A) \[\sqrt{3/2}\]
B) \[\sqrt{2}\]
C) \[\sqrt{2/3}\]
D) None of these
Correct Answer: C
Solution :
Let p be the length of the perpendicular from the vertex (2, ?1) to the base \[x+y=2\]. Then \[p=\left| \frac{2-1-2}{\sqrt{{{1}^{2}}+{{1}^{2}}}} \right|=\frac{1}{\sqrt{2}}\] If 'a' be the length of the side of triangle, then\[p=a\sin {{60}^{o}}\Rightarrow \frac{1}{\sqrt{2}}=\frac{a\sqrt{3}}{2}\Rightarrow a=\sqrt{\frac{2}{3}}\].
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