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J & K CET Engineering J and K - CET Engineering Solved Paper-2012

  • question_answer
    In   a   \[\Delta \,\,ABC,\] let \[\angle A=\frac{\pi }{2}\]  and \[(a+b+c)\,(b+c-a)=\lambda bc,\] then \[\lambda \] equals to

    A)  \[0\]                

    B)  \[1\]

    C)  \[2\]      

    D)  \[-2\]

    Correct Answer: C

    Solution :

    Since, \[\Delta \,ABC\] is a right angled triangle at A. \[\therefore \] \[{{b}^{2}}+{{c}^{2}}={{a}^{2}}\] LHS \[=(a+b+c)\,(b+c-a)\] \[={{(b+c)}^{2}}-{{a}^{2}}\] \[={{b}^{2}}+{{c}^{2}}+2bc-{{a}^{2}}\] \[={{a}^{2}}+2bc-{{a}^{2}}\] \[=2bc\] But \[RHS=\lambda bc\] \[\therefore \]\[\lambda =2\]


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