A) \[2{{a}^{2}}={{h}^{2}}\]
B) \[2{{h}^{2}}={{a}^{2}}\]
C) \[3{{a}^{2}}=2{{h}^{2}}\]
D) \[2{{h}^{2}}=3{{a}^{2}}\]
Correct Answer: B
Solution :
Let\[ABCD\]be a square of each side of length\[a\]. It is given that\[\angle OCP={{60}^{o}}\]. Length of diagonal \[AC=\sqrt{{{a}^{2}}+{{a}^{2}}}=a\sqrt{2}\] \[\therefore \] \[PC=\frac{AC}{2}=\frac{a}{\sqrt{2}}\] In \[\Delta \,\,OCP,\,\,\tan {{60}^{o}}=\frac{OP}{PC}\] \[\Rightarrow \] \[\sqrt{3}=\frac{h}{a/\sqrt{2}}\] \[\Rightarrow \] \[\sqrt{3}a=\sqrt{2}h\] \[\Rightarrow \] \[3{{a}^{2}}=2{{h}^{2}}\]You need to login to perform this action.
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