资源简介 姓名淮考证号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0D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m3.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.75B.76C.104D.1057.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)a60B.-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%=4r14x=3y3x=4yA.-3B.-1cD.28.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=CCB.∠A=LB1案既是轴对称图形又是中心对称图形的是C.B'C=2BDID.∠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BC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=5m2B.(4m)2=8mC.甲、乙饮料含甜味剂质量相同但比丙、丁的多C.x2y-1j=2cy·xD.(3a-2)2=9a2+4D.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)111. 3 3 12. (3 + 1) 13. 14. 65 15. 4 + 3 34三、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分或(3)方法不唯一,如图,点 H即为所求.······················································· 6分或或 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) 2002 2 最大 =4×( 100)×32000 (1600 ) = 32000 + (1600 ) ,4×( 100) 4002 (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) 展开更多...... 收起↑ 资源列表 2025年山西省太原市中考一模数学试卷(PDF版,含答案).pdf 数学答案.pdf