山东省济南市 高新区2019-2020学年第一学期九年级期中测试数学卷参考答案及评分标准.pdf

第 1页（共 4页） 2019 至 2020 学年第一学期期中学业水平测试 高新初中数学九年级参考答案及评分标准 一、选择题 题号 1 2 3 4 5 6 7 8 9 10 11 12 答案 B A D A A C C B B D D A 二、填空题:（本大题共 6 个小题，每小题 4 分，共 24 分．） 13．x1＝0，x2＝ 14．15 15．200（1+x）2 16．∠ABP＝∠C 或∠APB＝∠ABC 或 AB2＝AP•AC 17． 22 18．16 三、解答题:（本大题共 12 个小题，共 78 分．解答应写出文字说明、证明过程或演算步骤．） 19．（本题 6 分） 解：x2+2x﹣5＝0， （x+1）2＝6················································································································4分 ∴x1＝﹣1+ ；x2＝﹣1﹣ ·······················································································6分 20．（本题 6 分） 解：（1）根据题意得：（﹣2）2﹣4（m﹣1）＞0··································································4分 解得：m＜2················································································································6分 21．（本题 6 分） 解：∵AB∥CD， ∴∠A=∠D，∠B=∠C ∴△DOC∽△AOB········································································································2分 ∴ ＝ ··················································································································4分 ∵OA＝2，OD＝4，AB＝3． ∴CD＝6·····················································································································6分 22．（本题 8 分） 解：①BC＝（80﹣2x）米···································································································2分 ②依题意，得：x（80﹣2x）＝750·····················································································4分 整理，得：x2﹣40x+375＝0， 解得：x1＝15，x2＝25····································································································6分 ∴80﹣2x＝50 或 30． ∵80﹣2x≤45，第 2页（共 4页） ∴x＝25． 答：矩形的长为 30 米，宽为 25 米 ····················································································8分 23．（本题 8 分） （1）证明：由题意可知： ∠B＝∠C＝∠ADE＝60° ·································································································1分 ∵∠ADC＝∠B+∠BAD， ∠ADC＝∠ADE+∠CDE， ∴∠BAD＝∠CDE ·······································································································3 分 ∴△ABD∽△DCE·········································································································4 分 （2）∵△ABD∽△DCE ∴ ··················································································································6 分 设 AB＝x，解得：x＝6 ·································································································8 分 24．（本题 10 分） 解：（1）设 p 与 V 的函数的解析式为 p＝ ， 把点 A（1.5，64）代入··································································································· 1分 解得 k＝96··················································································································· 3分 ∴这个函数的解析式为 p＝ ························································································· 4 分 （2）把 v＝20 代入 p＝ 得：p＝4.8················································································ 5 分 当气球的体积为 20 立方米时，气球内的气压是 4.8 千帕 ·······················································6 分 （3）把 p＝144 代入 p＝ 得，V＝ ··············································································· 8 分 故 p≤144 时，v≥ ，气球的体积至少是 立方米 ································································10 分 25．（本题 10 分） 解：（1）答案：7、30%······································································································ 2 分 （2）补全条形图如下： ······························································································3 分第 3页（共 4页） （3）该校学生共 1600 人，则参加棋类活动的人数约为 1600× ＝280·····································5 分 （4）画树状图如下： ·········································································7 分 共有 12 种情况，选中一男一女的有 6 种············································································9 分 则 P（选中一男一女）＝ ＝ ····························································································10 分 26．（本题 12 分） 解：（1）①答案为： ··································································································2 分 ②如图（1），∵四边形 ABCD 是正方形， ∴∠BCA＝45°， ∵GE⊥BC， ∴∠CEG＝90°， ∴∠CGE＝∠ECG＝45°································································································· 3 分 ∴EG＝EC， ∴△CEG 是等腰直角三角形····························································································4 分 （2）线段 AG 与 BE 之间的数量关系为 AG＝ BE······························································5 分 连接 CG， 由旋转性质知∠BCE＝∠ACG＝α， 在 Rt△CEG 和 Rt△CBA 中， ＝ ， ∴△ACG∽△BCE·········································································································7 分 ∴ ＝ ，线段 AG 与 BE 之间的数量关系为 AG＝ BE····················································8 分 （3）∵延长 CG 交 AD 于点 H ∴∠AGH＝∠CAH＝45°， ∵∠CHA＝∠AHG， ∴△AHG∽△CHA········································································································ 9 分 作 HM⊥AF 于 M 若 AG＝6，GH＝2 易得 MG=2，AM=4，AH= 52 ······················································································10 分 易得 GC= 23 ，则正方形 CEGF 的边长为 3·····································································11 分第 4页（共 4页） 在 Rt△HDC 中，令 CD=x，则 HD=x- 52 由勾股定理可得正方形 ABCD 的边长为 3 ······································································12 分 27．（本题 12 分） 解：（1）答案为：（﹣4，0）；（0，2）；（2，3）········································································ 3分 （2）如图 1，作点 Q 关于 x 轴的对称点 Q′，连接 PQ′与 x 轴交于点 M，连接 QM，此时△PQM 的周长 最小． ∵点 P（2，3）在反比例函数 y＝ 图象上， ∴k＝2×3＝6，即反比例函数解析式为 y＝ ········································································4 分 ∴点 Q 的坐标为（6，1），点 Q′的坐标为（6，﹣1）····························································5 分 设直线 PQ′的解析式为 y＝mx+n（m≠0）， 将 P（2，3），Q（6，﹣1）代入 y＝mx+n，得： ， 解得： ， ∴直线 PQ′的解析式为 y＝﹣x+5·······················································································7 分 当 y＝0 时，﹣x+5＝0， 解得：x＝5， ∴点 M 的坐标为（5，0）·······························································································8 分 ∴当△PQM 的周长最小时，点 M 的坐标为（5，0）． （3）设点 E 的坐标为（x， ）（x＞2），则点 F 的坐标为（x，0）． 分两种情况考虑（如图 2）： ①当△EFB∽△AOC 时， ＝ ，即 ＝ ， 解得：x1＝3，x2＝﹣1（舍去）， ∴点 E 的横坐标为 3 ···································································································10 分 ②当△BFE∽△AOC 时， ＝ ，即 ＝ ， 解得：x1＝1+ ，x2＝1﹣ （舍去）， ∴点 E 的横坐标为 1+ ·····························································································12 分 综上所述：当△BEF 和△AOC 相似时，动点 E 的坐标为（3，2）或（1+ ， ）．