福建漳州市数学试题及答案实验区华师大版Word可编辑版.docx
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福建漳州市数学试题及答案实验区华师大版Word可编辑版
2021年福建漳州市数学试题及答案（实验区）华师大版
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2021年漳州市初中毕业暨高中阶段招生考试
数学试题（实验区）
（满分：
150分；考试时间：
120分钟）
一、填空题（本大题有14题，每小题3分，共42分）请将正确答案直接填写在横线上．
1．计算：
．
2．写出一个大于2的无理数________．
3．平方根等于它本身的数是________．
4．若，则．
5．根据天气预报，明天降水概率为，后天降水概率为，假如你准备明天或后天去放风筝，你选择________天为佳．
6．不等式组的解集是________．
7．若，则．
8．如图，点在直线上，，若，则度．
9．正多边形的一个外角等于，则这个正多边形的边数是________．
10．若方程无解，则．
11．为了解家庭丢弃塑料袋对环境造成的影响，某班研究性学习小组的六位同学记录了自己家中一周内丢弃塑料袋的数量．结果如下（单位：
个）30，28，23，18，20，31．若该班有50名学生，请你估算本周全班同学的家共丢弃塑料袋________个．
12．某礼堂的座位排列呈圆弧形，横排座位按下列方式设置：
排数
1
2
3
4
…
座位数
20
24
28
32
…
根据提供的数据得出第排有________个座位．
13．如图，已知中，是直径，是弦，，垂足为，
由这些条件可推出结论________（不添加辅助线，只写出1个结论）．
14．学校有一个圆形花坛，现要求将它三等分，以便在上面种植三种不同的花，你认为符合设计要求的图案是________（将所有符合设计要求的图案序号填上）．
二、选择题（每小题4分，共24分）本大题有6题，每小题有Ａ，Ｂ，Ｃ，Ｄ四个答案，其中有且只有一个答案是正确的．请把正确答案的代号填写在相应括号内，答对得4分，答错、不答或答案超过一个的一律得零分．
15．200粒大米重约4克，如果每人每天浪费1粒米，那么约458万人口的漳州市每天浪费大米约（ ）克（用科学记数法表示）
Ａ． Ｂ． Ｃ． Ｄ．
16．一个钢球沿坡角的斜坡向上滚动了米，此时钢球距地面的高度是（ ）米
Ａ． Ｂ． Ｃ． Ｄ．
17．下列运算正确的是（ ）
Ａ． Ｂ．
Ｃ． Ｄ．
18．经过折叠________围成一个正方体的图形是（ ）
19．已知内接于，于，如果，那么的度数为（ ）
Ａ． Ｂ． Ｃ．或 Ｄ．或
20．近一个月来漳州市遭受暴雨袭击，九龙江水位上涨．小明以警戒水位为点，用折线统计图表示某一天江水水位情况．请你结合折线统计图判断下列叙述________的是（ ）
Ａ．8时水位最高
Ｂ．这一天水位均高于警戒水位
Ｃ．8时到16时水位都在下降
Ｄ．点表示12时水位高于警戒水位0.6米
三、解答题（本大题有7题，共84分）
21．（10分）
先化简，再求值：
，其中．
解：
22．（10分）
小敏有红色、白色、黄色三件上衣，又有米色、白色的两条裤子．如果她最喜欢的搭配是白色上衣配米色裤子，那么黑暗中，她随机拿出一件上衣和一条裤子，正是她最喜欢搭配的颜色．请你用列表或画树状图，求出这样的巧合发生的概率是多少？
解：
23．（12分）
如图，已知是的直径，是弦，过点作于，连结．
（1）求证：
；
（2）若，求的度数．
（1）证明：
（2）解：
24．（12分）
2021年世界杯足球赛在德国举行．你知道吗？
一个足球被从地面向上踢出，它距地面高度可以用二次函数刻画，其中表示足球被踢出后经过的时间．
（1）方程的根的实际意义是________；
（2）求经过多长时间，足球到达它的最高点？
最高点的高度是多少？
解：
25．（12分）
根据十届全国人大常委会第十八次全体会议《关于修改＜中华人民共和国个人所得税法＞的决定》的规定，公民全月工资、薪金所得不超过1600元的部分不必纳税，超过1600元的部分为全月应纳所得额，月个人所得税按如下方法计算：
月个人所得税＝（月工资薪金收入－1600）×适用率－速算扣除数
注：
适用率指相应级数的税率．
月工资薪金个人所得税率表：
级数
全月应纳税所得额
税率％
速算扣除数（元）
1
不超过500元
5
2
超过500元至2021元的部分
10
25
3
超过2021元至5000元的部分
15
125
…
…
…
…
某高级工程师2021年5月份工资介于3700～4500元之间，且纳个人所得税235元，试问这位高级工程师这个月的工资是多少？
解：
26．（14分）
已知，是边上的中线，分别以所在直线为轴，轴建立直角坐标系（如图）．
（1）在所在直线上找出一点，使四边形为平行四边形，画出这个平行四边形，并简要叙述其过程；
（2）求直线的函数关系式；
（3）直线上是否存在点，使为等腰三角形？
若存在，求点的坐标；若不存在，说明理由．
27．（14分）
如图，已知矩形，在上取两点（在左边），以为边作等边三角形，使顶点在上，分别交于点．
（1）求的边长；
（2）在不添加辅助线的情况下，当与不重合时，从图中找出一对相似三角形，并说明理由；
（3）若的边在线段上移动．试猜想：
与有何数量关系？
并证明你猜想的结论．
2021年漳州市初中毕业暨高中阶段招生考试
数学试题参考答案及评分标准（实验区）
一、填空题（本大题有14题，每题3分，共42分）
1．； 2．如； 3．； 4．； 5．明； 6．； 7．；
8．； 9．； 10．； 11．； 12．； 13．如；
14．②③④．
二、选择题（本大题有6题，每题4分，共24分）
15．Ｃ； 16．Ａ； 17．Ｄ； 18．Ｂ； 19．Ｄ； 20．Ｃ．
三、解答题（本大题有7题，共84分）
21．（10分）
································································（4分）
··············································································（6分）
当时，
原式 ·································································（8分）
···································································（10分）
22．（10分）
上衣
裤子
红色
白色
黄色
米色
（米，红）
（米，白）
（米，黄）
白色
（白，红）
（白，白）
（白，黄）
（6分）
由上表（或图）可知，所有等可能结果共6种，其中正好是白色上衣配米色裤子的只有1种，所以所求概率是 ··························································································································（10分）
23．（12分）
（1）（6分）
证法一：
是的直径
··················································································（2分）
又
·················································································（4分）
··············································································（6分）
证法二：
是的直径
···························································（2分）
即
又 ·············································································（3分）
····································································（4分）
·········································································（5分）
··············································································（6分）
（2）（6分）
解法一：
是的直径，
·················································································（3分）
的度数为：
·······································（6分）
解法二：
是的直径，
·················································································（3分）
·················································································（4分）
的度数为 ·····································································（5分）
的度数为 ··································································（6分）
24．（12分）
（1）（4分）足球离开地面的时间 ·······························································（2分）
足球落地的时间 ···················································································（4分）
（2）（8分）
································································（1分）
·····················································（3分）
························································（5分）
当时，最大值 ··············································（7分）
经过，足球到达它的最高点，最高点的高度是．·······（8分）
25．（12分）
， ···············································（2分）
该工程师应纳税所得额在2021～5000元的部分，其税率为，
速算扣除数为125元． ················································································（4分）
设这位高级工程师这个月的工资是元，依题意，得 ··································（5分）
·····································································（8分）
解得：
·························································································（11分）
答：
这位高级工程师这个月的工资是4000元． ···········································（12分）
注：
正确列出方程，但未判断税率为得6分．
26．（14分）
（1）（4分）
正确画出平行四边形 ·································（2分）
叙述画图过程合理 ················································（4分）
方法一：
在直线上取一点，使
连结 ······················································（1分）
所以四边形是所画的平行四边形 ···············（2分）
方法二：
过画，交直线于
连结 ···························································（1分）
所以四边形是所画的平行四边形 ····································（2分）
（2）（4分）
是边上的中线
·····················································································（2分）
设直线的函数关系式：
，得
解得 ····································································（3分）
直线的函数关系式：
·····················································（4分）
（3）（6分）设 ···································································（2分）
分三种情况：
①
解得
··································································（3分）
②
解得
···································································（4分）
③
解得
，这时点在上，构不成三角形，舍去． ·························（5分）
综上所述，在直线上存在四点，即，，，，符合题意 （6分）
注：
观察图形，能直接得出两点坐标即和可得2分．
27．（14分）
（1）（4分）
过作于 ···············································································（1分）
矩形
，即，又
·······················································································（2分）
是等边三角形
在中
的边长为． ···············································································（4分）
（2）（4分）
正确找出一对相似三角形 ············································································（2分）
正确说明理由 ······························································································（4分）
方法一：
·······································································（2分）
理由：
矩形
································································································（3分）
···················································································（4分）
方法二：
······································································（2分）
理由：
矩形
································································································（3分）
又
··················································································（4分）
（3）（6）猜想：
与的数量关系是：
证法一：
在中，
·································································································（2分）
是等边三角形
·······································································（3分）
·······························································································（4分）
··························································································（6分）
证法二：
在中，
·································································································（2分）
是等边三角形，
·············································································（3分）
在中，
，即 ··························································（4分）
在中，
··························································································（6分）
证法三：
在中，
，
·············································································（2分）
是等边三角形
·······················································································（3分）
①················································································（4分）
即
②·················································································（5分）
把②代入①得，
··························································································（6分）
注：
猜想出，未给出正确的证明得2分．
升级会员 