2021年宁德初中数学第一次质检数学答案练习题

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