2. Sample Papers
  3. SSC

SSC Sample Paper SSC CHSL (10+2) Sample Test Paper-5

  • question_answer
    The perimeter of two squares are \[40\,\,cm\]and\[24\,\,cm\]. The perimeter of a third square, whose area is equal to the difference of the areas of these squares is

    A) \[34\,\,cm\]                   

    B) \[32\,\,cm\]  

    C) \[38\,\,cm\]       

    D)        \[30\,\,cm\]

    Correct Answer: B

    Solution :

     Side of the first square\[=\frac{40}{4}=10\,\,cm\] Side of the second square\[=\frac{24}{4}=6\,\,cm\] Difference of the areas of these squares\[=(10\times 10-6\times 6)c{{m}^{2}}\]             \[=(100-36)c{{m}^{2}}=64c{{m}^{2}}\] \[\therefore \]Area of the third square\[=64\,\,c{{m}^{2}}\] \[\therefore \]Side of Third Square\[=64\,\,c{{m}^{2}}\] \[\therefore \]Perimeter of this square             \[=(4\times 8)cm=32\,\,cm\]


You need to login to perform this action.
You will be redirected in 3 sec spinner