數(shù)學(xué)九上期中考試卷.ppt

期中考試卷,C,B,D,D,A,A,3,6,B,C,7,13,5,5,12,0,-2或1,對(duì)角線互相垂直的四邊形,17.,ADEBGE,S梯形ABCD=SDGC=4S DEF=32,32,18.,變小,解：,原式=,解：,原式=,3分,4分,3分,4分,解：,解：,2分,3分,5分,2分,3分,5分,證明：,矩形ABCD矩形BEDF,A=E=90,AB=ED,在ABM和EDM中, A=E=90, AMB=EMD, AB=ED, ABMEDM,在矩形ABCD和矩形BEDF中,ADBC,BEFD, 四邊形BNDM是平行四邊形, ABMEDM, BM=DM, BNDM是菱形.,2分,4分,6分,7分,8分,0,2,從平均數(shù)來(lái)看，平均數(shù)相等，說(shuō)明兩條生產(chǎn)線產(chǎn)品誤差相同； 從方差來(lái)看，乙的方差小，說(shuō)明乙生產(chǎn)線產(chǎn)品尺寸較穩(wěn)定.所以乙生產(chǎn)線工藝水平較高. 8分,若僅從眾數(shù)和方差分析，無(wú)平均數(shù)分析，則扣一分,2分,4分,在ABC中， AB=AC,B=C,作ADBC于D,在RtABD和Rt ACD中,ADB=ADC=90,AB=AC,AD=AD, RtABDRt ACD, B=C,2分,3分,4分,7分,8分,13-2x,x(13-2x)=20,x1=2.5,x2=4,當(dāng)x1=2.5時(shí),13-2x1=86， x1=2.5舍去,垂直于墻的邊為4m，另一邊為5m時(shí)，舉行場(chǎng)地面積為20m2.,1分,2分,3分,4分,5分,解：,設(shè)垂直于墻的邊為xm,x(13-2x)=30,2x2-13x+30=0,b2- 4ac=(-13)2- 4230=-710, 原方程無(wú)解, 不能使圍成的矩形場(chǎng)地面積為30m2.,6分,7分,8分,解：,長(zhǎng)方體底面正方形的面積為(10-x)2cm2,長(zhǎng)方體底面正方形的邊長(zhǎng)為,且0x10,2分,4分,6分,證明：,由題意得， BF=BF , BFE=BFE,在矩形ABCD中 ADBC, B EF=BFE, B EF=BFE, B E=BF, B F=BF, B E=BF,1分,2分,3分,4分,5分,解：,矩形紙片ABCD沿EF折疊,A=A=90,AE=AE,AB=AB=4,BC=10,F是BC的中點(diǎn), BF=5, B E=BF=5,在RtABE中,AE=AE=3,8分,9,16,2分,4分,6分,8分,10分,證明：過(guò)F點(diǎn)作FMBC于M,在正方形AEFG中 ,AE=EF,AEF=90,AEB+FEC=90,在正方形ABCD中 , B=90,AEB+BAE=90,BAE=FEC, M=90,在ABE和EMF中,B=M=90,BAE=FEC,AE=EF, ABEEMF(AAS), BE=MF,B+M=180,ABFM, HFBC, 四邊形BMFH是平行四邊形, BH=MF, BE=MF, BH=BE,又 BA=BC, AH=EC,在AHE和ECF中,證明：,在AHE和ECF中,BAE=FEC, AB=EF,AH=EC, AHEECF（SAS）,4分,四邊形CFHE是等腰梯形,證明：, HFBC，HE和CF不平行,四