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27、解:  BD=AD=300    (1′)      CD=100,BC=200 (3′)

     1号救生员:100÷6+300÷2≈178.9    (5′)

     2号救生员:200÷2≈173.2        (7′)

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26、解:(1)证得△PCD∽△PAB     (3′)

    (2) 由相似得=     (4′)

      由圆周角定理得∠ADB=90°  (5′)

      =cos60°=1/2       (6′)

      ∴=1/2          (7′)

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25、设道路宽为xm,(32-x)(20-x)=540  (3′)

          x1=50(舍去),x2=2  (5′)

  答:道路宽为2m。           (6′)

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24、略(每个三角形各2分)

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23、解:∵△ABC,△DEF为等边三角形。

     ∴∠B=∠C=∠DEF=60°     (1′)

     ∴∠BED+∠FEC=120°

     又∠FEC+∠EHC=120°

     ∴∠BED=∠EHC        (3′)

     ∴△DBE∽△ECH        (5′)

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22、(1)原式=  (2′)       (2) 原式=2-1+2+2 +1  (2′)

       =     (4′)           =4+2       (4′)

(3)解:x1=0,x2=5  (4′)       (3)   (4′)

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29、解:(1)∵∠B1CE=∠B1A1O=90°,∠CB1M=∠OB1A

     ∴△B1CM∽△B1A1O        (1′)

(2)B1(0,5),B1C=1          (3′)

  图2中,同理△B2CM1∽△B2A2P

  ∴

  ∴  , 当时,         (5′)

  图3中, 当时,(7′)

  (3)图2中,S=,不成立,舍去。(9′)

     图3中,S=,x=3    (11′)

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28、解:(1)P6(0,-64).··················································································· 2分

(2)······································································ 4分

(3)由题意知,旋转次之后回到轴正半轴,在这次中,点分别落在坐标象限的平分线上或轴或轴上,但各点绝对坐标的横、纵坐标均为非负数,因此,点的坐标可分三类情况:

令旋转次数为

①当时(其中为自然数),或当点落在轴上,

此时,点的绝对坐标为;········································································· 6分

②当时(其中为自然数),或当点落在各象限的平分线上,

此时,点的绝对坐标为···························································· 8分

③当时(其中为自然数),或当点落在轴上,

此时,点的绝对坐标为.······································································· 10分

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178.9﹥173.2,  2号救生员先到达B点。      (8′)

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27、解:  BD=AD=300    (1′)

      CD=100,BC=200 (3′)

     1号救生员:100÷6+300÷2≈178.9    (5′)

     2号救生员:200÷2≈173.2        (7′)

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