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老师你给我一份昌平2012--2013年度第二学期期末检测七年级数学试卷的答案

解题思路： 昌平区2012—2013学年第二学期初一年级期末
解题过程：
昌平区2012—2013学年第二学期初一年级期末 　　　 数　学　试　卷 2013．6 考生须知 1．本试卷共6页，共五道大题，满分120分。考试时间120分钟。 2．在答题卡上认真填写学校、班级、姓名和考试编号。 3．试题答案一律填涂或书写在答题卡上，在试卷上作答无效。 4．考试结束，请将本试卷和答题卡一并交回。 一、选择题（本题共32分，每小题4分） 下列各题均有四个选项，其中只有一个是符合题意的． 1．16的平方根是 A．4 B．-4 C．±4 D．8 2．下列各式计算正确的是 A． B． C． D． 3．已知AD是△ABC的中线，则下列结论不正确的是 A． B． C． D． 4．要清楚地了解某日的气温变化情况，应选用的最恰当的统计图是 A．扇形图 B．条形图 C．直方图 D．折线图 5．如图，有一块含有45°角的直角三角板的两个顶点放在 直尺的对边上．如果∠1＝20°，那么∠2的度数是 A．15° B．20° C．25° D．30° 6．下列变形中不正确的是 A．由得 　　 　 B．由得　　 C．由得 　　　 D．由得 7．已知二元一次方程2x＋y＝8，当y＜0时，x的取值范围是 A．x＜4 B．x＞4 C．x＞－4 D．x ＜- 4 8．若关于x的不等式组有2个整数解，则m的取值范围是 A． B． C． D． 二、填空题（本题共16分，每小题4分） 9．计算： = ． 10．已知三角形的两边的长分别为2和7，设第三边的长为x ，则x的取值范围是 ． 11．若代数式是完全平方式，则的值为 ． 12．图1是二环三角形, 设，请你写出_______度； 图2是二环四边形，设，图3是二环五边形……二环（，且为整数）边形中，________度（用含的式子表示最后结果）． 三、解答题（本题共30分，每小题5分） 13．计算： ． 14．解方程组： 15．解不等式：，并在数轴上表示不等式的解集． 16．解方程组： 17．先化简，再求值：，其中． 18．求不等式组的正整数解． 四、解答题（本题共20分，每小题5分） 19．按图填空． 已知：如图, ∠1=∠2, ∠3=∠E. 求证： AD∥BE. 证明：∵∠1 = ∠2 (已知)， ∴ ∥ ( 内错角相等，两直线平行 ) ． ∴ ∠E = ∠ ( ) ． 又∵ ∠E = ∠3 ( 已知 )， ∴ ∠3 = ∠ ( 等量代换 ) ． ∴ AD∥BE ( ) ． 20．某地教研室对当地初一年级学生周一至周五每天完成课外作业的大致平均时间进行了抽样调查，下面是根据调查所得数据制作的不完整的统计表和扇形统计图． 学生周一至周五每天完成课外作业的大致平均时间统计表 组别序号 第1组 第2组 第3组 第4组 第5组 时间 （单位：分钟） 人数 50 125 275 30 请你根据统计表、图中所提供的信息解答下列问题： （1）求本次调查的学生人数； （2）请把表和图中的缺项补全； （3）该地区共有初一学生约8000人，请你根据抽样调查所得数据，估计该地区初一学生中，有多少人完成当天课外作业所需时间不少于90分钟？ 21．如图，AD、AE分别是△ABC的高线与角平分线，且∠BAC=90°，∠B=30°，求∠DAE的度数. 22．我们都知道等底等高的三角形面积相等，请你利用这个结论解决下面的有关面积的问题． (1)如图1，∥，点为线段与的交点，则与的关系是 ； (2)如图2，点是大正方形中边上的中点，以为边作小正方形，并且小正方形的边长为，连结、、，则= ； (3) 如图3，∥，与相交于点，过点作∥交于，连结、，则与的关系是 ．
五、解答题（本题共22分，23小题8分，24小题7分，25小题7分） 23．列方程（组）、不等式（组）解应用题． 某中学计划购买A、B两种型号的小黑板，经市场调查，购买一块A型小黑板比买一块B型小黑板多用20元，且购买5块A型小黑板和4块B型小黑板共需820元． （1）求购买一块A型小黑板、一块B型小黑板各需要多少元? （2）根据学校实际情况，需购买A、B两种型号的小黑板共60块，要求购买A、B两种型号小黑板的总费用不超过5240元，并且购买A型小黑板的数量应大于购买B型小黑板数量的．请你通过计算，求出该中学购买A、B两种型号的小黑板有哪几种方案? 24．已知：关于x、y的二元一次方程组 （1）x + y= ； （2）当方程组的解＞时，求的取值范围； （3）在（2）的条件下,若0，试判断代数式的符号，并说明理由. 25. 在△ABC中，AB=AC，∠A=，点E是边AB上一点，点D是射线BC上一点，作 ∠CDF与∠AED互补，交直线AC于点F（点D、E、F均不与△ABC的顶点重合）. （1）如图，当=90°且DE⊥AE时，请你补全图形，并直接写出∠EDF的度数； （2）当0°＜＜180°时，求∠EDF的度数（用含的式子表示）. 昌平区2012—2013学年第二学期初一年级期末 数学试卷答案及评分参考 2013．6 一、选择题（本题共32分，每小题4分） 题号 1 2 3 4 5 6 7 8 答案 C D A D C B B C 二、填空题（本题共16分，每小题4分） 题 号 9 10 11 12 答 案 三、解答题（本题共30分，每小题5分） 13．解： …………………………………………2分 ………………………………………………………………5分 14． 解：① + ②，得 ························································································· 1分 ·························································································· 2分 把代入①，得 ，··········································································· 3分 ．········································································ 4分 所以，原方程组的解是·········································································· 5分 15． 解： ．··························································································· 2分 ．········································································································· 3分 ．········································································································· 4分 不等式的解集在数轴上表示为 ··································· 5分 16． 解：由 ① 得 ③····················································································· 1分 由 ② 得 ④················································································· 2分 ③×2 + ④，得 ························································································· 3分 把代入③，得 .·················································································· 4分 所以，原方程组的解为··············································································· 5分 17．解： ················································································ 2分 ········································································································ 3分 ∵， ∴．······································································································· 4分 ∴············································································ 5分 18． 解：解不等式①，得 ·························································································· 1分 解不等式②，得 ························································································ 3分 ∴原不等式组的解集为········································································· 4分 ∴其正整数解为····························································································· 5分 四、解答题（本题共20分，每小题5分） 19．证明：∵∠1 = ∠2 (已知)， ∴ DB ∥ EC ( 内错角相等，两直线平行 ) ．··························· 1分 ∴ ∠E = ∠ 4 ( 两直线平行，内错角相等 ) ． …………………3分 又∵ ∠E = ∠3 ( 已知 )， ∴ ∠3 = ∠ 4 ( 等量代换 ) ．···················································· 4分 ∴ AD∥BE ( 内错角相等，两直线平行 ) ． ……………………………5分 20．解：（1） 答：一共调查了500名学生．·································································· 1分 （2）表中所填数据为20. ………………………………………………………2分 图中所填数据为：第3组25%，第4组55%．……………………………4分 （3） 答：该地区有4880人完成当天课外作业所需时间不少于90分钟．············ 5分 21．解：∵∠BAC = 90°, AE是角平分线, ∴∠EAC =∠BAC = 45°. ………………………………………………………1分 ∵∠B = 30°, ∴∠C = 60°. ………………………………………………………………………2分 ∵AD是高， ∴∠CDA = 90°. …………………………………………………………………3分 ∴∠DAC = 30°. ………………………………………………………………4分 ∴∠DAE =∠EAC -∠DAC = 45°-30°= 15°. …………………………5分 22．解：（1） 相等 ； ……………………………………………………………1分 （2）； ……………………………………………………………………3分 （3）．……………………………………………………………5分 五、解答题（本题共22分，23小题8分，24小题7分，25小题7分） 23．解：（1）设购买一块A型小黑板需要元，一块B型小黑板元．························ 1分 依题意，得····································································· 3分 解之得，·················································································· 4分 答：购买一块A型小黑板需要100元，一块B型小黑板80元． （2）设购买A型小黑板块. 依题意，得，······················································· 6分 解之得，．··················································································· 7分 所以，有两种方案： ①购买21块A型小黑板，39块B型小黑板； ②购买22块A型小黑板，38块B型小黑板．···················································· 8分 24．解：（1）································································································ 1分 （2） ②（1）， 得③·················································································· 2分 把③代入（1）中，得··········································································· 3分 ∵＞ ∴，即·················································································· 4分 （3）······························································································ 5分 ∵0， ∴． ····························································································· 6分 ∵． ···································· 7分 又∵， ∴． 25．解：（1）如图1，……………………………1分 ∠EDF = 45°. ………………………2分 （2）∵∠A＝， AB=BC， ∴…………………………3分 ∵∠CDF与∠AED互补， ∠BED与∠AED互补， ∴∠CDF = ∠BED. ……………………4分 ①当点D在边BC上且0°＜＜90°时，如图2， ∵∠EDC=∠B +∠BED =∠EDF +∠CDF， ∴. ………………5分 ②当点D在BC的延长线上且0°＜＜90°时，如图3， 在△中， ∴ ……………………6分 当时，上述①②结论仍成立. ……………………………………………………………………………7分 所以，当0°＜＜180°时，或
最终答案：略