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.
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