A) \[2/\sqrt{15}\]
B) \[4/3\sqrt{3}\]
C) \[1/\sqrt{5}\]
D) \[4/\sqrt{15}\]
Correct Answer: A
Solution :
Key Idea: In equilateral triangle all sides of a triangle are equal. Given equation of \[BC\] is\[x+2y=1\] The perpendicular distance from a point\[A(2,\,\,-1)\]is \[AD=\frac{|2+2(-1)-1|}{\sqrt{1+4}}=\frac{|-1|}{\sqrt{5}}\] \[=\frac{1}{\sqrt{5}}\] In\[\Delta ABD\], \[\sin {{60}^{o}}=\frac{AD}{AB}\] \[\Rightarrow \] \[AB=\frac{1}{\sqrt{5}}\times \frac{2}{\sqrt{3}}\] \[=\frac{2}{\sqrt{15}}\] Note: In equilateral triangle, median of a triangle is the mid poin.t of base.You need to login to perform this action.
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