九年级数学学科试卷 参考答案 一、选择题：本大题有 10 个小题，每小题 3 分，共 30 分． 题号 1 2 3 4 5 6 7 8 9 10 答案 C C D C D A C A B A 二、填空题：本大题有 6 个小题，每小题 4 分，共 24 分． 1 11． (0 , 5) ； 12． 60° ； 13． ； 4 14． 75° ； 15． 5或 ﹣1 ； 16． ﹣12 m<﹣2 ． 三、解答题：本大题有 7 个小题，共 52 分． 17． (本题 6分) (1) y 2x2 4x 6 ················ ············ ··········· ············ ·(3分) (2)不在 ·········································· ············ ·········· (结论 1分，过程 2分) 18．(本题 6分) （1）卡片上的数是 0的概率是 ···························· ··········· ····· (2分) （2）列表格如下：（设正数为 A,负数为 B） A B 0 A A，A B，A 0，A B A，B B，B 0，B 0 A，0 B，0 0，0 由表可知，共有 9种等可能的结果，其中两次抽取的数字之积为 0有 5种可能， 5 所以两次抽取的数字之积为 0的概率为 ．················ ········ (表格/树状图 2分，概率 2分) 9 19．(本题 6分) (1)如图，△AB′C′即为所求． ······· (3分) (2)如图，点 O即为所求．·············· (3分) 1 20．(本题 8分) （1）∵点 D（3，0）在抛物线 y＝x2﹣2x+c ∴9﹣6+c＝0，∴c＝﹣3． ············ (1分) 由 y＝x2﹣2x﹣3＝（x﹣1）2﹣4得，顶点 A为（1，﹣4） ············ ············ ·····（1分） ∵顶点 A在直线 y＝x﹣a上，∴当 x＝1时，∴y＝1﹣a＝﹣4，∴a＝5； ············ (2分) （2）△ABD是直角三角形； ···························· ············ ··········· ············ (1分) 由（1）可知，y＝x2﹣2x﹣3，∴B（0，﹣3）， ∴BD2＝OB2+OD2＝18， ·························· ··········· ··········· ··········· ············· (1分) AB2＝（4﹣3）2+12＝2， AD2＝（3﹣1）2+42＝20， ∴BD2+AB2＝AD2， ∴∠ABD＝90°，即△ABD是直角三角形． ················ ··········· ··········· ··········· (2分) 21．(本题 8分) (1)证明：∵AB是⊙O的直径， ∴∠ACB＝90°， ∵OD∥BC， ∴∠OFA＝90°，·························· ··········· ··········· ··········· ······· ············· (2分) ∴OF⊥AC， ∴ = 即点 D为 的中点；·························· ··········· ··········· ········· ·············· (2分) (2)设圆半径为 r ∵OD⊥AC，OD是半径 1 ∴AF=CF= AC=8························· ··········· ··········· ··········· ······· ··········· (1分) 2 2 ∵DF=4∴OF=r-4 Rt OAF r 4 2 82 r 2在 中，有 ，·············· ··········· ··········· ··············· (1分) 解的 r=10，∴⊙O的直径等于 20．·············· ··········· ·· ... ...
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