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SSC Sample Paper Mock Test-10 SSC CGL Tear-II Paper-1

  • question_answer
           In figure, ABCD is square. F is the mid-point of AB, BE is one-third of BC. If the area of the\[\Delta \,\,FBE\] is 108 sq cm. Then, the length of AC is      

    A)  \[36\sqrt{3}\,\,\text{cm}\]

    B)  \[35\sqrt{2}\,\,\text{cm}\]  

    C)  \[12\sqrt{3}\,\,\text{cm}\]

    D)  \[36\sqrt{2}\,\,\text{cm}\]

    Correct Answer: D

    Solution :

       \[BF=\frac{1}{2}AB=\frac{1}{2}\times 6x=3x\] Area of \[\Delta FBE=\frac{1}{3}\times 3x\times 2x=3{{x}^{2}}\] \[\therefore \]\[3{{x}^{2}}=108\] \[{{x}^{2}}=36\]\[\Rightarrow \]\[x=6\,\,\text{cm}\] \[\therefore \]\[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}=2A{{B}^{2}}=2\,\,{{(36)}^{2}}\]\[(\because AB=36)\] \[AC=36\sqrt{2}\,\,\text{cm}\]


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