A) \[8\sqrt{2}\,\,cm\]
B) 16 cm
C) \[\sqrt{32}\,\,cm\]
D) 8 cm
Correct Answer: D
Solution :
Side of the first square \[=\frac{1}{\sqrt{2}}\times 4\sqrt{2}=4\,\,cm\] Its area \[={{(4)}^{2}}=16\,\,c{{m}^{2}}\] \[\therefore \] Area of second square \[=2\times 16=32\,\,c{{m}^{2}}\] Its side \[=\sqrt{32}=4\sqrt{2}\,\,cm\] \[\therefore \]Required diagonal \[=\sqrt{2}\times 4\sqrt{2}=8\,\,cm\]You need to login to perform this action.
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