包头年中考数学模拟题四答案

1包头2014年中考数学模拟题四答案一、1.A2．A．3．C4.C5．B6.C7.B8．C9.B10.B11.C12B二、填空题（每题3分，共24分）13.x≥3,14.-1,0,115.11aa.16.3217.418.4n+419.220.1三.解答题(共60分）21解：（1）由题意可知；50次摸球实验活动中，出现红球20次，黄球30次，所以红球所占百分比为205040％黄球所占百分比为305060％答：红球占40％黄球占60％（2）由题意可知，50次摸球实验活动中，出现有记号的球4次，所以总球数为5081004。所以红球数为1004040％。答：盒中红球有40个。22.解：在Rt△ABC中，∵AB=5，∠ABC=45°，∴AC=ABsin45°=5×=，在Rt△ADC中，∠ADC=30°，∴AD==5=5×1.414=7.07，AD﹣AB=7.07﹣5=2.07（米）．答：改善后滑滑板会加长2.07米．23.解：设第一批玩具每套的进价是x元，则第二批每套进价是(x＋10)元，由题意得：2500x×1.5＝4500x＋10，解得x＝50，经检验x＝50是分式方程的解．故第一批玩具每套的进价是50元．(2)设每套售价至少是y元，250050×(1＋1.5)＝125(套)．125y－2500－4500≥(2500＋4500)×25%，y≥70，那么每套售价至少是70元．24.（1）证明：连接OB。∵OA=OB，∴∠A=∠OBE。∵CE=CB，∴∠CEB=∠EBC，∵∠AED=2∠EBC，∴∠AED=∠EBC，又∵CD⊥OA∴∠A+∠AED=∠OBA+∠EBC=90°，∴BC⊙O是的切线；（2）∵CD垂直平分OA，∴OF=AF，又OA=OF，∴OA=OF=AF，∴∠O=60°，∴∠ABF=30°；（3）作CG⊥BE于G，则∠A=∠ECG。∵CE=CB，BD=10，∴EG=BG=5，∵sinECG=sinA=135，∴CE=13，CG=12.又CD=15，∴DE=2。∵ADE∽△CGE，∴EGDECGAD，即5212AD，∴AD=524，∴OA=548，即⊙O的半径是548。25解：（1）RtA，6AB，8AC，10BC．点D为AB中点，132BDAB．90DHBA，BB．BHDBAC△∽△，DHBDACBC，∴5128103ACBCBDDH（2）QRAB∥，90QRCA．CC，RQCABC△∽△，RQQCABBC，10610yx，即y关于x的函数关系式为：365yx．（3）存在.按腰相等分三种情况：①当PQPR时，过点P作PMQR于M，则QMRM．1290，290C，1C．84cos1cos105C，45QMQP，1364251255x，185x．②当PQRQ时，312655x，6x．ABCDERPHQM21ABCDERPHQ3③当PRQR时，则R为PQ中垂线上的点，于是点R为EC的中点，11224CRCEAC．tanQRBACCRCA，366528x，152x．综上所述，当x为185或6或152时，PQR△为等腰三角形．26解：（1）由t2＋2t－24＝0，解得t1＝－6，t2＝4．··························································1分∵t1＜t2，∴A（－6，0），B（0，4）．······························································2分∵抛物线y＝32x2＋bx＋c的图象经过点A，B两点∴4062＝＝＋4－ccb解得4314＝＝cb∴这个抛物线的解析式为y＝32x2＋314x＋4．···········································································································4分（2）∵点P（x，y）在抛物线上，且位于第三象限，∴y＜0，即－y＞0．又∵S＝2S△APO＝2×21×|OA|·|y|＝|OA|·|y|＝6|y|∴S＝－6y．··········································································································6分＝－6(32x2＋314x＋4)＝－4(x2＋7x＋6)＝－4(x＋27)2＋25．····················································································7分令y＝0，则32x2＋314x＋4＝0，解得x1＝－6，x2＝－1．∴抛物线与x轴的交点坐标为（－6，0）、（－1，0）∴x的取值范围为－6＜x＜－1．·······································································8分（3）当S＝24时，得－4(x＋27)2＋25＝24，解得：x1＝－4，x2＝－3．············9分代入抛物线的解析式得：y1＝y2＝－4．∴点P的坐标为（－3，－4）、（－4，－4）．当点P为（－3，－4）时，满足PO＝PA，此时，□OPAQ是菱形．当点P为（－4，－4）时，不满足PO＝PA，此时，□OPAQ不是菱形．4·········································································································10分要使□OPAQ为正方形，那么，一定有OA⊥PQ，OA＝PQ，此时，点的坐标为（－3，－3），而（－3，－3）不在抛物线y＝32x2＋314x＋4上，故不存在这样的点P，使□OPAQ为正方形．····························································12分