∴△EOA≌△EOD······························································〔1分〕∴∠EDO=∠EAO=90°·······················································〔1分〕∵OD過圓心O∴為⊙的切線····························································〔1分〕〔利用其它方法，可以參照上述評分標準給分〕24.(此題總分值14分〕解：〔1〕過D點作DH⊥AB于H那么四邊形DHBC為矩形，∴HB=CD=6∴AH=AB-CD=2，················〔1分〕∵AP=x，∴PH=x-2,再證明：△DPH∽PEB·····································〔1分〕∴，∴················〔1分〕整理得：y=(x-2)(8-x)=-x2+x-4············〔1分〕(2)先求出AD=2················································································〔1分〕要使△APD是等腰三角形，那么情況①：當AP=AD=2，即x=2時：BE=y=-×(2)2+×2-4=5-9