A) \[48\sqrt{2}\,\,{{m}^{2}}\]
B) \[12\sqrt{2}\,\,{{m}^{2}}\]
C) \[48\,\,{{m}^{2}}\]
D) \[12\sqrt{3}\,\,{{m}^{2}}\]
Correct Answer: A
Solution :
Let\[AE\]is a vertical lamp post In\[\Delta \,\,AEC\], \[\tan {{45}^{o}}=\frac{AE}{AC}\] \[\Rightarrow \] \[AC=AE=12\,\,m\] and in\[\Delta \,\,ABE\], \[\tan {{60}^{o}}=\frac{AE}{AB}\] \[\Rightarrow \] \[AB=\frac{AE}{\sqrt{3}}=4\sqrt{3}\] Now, \[BC=\sqrt{A{{C}^{2}}-A{{B}^{2}}}\] \[=\sqrt{144-48}=\sqrt{96}\] \[=4\sqrt{6}\] \[\therefore \]Area of rectangle ABCD \[=AB\times BC\] \[=4\sqrt{3}\times 4\sqrt{6}\] \[=48\sqrt{2}\,\,{{m}^{2}}\]You need to login to perform this action.
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