2025年山西省太原市中考一模数学试卷(图片版,含答案)

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2025年山西省太原市中考一模数学试卷(图片版,含答案)

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5,铜砝码作为古代计量工具,见证历史的变迁和计量技术的发展.如图是一个请代铜砝码的
示意图及其俯视图,则它的左视图为
太原市2025年初中学业水平模拟考试(一)
数学
0
D
正面
俯视图
(第5蓝明)
1考试时间:上午8:30—10:30)
6.如图,一个平面锁EF放置在两个互相平行的挡板m和
的版2
注意享项
之间,平面镜EF与挡板形成的锐角为23°.一支微光笔
1.本试卷分第I卷和第I卷两部分.全卷共8面,满分120分,考诚时罚120分钟
从点A处发出的光束投射到平而镜上的点B处,反射光束
华面镜
2客题前,芳生务必将自己的姓名、准芳证号填写在太试卷相应的住置
挡程n
投射到挡板m上的点C处.设光束AB所在直缇与挡板m
3.答案全部在签题卡上完成,答在本试卷上无效
的交点为D,若∠DBF=∠CBE=52°,则∠BCD的度数为
(第6恩国)
4.芳试结来后,本试卷和答题卡一并变回
A.75
B.76
C.104
D.105
7.晋祠天龙山新晋为太原首个国家5A级景区.这是太原旅游业发展的一个重要里程碑.已
第I卷选择题(共30分)
知天龙山门票的单价旺季比淡季贵20元,旺季3张门票的总阶和淡季4张门票的总价相
一、选择题〔本大题共10个小题,每小题3分,共30分.在每个小题给出的四个远项中,只有
同.设旺季门票的单价为x元张,淡季门票的单价为y元张,则x,y满足的方程组是
一项符合题目婴求,诗选出并在答题卡上将该项涂黑)
a60
B.-y=20,
c.P-x=20,
D.x=20.
1.下列各数在数轴上对成的点,离原点最近的是
3%=4r
14x=3y
3x=4y
A.-3
B.-1
c
D.2
8.如图,将△ABC沿着AB的方向平移得到ǜA'BC,其中A'C与BC交
2.中国红十字会是中华人民共和国统一的笑十字组织,以保护人的生命和健康,维铲人的连
于D,连接CC',则下列结论一定成立的是
严,发扬人道主义特神,促进和平进步事业为宗常.下列红十字会的图标中,文字上方的图
A.A'B=CC
B.∠A=LB1
案既是轴对称图形又是中心对称图形的是
C.B'C=2BD
ID.∠B=∠BCC
《第8缆图)
9.“无糖饮料”逋常使用糖醇和低璨鞋等不升高血椨浓度的甜味剂作为神的替代品,但并
非其正意义的无糖.现有甲、乙、丙、丁四种无糖饮料,它们的含德浓度y(含糖浓度
户回红十字会会微
二字将爱家因
造拉千如胞扔献
中回红十学基余会
B
C
雄味剂质至×1%)与饮料质量x(g)之间的关系,可近似地用如图的反比例函数图象
饮料:董
3.为了解某校学生每天体育活动的情况,下列抽样调查的方式中最合造的是
A.随机抽取某一个班的全体同学
表示,其中甲、乙仗料y与无的关系满足y=上〔x>0),丙、丁饮料y与x的关系满是
B.每个年级随机加取15名女生
C.课外活动时间.在操场上随机取20名同学
y一兰(x>0).根据周象.下列结论正确的是
D.将全校学生姓名输人电暂程序,由电脑随机抽取150名学生
A,甲饮料含甜味剂质址比乙饮料的多
4.下列运算正确的是
B.内饮料含甜味剂质量比丁饮料的多
A.2m+3r=5m2
B.(4m)2=8m
C.甲、乙饮料含甜味剂质量相同但比丙、丁的多
C.x2y-1j=2cy·x
D.(3a-2)2=9a2+4
D.丙、丁饮料含甜咴剂质量相同但比甲、乙的多
(幕9题图)
数学试卷(一》第1页(共8页)
数学试卷(一)第2页(共8页)2025 年太原市初中学业水平模拟考试(一)
数学参考答案及评分细则
一、选择题(本大题共 10个小题,每小题 3分,共 30分)
题号 1 2 3 4 5 6 7 8 9 10
选项 C C D C B A B D D B
二、填空题(本大题共 5个小题,每小题 3分,共 15分)
1
11. 3 3 12. (3 + 1) 13. 14. 65 15. 4 + 3 3
4
三、解答题(本大题共 8个小题,共 75分)
16.(每小题 5 分,共 10 分)
1 1 1( )解:原式=2 + ( 3) ÷ × ····························································· 2分
4 3
=2 + ( 3) × 4 × 1 ··························································3分
3
=2 + ( 4) ··········································································· 4分
= 2. ·················································································· 5分
(2)解:原方程可变形为 2 9 2( + 3) = 0, ····································· 1分
( + 3)( 3) 2( + 3) = 0,
( + 3)( 5) = 0, ····························································· 3分
∴ + 3 = 0 或 5 = 0, ······················································ 4分
∴ 1 = 3, 2 = 5. ··························································· 5分
17.(本题 7分)
解:(1)如图,∠ 、射线 、点 即为所求.
·············································· 2分
(2)四边形 是矩形. ········································································· 3分
证明:方法 1:
∵∠ = ∠ ,
∴ // . .................................................................................4分
∵ , 分别是 , 的中点,
九年级数学答案第 1页(共 5 页)
∴ 是△ 的中位线,
∴ ∥ . ..............................................................................5分
∵ ∥ , ∥ ,
∴四边形 是平行四边形. .........................................................6分
∵∠ = 90° ,
∴ 是矩形. ........................................................................7分
证明:方法 2:
∵∠ = ∠ ,
∴ // . .................................................................................4分
∴∠GAB+∠B=90°.
∵∠B=90°,
∴∠GAB=90°..............................................................................5分
∵ , 分别是 , 的中点,
∴ 是△ 的中位线,
∴ ∥ . ...............................................................................6分
∴∠B+∠BDF=90°,
∴∠BDF=90°.
∴四边形 是矩形. ..................................................................7分
18.(本题 9分)
解:(1)35.5;32;2;··················································································6分
(2)答案不唯一,选择两个统计量说明理由即可,如:
①从平均数看,乘坐地铁的平均用时 32分低于开私家车平均用时 34分,即乘坐地铁用时更
短,所以选择乘坐地铁;
②从中位数看,乘坐地铁用时的中位数 32分低于开私家车用时的中位数 35.5分,即乘坐地
铁用时更短,所以选择乘坐地铁;
③从众数看,乘坐地铁用时的众数 32分低于开私家车用时的众数 40分,即乘坐地铁用时更
短,所以选择乘坐地铁;
④从方差看,乘坐地铁用时的方差 2低于开私家车用时的方差 50.75,乘坐地铁所用时间更
九年级数学答案第 2页(共 5 页)
稳定,所以选择乘坐地铁.
················································································································· 9分
【评分说明】写出第一条理由得 2分,第二条再得 1分,共 3分.
19.(本题 7分)
解:设直角书架的单价为 元/个,弧形书架的单价为(1 + 20%) 元/个. ················· 1分
18000 = 9000由题意,得 + 6. ····························································· 3分
(1+20%)
解,得 = 1000. ······································································· 4分
经检验, = 1000是原方程的解. ·······························································5分
当 = 1000时,
(1 + 20%) = 1200. ················································································· 6分
答:弧形书架的单价是 1200元/个,直角书架的单价是 1000元/个. ·····················7分
20.(本题 10 分)
解:(1)设路灯 的高度为 米. ·····································································1分
在 Rt△ 中,∠ = 90°,∠ = 26.6°,
∴ tan∠ = = tan26.6° ≈ 0.50 ,即 ≈ 0.5,

∴ ≈ = 2 . ······················································································· 2分
0.50
在 Rt△ 中,∠ = 90°,∠ = 38.7°,
∴ tan∠ = = tan38.7° ≈ 0.80,

∴ ≈ 0.80,

∴ ≈ = 5 .·························································································· 3分
0.80 4
∵ = = 6.7米,
∴ 2 5 = 6.7. ························································································ 5分
4
解,得 ≈ 8.9. ························································································· 6分
答:路灯 的高度为 8.9米. ··········································································7分
(2)线段 ;···························································································· 8分
+
或( + ) ;············································································ 10分

或:∠ ;··························································································8分
( tan + ). ························································································10分
九年级数学答案第 3页(共 5 页)
21.(本题 6分)
解:(1)平行线分线段成比例基本事实的推论:平行于三角形一边的直线与其他两边相交,
截得的对应线段成比例;········································································· 2分
【评分说明】依据写为:平行线分线段成比例基本事实(两条直线被一组平行线所截,
所得的对应线段成比例)也可得分.
(2)如图,点 M即为所求;·······································································4分

(3)方法不唯一,如图,点 H即为所求.······················································· 6分

或 或
22.(本题 13 分)
解:(1)一次; = 50 + 1200;··································································· 3分
(2)设销售总额为 元,由题意,得
= (80 2 )
= (50 + 1200)(80 2 )
= 100 2 + 1600 + 96000. ·······························································5分
根据题意, ≥0 且 80 2 >0,所以 0≤ <40.
因为 a = -100<0,所以 w有最大值,
= 1600当 = 8 时,销售总额最大. ····················································· 6分
2×( 100)
九年级数学答案第 4页(共 5 页)
答:若要使这批马铃薯全部售完的销售总额最大,应储存 8 个星期. ······················ 7分
(3)设全部售完的销售利润为 元,由题意,得
= 64000
= 100 2 + 1600 + 96000 64000
= 100 2 + (1600 ) + 32000. ························································ 9分
根据题意, ≥0 且 80 2 >0,所以 0≤ <40.
因为 a = -100<0,所以 m有最大值,
= 1600 = 1600 由题意,得 当 时,
2×( 100) 200
2 2
最大 =
4×( 100)×32000 (1600 ) = 32000 + (1600 ) ,
4×( 100) 400
2
(1600 )因为 最大 = 35600,所以 32000 + = 35600,····························· 10分400
解,得 1 = 400, 2 = 2800,····························································· 12分
= 400 = 1600 400当 时, = 6.
2×( 100)
当 = 2800 1600 2800时, = = 6(不符合题意,舍去).
2×( 100)
所以, = 400, = 6.
答: 的值是 400,相应的存储星期数为 6星期.·················································13分
23.(本题 13 分)
解:(1)AE=CF,AE//CF. ··········································································· 1分
证明:∵线段 AB绕点 D逆时针旋转 90°得到线段 EF,
∴AB=EF,CA=CE,CB=CF,∠ACE=90°.
∴△ABC≌△EFC, .................................................................................................2分
∴∠BAC=∠FEC.
∵∠BAC=90°,
∴∠FEC=90°=∠ACE. .......................................................................................3分
∴EF//AC. .................................................................................................................4分
∵AB=AC,
∴EF=AC,
∴四边形 AEFC是平行四边形, .............................................................................5分
∴AE=CF,AE//CF.
(2)解:①AE=CF,AE//CF.
九年级数学答案第 5页(共 5 页)
理由:连接 AD,ED,FD,并延长 FD交 AC于点 G. ............................................6分
∵AB=AC,∠BAC=90°,
∴∠B=∠ACB= 1 (180°-∠BAC)=45°. ............................................................7分
2
∵线段 AB绕点 D逆时针旋转 90°得到线段 EF,
∴AB=EF,DA=DE,DB=DF,∠BDF=90°,
∴△ABD≌△EFD,∴∠B=∠EFD=45°.
∵∠CDG=∠BDF=90°,
∴∠DCG+∠DGC=90°,
∴∠DGC=45°,..........................................................8分
∴∠DGC=∠EFD,
∴EF//AC.
∵EF=AB=AC,
∴四边形 AEFC是平行四边形.
∴AE=CF,AE//CF. .........................................................................................9分
②存在;B,D两点之间的距离为 4 2 +4或4 2 -4. ..........................................13分
【说明】以上解答题的其他解法,请参照此标准评分.
九年级数学答案第 6页(共 5 页)

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