4月鞍山市初中第一次質(zhì)量檢測(cè)（數(shù)學(xué)）參考答案及評(píng)分標(biāo)準(zhǔn)

參考答案一、選擇題：DDCCBABC二、填空題：9、10、11、112、13、14、2或15、416、三、解答題：17、…………..….………….3……………………….……….5當(dāng)時(shí)…….…………………….………………..………….6原式…….……….…………….818、證明：∵四邊形是菱形∴∥∥…….……...……….3∴…….……………..…….6∴…….………………….….819、解：（1）200…….……………...............................….2（2）（人）；（人）…….……………..…….……….4畫(huà)圖正確…….……………….…………….…….6（3）（人）…….………………..….……….9答：約有580人?！?….……….1020、解：（1）………………………..…….….……….2（2）…………….…….……….5此事件共有9種可能情況，且可能性相等，其中兩人選擇不同電影有6種可能情況，（刺，唐）（刺，你）（唐，刺）（唐，你）（你，刺）（你，唐）……………….….……….9……………….…….….……….1021、解：過(guò)作交于，設(shè)為米….……..…...................…...…….2∵∴四邊形為矩形∴米∵∥∥∴….…….….……….3在中….…….….…...............................................…….5….…..............................….….……….6在中….…….….…................................................…….8米….………………..........................................9答：高度約為30米….…….….……....................................................................….1022、解：（1）∵過(guò)（2，4）∴∴…....................................................................................................................….2∵點(diǎn)橫坐標(biāo)為4時(shí)（4，2）…...............................................................................................................….3∵一次函數(shù)的圖象過(guò)、兩點(diǎn)∴…............................................................................................................….5（2）過(guò)作∥軸交于…................................................................................….6設(shè)（，0）（，）…......................................................................................................….7…...............................................................................….9（4，0）…......................................................................................................….1023、證明：（1）連接...........................................................................................................................….1∵∴∵∥∴.......................................................................................................….3∵為直徑∴∴∴........................................................................................….4（2）連接，過(guò)作交于.........................................................................….5∵∴∵在中設(shè)∴.............................................................................................….7∵∴在中設(shè)∴........................................................................................….9∵四邊形為圓內(nèi)接四邊形∴∴在中∴.....................................................................................….1024、解：（1）……….….…...............................................................…….2………...........................................................................….……….3∵∴有最大值當(dāng)……….….…........................................................................................…….5（元）……….….……........................….6答：略（2）設(shè)購(gòu)進(jìn)煙花的價(jià)格為元/件……….….…...................................................................…….7（件）（天）……….….................................................................……….8……….….….....................................................................................…….10答：略25、證明：（1）證出..........................................................................................…….2證出........................................................................................…….3（2）證出證出為斜邊中線............................................................................…….5證出∴.....................................................................…….7（3）方法1：過(guò)作交于，過(guò)作交延長(zhǎng)線于.….8證出：證出：∴證出：∵∴∥...................................................................................…….9證出：證出：∵∥∴.................................................................................…….10證出：....................................................…….11∵∥∴∴...............................................................................…….12方法2:延長(zhǎng)至，使得，連接.........................................................…….8證出：∴證出：后續(xù)證明同方法126、解：（1）求出：（3，0）（0，4）.........................................................…..................….2求出：解析式為....................................................................…….4（2）（0，）；（0，）..............................................................................…….8（3）方法1：過(guò)作軸交于.....................................................................…….9設(shè)證出：∴....................................................…….10證出：即...............................................................................…….11當(dāng)方程有唯一實(shí)數(shù)根時(shí)，滿足條件的只有一個(gè)..................................................…….12解得：（舍去）求出：（，）.........................................................…….13設(shè)平移后的拋物線為：將點(diǎn)坐標(biāo)代入平移后解析式