初中八年級(jí)數(shù)學(xué)下冊(cè)期末附答案

新課標(biāo)人教版初中八年級(jí)數(shù)學(xué)下冊(cè)期末精品試題（附答案）本試卷滿分120分考試時(shí)間90分鐘命題人：題號(hào)一二三總分核分人得分21222324252627得分評(píng)分人選擇題（每小題3分，共36分．下列各題所附的四個(gè)選項(xiàng)中，有且只有一個(gè)是正確的,請(qǐng)你把答案填在答題卡上）(選擇題答題卡)題號(hào)123456789101112答案1．由分式SKIPIF1<0得到分式SKIPIF1<0，所需的條件是（）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<02．下列各式的從左到右的變形，正確的是（）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<03．已知等腰梯形的的腰等于中位線的長(zhǎng)，周長(zhǎng)為SKIPIF1<0，則腰長(zhǎng)為（）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．以上結(jié)果都不正確4．已知點(diǎn)SKIPIF1<0是雙曲線SKIPIF1<0上一點(diǎn)，則下列各點(diǎn)中在該圖象上的點(diǎn)是（）．A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<05．在SKIPIF1<0的圖象中，陰影部分面積不為SKIPIF1<0的是（）．6.下列說(shuō)法中正確的個(gè)數(shù)有①對(duì)角線互相平分且相等的四邊形是菱形；②有一組對(duì)邊平行的四邊形是梯形；③如果四邊形的兩條對(duì)角線互相垂直，則它的面積等于兩條對(duì)角線長(zhǎng)的積的一半；④如果一個(gè)四邊形繞對(duì)角線的交點(diǎn)旋轉(zhuǎn)900后，所得圖形及原來(lái)的圖形重合，則這個(gè)四邊形是正方形.其中正確的個(gè)數(shù)有（）．（A）1個(gè)（B）2個(gè) （C）3個(gè)（D）4個(gè)7.如圖，已知梯形ABCD中，AD∥BC，AB=CD=AD，AC，BD相交于O點(diǎn)，∠BCD=60°，則下列說(shuō)法錯(cuò)誤的是（）．(A)梯形ABCD是軸對(duì)稱圖形；(B)BC=2AD；(C)AC平分∠DCB；(D)梯形的面積是△BOC的面積的2倍。ABCDFEOABCD8.ABCDFEOABCD（A）1(B)SKIPIF1<0（C）SKIPIF1<0（D）29．若直角三角形的兩直角邊分別是１和SKIPIF1<0，則斜邊上的高為（）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<010.在□ABCD中，AB=5cm，BC=4cm，則□ABCD的周長(zhǎng)為cm。（）A．17B．9C．20D．1811．化簡(jiǎn)SKIPIF1<0的結(jié)果是A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<012．已知：一組數(shù)據(jù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的平均數(shù)是2，方差是SKIPIF1<0，則另一組數(shù)據(jù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0－2，SKIPIF1<0的平均數(shù)和方差分別是（）A．2，SKIPIF1<0B．2，1C．4，SKIPIF1<0D．4，3得分評(píng)分人二、填空題（每小題3分，共24分）13．．直線SKIPIF1<0及雙曲線SKIPIF1<0交于點(diǎn)SKIPIF1<0，則SKIPIF1<0；SKIPIF1<0。14.菱形ABCD中，對(duì)角線AC及BD相交及點(diǎn)O，點(diǎn)E為AB的中點(diǎn)，若AD=8cm，則OE的長(zhǎng)為_________cm。15．已知如圖，在正方形ABCD中，E是CD邊上的一點(diǎn)，連接BE，將△BCE繞點(diǎn)C順時(shí)針方向旋轉(zhuǎn)900，得到△DCF，連接EF，若∠BEC=600，則∠EFD的度數(shù)是_________0。SKIPIF1<016．如圖，有一圓柱，其高為12cm，它的底面半徑為3cm，在圓柱下底面A處有一只螞蟻，它想得到上面B處的食物，則螞蟻經(jīng)過(guò)的最短距離為________cm。（π取3）17．當(dāng)SKIPIF1<0時(shí)，分式方程SKIPIF1<0的解為SKIPIF1<0。18一組數(shù)據(jù)同時(shí)減去99，得到新數(shù)據(jù)的平均數(shù)是1，方差是11，極差是21，則原數(shù)據(jù)的平均數(shù)是_____；方差是_____；極差是_____19．如圖所示，將兩條等寬的紙條重疊在一起，則四邊形SKIPIF1<0是SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0。20．已知直角三角形兩邊x、y的長(zhǎng)滿足SKIPIF1<0，則第三邊的長(zhǎng)是_______。得分評(píng)分人三、解答題（本大題共7小題，共60分）21．（10分）計(jì)算：（1）SKIPIF1<0（2）SKIPIF1<022.（10分）解方程（1）SKIPIF1<0（2）SKIPIF1<023.(6分)如圖，一次函數(shù)SKIPIF1<0的圖象及反比例函數(shù)SKIPIF1<0的圖象交于A、B兩點(diǎn)．（1）利用圖中的條件求反比例函數(shù)和一次函數(shù)的解析式．（2）根據(jù)圖象寫出使一次函數(shù)的值大于反比例函數(shù)的值的x的取值范圍．24．（8分）如圖，在梯形SKIPIF1<0中，SKIPIF1<0，對(duì)角線SKIPIF1<0于SKIPIF1<0，若SKIPIF1<0，求梯形的高和梯形的面積。25.(8分)如圖，在平行四邊形ABCD中，SKIPIF1<0分別是邊SKIPIF1<0和SKIPIF1<0上的點(diǎn)且SKIPIF1<0。①線段SKIPIF1<0及線段SKIPIF1<0有怎樣的數(shù)量關(guān)系和位置關(guān)系？并證明你的結(jié)論．（5分）②若AE⊥BC，則四邊形AECF是下列選項(xiàng)中的（）．（3分）A．梯形B．菱形C．正方形D．矩形AABCDFE26（8分）如圖所示，有兩種形狀不同的直角三角形紙片各兩塊，其中一種紙片的兩條直角邊長(zhǎng)分別為1和2，另一種紙片的兩條直角邊長(zhǎng)都為2．圖a、圖b、圖c是三張形狀、大小完全相同的方格紙，方格紙中的每個(gè)小正方形的邊長(zhǎng)均為1．請(qǐng)用三種方法將圖中所給四塊直角三角形紙片拼成平行四邊形（非矩形），每種方法要把圖中所給的四塊直角三角形紙片全部用上，互不重疊且不留空隙，三種方法所拼得的平行四邊形（非矩形）的周長(zhǎng)互不相等，并把你所拼得的圖形按實(shí)際大小畫在圖a、圖b、圖c的方格紙上。要求：（l）所畫圖形各頂點(diǎn)必須及方格紙中的小正方形頂點(diǎn)重合。（2）畫圖時(shí)，要保留四塊直角三角形紙片的拼接痕跡。27.（10分）如圖，已知平行四邊形ABCD中，對(duì)角線AC、BD交于點(diǎn)O,E是BD延長(zhǎng)線上的點(diǎn)，且△AEC等邊三角形．①四邊形ABCD是菱形嗎？并說(shuō)明你的理由；②若∠AED=2∠EAD，求證：四邊形ABCD是正方形．EECDBAO初二年級(jí)數(shù)學(xué)試題答案一、DDABBCDBDB二、13b=1n=31441515度161517m=318100；11；2119菱形SKIPIF1<0205三、21①SKIPIF1<0；②SKIPIF1<022①SKIPIF1<0②x=223①反比例函數(shù)：SKIPIF1<0······································2分一次函數(shù)：y=x-1··········································4分②-1<x<0且x>2············································6分24高6··························································4分面積36··························································8分25①AE及CF平行且相等證明∵四邊形ABCD是平行四邊形∴AF∥CEAD=BC··················································1分又∵DF=BE∴AD-DF=BC-BE即AF=CE··················································3分又∵AF∥CE∴四邊形AECF是平行四邊形∴AE平行且等于CF············································5分②D·······················································3分263分2分3分①是菱形證明：∵△AED是等腰三角形∴∠AEB=∠CEBAE=CE··················································1分在△AED及△CDE中AE=CE∠AEB=∠CEBED=ED(公共邊)∴△AED≌△CDE（SAS）∴AD=CD··················································4分又∵四邊形ABCD是平行四邊形∴四邊形ABCD是菱形··········································5分②證明：∵AB∥CD,AD∥BC∴