A) \[48\sqrt{2}sq.m\]
B) \[48\sqrt{3}sq.m\]
C) \[48sq.m\]
D) \[12\sqrt{2}sq.m\]
E) \[12\sqrt{3}sq.m\]
Correct Answer: A
Solution :
Let AE is a vertical lamp-post. Given, \[AE=12\]m \[\tan {{45}^{o}}=\frac{AE}{AC}\] \[AC=AE=12m\] \[\tan {{60}^{o}}=\frac{AE}{AB}\] \[AB=\frac{AE}{\sqrt{3}}=4\sqrt{3}\] \[BC=\sqrt{A{{C}^{2}}-A{{B}^{2}}}\]\[=\sqrt{144-48}\] \[=\sqrt{96}\] \[=4\sqrt{6}\] Area =\[AB\times BC=4\sqrt{3}\times 4\sqrt{6}\] =\[48\sqrt{2}\]sq. cm.You need to login to perform this action.
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