2021年宁德初中数学第一次质检数学答案练习题

这是一份2021年宁德初中数学第一次质检数学答案练习题，共9页。试卷主要包含了选择题，填空题，解答题等内容，欢迎下载使用。
2021年宁德市初中毕业班第一次质量检测数学试题参考答案及评分标准⑴本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可参照本答案的评分标准的精神进行评分．⑵对解答题，当考生的解答在某一步出现错误时，如果后续部分的解答未改变该题的立意，可酌情给分．⑶解答右端所注分数表示考生正确作完该步应得的累加分数．⑷评分只给整数分，选择题和填空题均不给中间分．一、选择题：（本大题有10小题，每小题4分，满分40分）1．C；2．A ；3．B；4．D；5．A；6．C；7．B；8．C；9．B；10．D．二、填空题：（本大题有6小题，每小题4分，满分24分）11．；12．；13．；14．兔子的只数（或兔子的数量等）；15．；16．16． 三、解答题（本大题共9小题，共86分．请在答题卡的相应位置作答）17．（本题满分8分）解法一：①②，得 ．解得 ，························································4分将代入①，得，解得 ．························································7分所以原方程组的解为···············································8分解法二：，由①得 ③，将③代入②，得 ，解得 ··························································4分把代入③，得 ．·················································7分所以原方程组的解为···············································8分18．（本题满分8分）证明：∵∠BAD=∠CAE，∴∠BAD+∠DAC=∠CAE+∠DAC．即∠BAC=∠DAE．·························3分∵AB=AD，∠C=∠E，∴△ABC≌△ADE．··························6分∴BC=DE．·······························8分19．（本题满分8分）解：····························································2分····························································4分． ·························································6分当时，原式．·························································8分20. （本题满分8分）解：设需要调用辆型车，根据题意，得 ································1分 ．····························································5分 解得 ．························································7分∵为正整数，∴的最小值为18．················································8分答：至少需要调用型车18辆．21．（本题满分8分）（1）解：如图所示．                                        或     ∴图中△FBC就是所求作的三角形．·································4分（注：仅作出垂直平分线给2分）（2）由（1）得 FB=FC=AB=5．设FG⊥BC于点G． ∴,∠FGB=90°．···············································5分在矩形ABCD中，∵∠ABC=90°，∴∠ABF+∠FBG=∠BFG+∠FBG=90°．∴∠ABF=∠BFG．·············································6分在Rt△FBG中,==．························································7分∴==．······················································8分22．（本题满分10分）（1）证明：连接OD． ∵DE⊥AC，∴∠DEC=90°．····················1分∵AB=AC,OB=OD， ∴∠B=∠C,∠B=∠ODB．············3分∴∠C=∠ODB． ∴OD∥AC． ∴∠ODE=∠DEC=90°．··············4分∴直线DE是⊙O的切线．···········································5分（2）连接AD．∵AB为⊙O直径，∴∠ADB=90°．··················································6分∵，∴．在Rt△ABD中，．····························································7分根据勾股定理,得．∴，AC=AB=10．·················································8分∵，∴．解得DE=4．·····················································9分在Rt△ODE中，根据勾股定理,得．····························································10分23. （本题满分10分）（1）每人每天平均加工零件个数的中位数为：=21.5（个）. ················1分平均数为：= =23（个），··················································4分答：每人每天平均加工零件个数的中位数是21.5个，平均数是23个.（2）①根据题意,得这30名工人每个月基本工资总额为：=84 800（元）.这30名工人所生产的零件计件工资总额为：=45 540.························································6分这30名工人每个月工资总额为：84 800+45 540=130 340（元）.因为130 340>130 000，所以该等级划分不符合工厂要求. ····································8分②方法1：将每天生产18个以下（含18个）的确定为普工，每天生产29个以上（含29个）的确定为技术能手.方法2：将每天生产19个以下（含19个）的确定为普工，每天生产28个以上（含28个）的确定为技术能手.方法3：将每天生产19个以下（含19个）的确定为普工，每天生产29个以上（含29个）的确定为技术能手.               10分 24．（本题满分12分）解：（1）证明：∵四边形ABCD是正方形，∴AB=BC，∠BAE=∠BCF=．      ·····································1分∵BE= BF，∴∠BEF=∠BFE．∴∠AEB=∠CFB．          ············································2分            ∴△ABE  ≌△CBF．∴AE=CF．         ··················································3分（2）∵∠BEC=∠BAE+∠ABE =+∠ABE，∠ABF=∠EBF+∠ABE=+∠ABE，∴∠BEC=∠ABF．·························4分∵∠BAF=∠BCE=，∴△ABF∽△CEB．   ·······················5分∴．∴=16．     ······················································7分（3）解法一：如图2∠EBF=∠GCF=45°,∠EFB=∠GFC,∴△BEF∽△CGF. ·························8分∴.即.∵∠EFG=∠BFC,∴△EFG∽△BFC. ························10分∴∠EGF=∠BCF=45°.∴∠EBF =∠EGF. ∴EB=EG. ·····················································12分解法二：如图3，过点E作交CD于点K，交AB于点H，连接BD，∵四边形ABCD是正方形，∴∠BAE=∠BDG=∠ABD=．∴∠ABD=∠EBF= ．∴∠ABE=∠DBG．∴△ABE ∽△DBG．      ······················8分∴．∴．在Rt△AHE中，∠HAE=∠AEH=，∴，AH=HE．∴．      ···································9分在四边形AHKD中，∵∠DAH=∠ADK=∠AHK=，∴四边形AHKD是矩形．∴DK=AH．∴KG=DG-DK=2AH-AH=AH．∴HE=KG．     ··················································10分在Rt△CEK中，∠KEC=∠KCE=，∴EK=CK．∵DK=AH，∴AB-DK=CD-AH．∴CK=BH．∴EK=BH．  ···················································11分∵HE=KG，∠BHE=∠EKC=，EK=BH，∴△BHE  ≌△EKG．∴BE=EG．     ··················································12分解法三：过点E作交AB于点H，交CD于点K，作交CD于点G′,连接EG′,∴∠BHE=∠EKG′=90°.∴∠BEH+∠EBH=90° ,∠BEH+G′EK=90°.∴∠EBH=∠G′EK.∵∠KHB=∠HBC=∠BCK=，∴四边形HBCK是矩形．∴HB=KC．∵∠KEC=∠KCE=，∴KE=KC=HB．∴△BEH≌△EG′K. ···············································9分∴BE=EG′.∵BE⊥EG′，∴∠EBG′=∠EG′B=45°.∴∠EBG′=∠EBG=45°.············································11分∵点G′与点G都在CD上，且在BE同侧，∴点G′与点G重合.∴BE=EG. ·····················································12分 25．（本题满分14分）（1）解：依题意，得，．∴点A的坐标为(-3, )． ············································2分当时，．∴点B的坐标为(0，) ．············································3分（2）∵四边形ABCD是平行四边形，∴CD∥AB．∵点A是抛物线的最高点，点D在抛物线上，∴点D在点A的下方．由平移的性质可得点C在点B的下方．∵点C在x轴上，点B的坐标为(0，) ，∴＞0．①如图1，过点A，D作AE⊥y轴于点E，DF⊥x轴于点F．∴∠AEB=∠DFC=90°．∴∠EAB+∠ABE=90°．∵四边形ABCD是矩形，∴∠ABC=90°,AB=DC．∴∠ABE+∠CBO=90°．∴∠EAB=∠CBO．同理可得∠DCF=∠CBO．∴∠DCF=∠EAB．∵∠AEB=∠COB=90°，∴△ABE∽△BCO,△ABE≌△CDF ．··································6分∴，CF=AE，DF=BE．∵AE=3,BE=，BO=c，∴CO=．∴点C的坐标为(，0)，点D的坐标为(，)．将点D(,)代入得．解得（舍去）， ．················································9分所以c的值为②如图2，设直线AB的表达式，将A(-3, )，B (0, )代入得 解得 ∴直线AB的表达式为． ···········································11分过点D作DG⊥x轴交AB于点G，设点D的坐标为（t，），则点G的坐标为（t，），=2=2××，         =2×（）×3，         =．∴当时，四边形ABCD的面积最大为．  ································14分解法二：连接AC，设抛物线的对称轴交x轴于点H，连接HB．∵四边形ABCD是平行四边形，∴CD∥AB．设点D的坐标为（t，），由平移的性质可得点C的坐标为（t+3，），∵点C在x轴上， ∴ =0∴c=     ························································11分∴======．∴当时，△ABC的面积最大为．∵，∴四边形ABCD的面积最大为．  ·····································14分