一、选择题（每小题2 分共14 分）

           题号        1       2       3       4       5       6       7

           答案       C       B       C       D       B       B       A

二、填空题 （每小题3 分共27 分）

    8．∠EAD=∠B (或∠CAD=∠C 或∠BAD+∠B=180°)；  9．2

                                     x  2

                                    ?

    10．(1.3)      11．270°      12．?             13．3        14．5, 5

                                     y  3

                                    ?

    15．2，1           16．2

三、解答题（本大题共有7 个小题，共59 分）

    17．解法一（代入消元法）：

        由②，得 y      22??6x ．③·······························································1 分

       把③代入①，得 x ?2(22?6x)          11 ．整理并解得x         3 ．························4 分

                                       x  3,

                                       ?

       把x     3代入③，得y        4 ．所以?

                                       y   4.

                                       ?     ·············································7 分

       解法二（加减消元法）：

        ①?6 ，得                  ．③························································2 分

                    6x ?1 2y  6 6

        ③?②，得                ．所以y      4 ．···············································4 分

                      1 1y  4 4

                                      x   3，

                                     ?

       将y    4 代入①，得x        3 ．所以?

                                      y   4．

                                     ?       ·············································7 分

   18．解：（1）原式=3+3=6；···································································3 分

                           七年级参考答案 共3 页 第1 页

----------------------- Page 2-----------------------

                              1

        （2）原式=       3(  3 ?  ) =3-1=2．·····················································7 分

                               3

 19．解：去分母，得 x ?5 ≤3x ?2?2. ····················································3 分

      移项、合并同类项，得 2x ≥1.                    ·······················································5 分

                             1

      系数化为1，得x ≥ ．··································································6 分

                             2

      解集表示略．················································································8 分

 20．解：设买一只猫x 元，买一只狗y  元．···············································1 分

                          x ?2y    70,

                         ?

          根据题意得 ?                         ························································4 分

                          2x ?y    50.

                         ?

                               x   10,

                              ?

          解这个方程组得 ?                     ··························································7 分

                               y   30.

                              ?

      答：买一只猫10 元，买一只狗30 元．··············································8 分

                         3

 21．（1）该班有15?                50 名学生；·······················································2 分

                        10

      （2）去敬老院服务的学生有10 人.                     ····················································4 分

     补全图形如下：

                                                    ············································6 分

       （3）若全年级有200 名学生，则估计去敬老院的人数为200×20 ％＝40                                        （人）.

      ···································································································9 分

22．解（1）平行．················································································1 分

           理由：因为∠CDB+∠2=180°, ∠1+∠2=180°,

                              所以∠1 =∠CDB ,  所以AE ∥FC ．························5 分

                              七年级参考答案 共3 页 第2 页

----------------------- Page 3-----------------------

       （2）平行．················································································6 分

         理由：由（1）知AE ∥FC ，所以∠DAE =∠FDA                   ，

                        又因为∠DAE =∠BCF          ，所以∠BCF =∠FDA         ，

                        所以AD ∥CB      ．··············································10 分

                                            3     3

23．解：（1）在△ABC  中，∵BC= yC             ?y B    ??0    , BC 边上的高=OB=2．

                                            2     2

                       1             1  3      3

     ∴△ABC  的面积为 ×BC ×OB=             ? ?2      .  ···································3 分

                       2             2  2     2

                            1            1

     （2）∵△ABO 的面积为 ×AO ×OB=                ?1?2   1.

                           2             2

   在△AOP 中，∵AO= yA           1,  AO 边上的高= xO  ?xP      0 ??a ??a.

                       1                 1            1

    ∴△AOP 的面积为 ×AO ×（-a）=                 ?1??(  a) ?? a.

                       2                 2            2

                                1

    ∴四边形ABOP  的面积为1? a ．····················································7 分

                                2

                      1

     （3）存在P     （-1， ）．求解过程如下：·············································8 分

                      2

                                 3

     由（1）知，△ABC  的面积为 ；

                                 2

                                          1

     由（2）知，四边形ABOP  的面积为1? a ．

                                          2

                    1    3                            1

    依题意，得 1? a =           ，解得a ??1．∴P          （-1， ）．······················10 分

                    2    2                           2

                         七年级参考答案 共3 页 第3 页

我班已建立班级博客圈，欢迎来访^_^

0

推荐