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SSC Sample Paper Mock Test-12 SSC CGL Tear-II Paper-1

  • question_answer
    In A ABC show in the figure \[\angle A=90{}^\circ .\]Let D be a point on BC such that BD : DC = 1 : 3. If DM and DL, respectively are perpendicular on AB and AC, then DM and LC are in the ratio of

    A) 1 : 3     

    B)  1 : 2   

    C)  1 : 1    

    D)  4 : 1

    Correct Answer: A

    Solution :

    Consider \[\Delta BMD\]and \[\Delta DLC\] As         \[\angle BMD=\angle DLC=90{}^\circ \]             [Each] Also, \[\angle BDM=\angle DCL\] [Corresponding angle] \[\therefore \]      \[\Delta BMD\sim \Delta DLC\] \[\therefore \]      \[\frac{BD}{DC}=\frac{DM}{LC}=\frac{BM}{DL}\] \[\frac{BD}{DC}=\frac{DM}{LC}=\frac{1}{3}\] \[\therefore \]      DM : LC = 1 : 3


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