CLAT Sample Paper CLAT Sample Paper-8

  • question_answer
    The perimeters of both, a square and a rectangle are each equal to 48 m and the difference between their areas is \[4\,\,{{m}^{2}}\]. The breadth of the rectangle is

    A)  10 m               

    B)  12 m   

    C)  14 m               

    D)  None of the above

    Correct Answer: A

    Solution :

    Let the length of rectangle \[=x\,m\] and              its breadth = y m Also, let the side of the square be z m. Then,       \[2(x+y)=4z=48\] \[\Rightarrow \]               \[x+y=24\] and \[z=12\] Also,                 \[{{z}^{2}}-xy=4\] \[\Rightarrow \]               \[xy={{z}^{2}}-4=144-4=140\] So,                   \[{{(x-y)}^{2}}={{(x+y)}^{2}}-4\,xy\] \[=576-560=16\] \[\therefore \]      \[x-y=4\] and \[x+y=24\] On solving the above equations, we get \[2y=20\] \[\therefore \]      \[y=10\,m\]

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