A) \[3\]
B) \[2\]
C) \[2\sqrt{2}\]
D) \[3\sqrt{2}\]
Correct Answer: D
Solution :
Let\[A=(1,\,\,0,\,\,0),\,\,B=(0,\,\,1,\,\,0)\]and\[C=(0,\,\,0,\,\,1)\] Now, \[AB=\sqrt{{{(0-1)}^{2}}+{{(1-0)}^{2}}+{{0}^{2}}}=\sqrt{2}\] \[BC=\sqrt{{{(0-1)}^{2}}+{{(1-0)}^{2}}+{{0}^{2}}}=\sqrt{2}\] and \[CA=\sqrt{{{(1-0)}^{2}}+{{0}^{2}}+{{(0-1)}^{2}}}=\sqrt{2}\] \[\therefore \]Perimeter of triangle\[=AB+BC+CA\] \[=\sqrt{2}+\sqrt{2}+\sqrt{2}\] \[=3\sqrt{2}\]You need to login to perform this action.
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