初中数学北京课改版七年级下册7.2 实验精品同步练习题

这是一份初中数学北京课改版七年级下册7.2 实验精品同步练习题，共6页。试卷主要包含了本试卷分为第一部分，已知，，则的值为等内容，欢迎下载使用。
试卷类型：A咸阳市实验中学2021~2022学年度第一学期第三次月考七年级数学试题注意事项：1．本试卷分为第一部分（选择题）和第二部分（非选择题）。全卷共4页，总分120分。考试时间120分钟。2．领到试卷和答题卡后，请用0.5毫米黑色墨水签字笔，分别在试卷和答题卡上填写姓名和准考证号，同时用2B铅笔在答题卡上填涂对应的试卷类型信息点（A或B）。3．请在答题卡上各题的指定区域内作答，否则作答无效。4．作图时，先用铅笔作图，再用规定签字笔描黑。5．考试结束，本试卷和答题卡一并交回。第一部分（选择题  共24分）一、选择题（共8小题，每小题3分，计24分．每小题只有一个选项是符合题意的）1．下列各等式中，是一元一次方程的是（    ）A． B． C． D．2．若单项式与是同类项，则m的相反数为（    ）A． B．3 C． D．23．下列说法正确的是（    ）A．射线OA和射线AO是同一条射线 B．延长线段AB和延长线段BA的含义是相同的C．经过两点可以画无数条直线 D．两点之间的所有连线中，线段最短4．下列方程中，解为的是（    ）A． B． C． D．5．将50.26°用度、分、秒表示，正确的是（    ）A． B． C． D．6．已知，，则的值为（    ）A．8 B． C．2 D．7．如图，A地和B地都是海上观测站，A地在灯塔O的北偏东30°方向，，则B地在灯塔O的（    ）A．东偏南30°方向  B．南偏东40°方向C．南偏东50°方向  D．南偏西50°方向8．用形状大小完全相同的正三角形和正方形按如图所示的规律拼图案，若第n个图案中三角形的个数比正方形的个数多802个，则n的值为（    ）A．266 B．268 C．256 D．258第二部分（非选择题  共96分）二、填空题（共5小题，每小题3分，计15分）9．若，则a的倒数是________．10．单项式的系数是m，多项式的次数是n，则的值为________．11．某地居民生活用水收费标准为：每月用水量不超过17立方米（含17立方米），每立方米a元；超过17立方米的部分每立方米（）元，该地区某用户上个月的用水量为20立方米，则应缴水费为________元．（用含a的代数式表示）12．如图，小明同学在参加“几何小能手”社团活动时，制作了一副（两个）与众不同的三角板，用它们可以画出一些特殊的角度．在①9°；②18°；③55°；④117°中，能用这副三角板画出的角度有________．（填序号）13．我国古代数学名著《算法统宗》中，有一道“群羊逐草”的问题，大意是：牧童甲在草原上放羊，乙牵着一只羊来，并问甲：“你这群羊有100只吗？”甲说：“如果在这群羊上加上同样的一群，再加上半群，又加上四分之一群，再加上你的一只，才满100只．”问牧童甲赶着多少只羊？若设牧童甲赶着x只羊，则可列方程为________________________．三、解答题（共13小题，计81分．解答应写出过程）14．（5分）解方程：．15．（5分）如图，已知点A、B和线段a，b，连接AB，利用尺规在BA的延长线上求作线段AC，使得．（不写作法，保留作图痕迹）16．（5分）体育课上全班学生进行了百米测验，达标成绩为18秒，下面记录的是第一小组8名学生的成绩（用正数表示超过18秒的部分，用负数表示不足18秒的部分）：，，，，0，，0，．求这一组学生的平均成绩．17．（5分）一个三角形的第一条边长为，第二条边长比第一条边长小，第三条边长是第二条边长的2倍，用含x的代数式表示这个三角形的周长．18．（5分）将一个半径为的圆分成三个扇形，其圆心角度数之比为1：3：5．（1）求三个扇形的圆心角度数；（2）求其中最小一个扇形的面积．（结果保留）19．（5分）规定这样一种新运算法则：．如：．若，求x的值．20．（5分）已知从n边形的一个顶点出发共有4条对角线；从m边形的一个顶点出发的所有对角线把m边形分成7个三角形；正t边形的边长为6，周长为48，求代数式的值．21．（6分）全国足球联赛规定：胜一场得3分，平一场得1分，负一场不得分，某球队比赛了8场，踢平的场数是负的场数的2倍，共得17分，求该球队胜的场数．22．（7分）已知，．（1）化简；（2）当，时，求的值．23．（7分）如图，点C在线段AB上，点D是线段AC的中点，点E是线段BC的中点．（1）若，，求DE的长；（2）若，求AB的长．24．（8分）已知方程与关于x的方程的解相同．（1）求a的值；（2）若a、b在数轴上对应的点在原点的两侧，且到原点的距离相等，c是最大的负整数，求的值．25．（8分）如图所示，OB是的平分线，OD是的平分线．（1）若，，求的度数；（2）若，，求的度数．26．（10分）如图，甲、乙两人（看成点）分别在数轴上和10对应的位置上，沿着数轴做东、西移动的游戏，移动游戏的规则为：用一枚硬币，先由乙抛掷后遮住，甲猜向上一面是正还是反，如果甲猜对了，甲向东移动3个单位，如果甲猜错了，甲向西移动3个单位；然后再由甲抛掷后遮住，乙猜向上一面是正还是反，如果乙猜对了，乙向西移动2个单位，如果乙猜错了，乙向东移动3个单位．两人各抛掷一次硬币并完成相应的移动算一次游戏．10次游戏结束后，甲猜对了m次，乙猜对了n次．（1）请用含m、n的代数式表示当10次游戏结束时，甲、乙两人在数轴上的位置所表示的数；（2）当甲、乙两人都猜对5次时，求甲、乙两人之间的距离；（3）10次游戏结束后，若甲10次都猜对了，且两人在数轴上的位置恰好相距10个单位，求乙猜对的次数． 试卷类型：A咸阳市实验中学2021~2022学年度第一学期第三次月考七年级数学试题参考答案及评分标准一、选择题（共8小题，每小题3分，计24分．每小题只有一个选项是符合题意的）1．C    2．A    3．D    4．B    5．A    6．D    7．C    8．B二、填空题（共5小题，每小题3分，计15分）9．    10．3    11．    12．①②④    13．（其他形式正确均可）三、解答题（共13小题，计81分．解答应写出过程）14．解：去分母，得，······························································（2分）去括号，得，·····································································（3分）移项、合并同类项，得，····························································（4分）系数化为1，得．···································································（5分）15．解：如图所示，AC即为所求．·····················································（5分）16．解：·········································································（3分）（秒），所以这一组学生的平均成绩是17.8秒．··················································（5分）17．解：因为第一条边长为，第二条边长比第一条边长小，所以第二条边长为，································································（2分）所以第三条边长为．································································（4分），所以这个三角形的周长为．···························································（5分）18．解：（1），，·································································（1分），，所以三个扇形的圆心角度数分别是40°，120°和200°．·······································（3分）（2）最小扇形的面积为：．··························································（5分）19．解：因为．所以．··········································································（2分）因为，所以，··········································································（4分）解得．··········································································（5分）20．解：因为从n边形的一个顶点出发共有4条对角线，所以．··········································································（2分）因为从m边形的一个顶点出发的所有对角线把m边形分成7个三角形，所以．··········································································（3分）因为正t边形的边长为6，周长为48，所以，··········································································（4分）所以代数式．·····································································（5分）21．解：设该球队负了x场，则踢平了场，根据题意，得，···································································（3分）解得，··········································································（5分）则．答：该球队胜的场数是5场．··························································（6分）22．解：（1）因为，，所以············································································（2分）．··············································································（4分）（2）当，时，原式．··········································································（7分）23．解：（1）因为点D是线段AC的中点，点E是线段BC的中点，，，所以，，·········································································（2分）所以．··········································································（3分）（2）因为点D是线段AC的中点，点E是线段BC的中点，所以，．·········································································（5分）所以．因为，所以，即．·······································································（7分）24．解：（1）解方程，去分母，得，·····································································（1分）去括号，得，移项、合并同类项．得．系数化为1，得．···································································（3分）将代入，得，·····································································（4分）解得．··········································································（5分）（2）由题意得，，·································································（7分）所以．··········································································（8分）25．解：（1）因为OB是的平分线，OD是的平分线，所以，，·········································································（2分）所以．··········································································（3分）（2）因为OD是的平分线，，所以．··········································································（5分）因为，所以．··········································································（7分）因为OB是的平分线，所以．··········································································（8分）26．解：（1）因为10次游戏结束后，甲猜对了m次，乙猜对了n次，所以甲在数轴上的位置所表示的数为：；················································（1分）乙在数轴上的位置所表示的数为：．····················································（2分）（2）当甲、乙两人都猜对5次时，，则甲在数轴上的位置所表示的数为······················································（3分）乙在数轴上的位置所表示的数为，······················································（4分）所以甲、乙两人之间的距离为（个）单位．···············································（5分）（3）因为甲10次都猜对了，所以10次游戏结束后，甲在数轴上的位置所表示的数为．····································（6分）因为两人在数轴上的位置恰好是10个单位，所以乙在数轴上的位置对应的数是或，··················································（7分）所以或，·········································································（9分）解得或，所以乙猜对的次数是2次或6次．·······················································（10分）