海陵区2013-2014学年度第一学期期末调研测试九年级数学试题答案

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1、初三数学期末调研参考答案（其他解法参照得分）一、选择题：BACCAB二、填空题：7.8.30°9.10.相交11.-4或712.13.12π14.15.16.三．解答题：17.（1）原式=（过程4分结论1分）（2）原式=3（过程4分结论1分）18.（1）4a+4··················3分·············4分(2)由题意的：·············5分此时方程为解得：,····················8分19.如图所示：点E即为所求，BE=DE（注：作一个角等于已知角和垂直平分线各3分）20.（1

2、）178··········2分178···········4分（2）··········6分············8分甲理由略··················10分21.（1）∵AC⊥CD，∴∠ACD=90°，∴AC2+CD2=AD2，而CD=60cm，AC=45cm，∴AD=75cm··················4分答：车架档AD的长为75cm；···············5分（2）过E作EF⊥AB于F点，如图，在Rt△AEF中，∠EAF=75°，AE=AC+CE=45+20=65，∴EF=AE•sin75°≈

3、65×0.9659≈63（cm），··················9分答：车座点E到车架档AB的距离为63cm．··················10分22.（1）下降0.2元后，卖出600个烧饼，每个利润为1.3元，每天利润为600×1.3=780元············4分（2）由题意得：，化简得：·············6分解得：,·············9分答：略··········10分23.(1)∵四边形ABCD是矩形，∴OA=OB=OC=OD且AC=BD·················2分∵AE=BF

4、=CG=DH，∴OE=OF=OG=OH，∴四边形EFGH是平行四边形··················4分又EG=FH∴四边形EFGH是矩形·················5分(2)∵OH=OG又HG=OG，∴三角形OGH是等边三角形·················7分∴∠CDO=∠GHO=60°，而CD=AB=2cm∴AD=CDtan=60°=(cm）△ADC面积为(cm2）················9分∵△ADO与△ODC等底等高，∴△ADO的面积是cm2·······10分24.(1)连接OD.∵CG是切线，∴

5、∠GCO=90°，即CG⊥CO·················2分∵OA=OD，△OAD是等腰三角形，又AF=FD，∴CF⊥AD，∴CG∥AD··················4分(2)连接AC，∵AB⊥CD∴弧AC=弧AD∴AC=AD，同理AC=CD，∴△ACD是等边△，∴∠FCD=30°∴，∴E是OB的中点···············8分（3）∵AB=8，∴OC=4,OE=2，在Rt△OCE中，∴，∴∴，∴··················12分25.（1）分别过点Q、D作QE⊥OC,DF⊥OC交OC与点E、F.由题

6、意得：D（2,4）··················1分∴BD=3，∵BC=4，∴可求出：CD=5，可证得：△CQE∽△CDF∴，∴··················3分∴··················4分（2）不存在··················5分∵，令解得··············7分，此时Q与C重合，不能构成三角形.··············8分（3）由△CQE∽△CDF，，，，，①当时，，解得··········10分②当时,，··················12分答：当或时，△DPQ是一个以D

7、P为腰的等腰三角形.26.（1）过O作OE⊥CF于E，∵∠OCF=30°∴OE=AB/2=2,又⊙O的半径是2，∴⊙O与CF相切············4分（2）连结QA、QB，OA=AC=2，△COE是直角三角形，故AE=CO/2=2········6分cos∠AEQ=3/4得cos∠ABQ=3/4，AB=4，∴BQ=3，由△AED∽△BQD得BD/DE=QB/AE=3/2·················9分（3）设AP=a（0

8、=4(4-2a+a2)>0，方程总有解，即不论P在何处，AD×BD=AP×PC总能成立·················12分又由△AED∽△BQD可得ED×QD=AD×BD，∴不论P在何处，总存在弦EQ使得ED×QD=AP×PC成立··············