A) \[4/\sqrt{15}\]
B) \[2/\sqrt{15}\]
C) \[4/3\sqrt{15}\]
D) \[1/\sqrt{15}\]
Correct Answer: B
Solution :
Let ABC be an equilateral triangle with base BC. So, \[AD\bot BC\] Now, length \[AD=\left| \frac{2+2(-1)-1}{\sqrt{{{(1)}^{2}}+{{(2)}^{2}}}} \right|=\left| \frac{2-2-1}{\sqrt{5}} \right|=\frac{1}{\sqrt{5}}\] Let \[AB=BC=AC=x\] So, in \[\Delta ABD,\] \[A{{D}^{2}}+B{{D}^{2}}=A{{B}^{2}}\] \[\Rightarrow {{\left( \frac{1}{\sqrt{5}} \right)}^{2}}+\frac{{{x}^{2}}}{4}={{x}^{2}}\] \[\left( \because \,\,BD=\frac{BC}{2} \right)\] \[\Rightarrow \,\,\,\frac{1}{5}={{x}^{2}}-\frac{{{x}^{2}}}{4}\Rightarrow \frac{3}{4}{{x}^{2}}=\frac{1}{5}\] \[\Rightarrow \,\,{{x}^{2}}=\frac{4}{15}\Rightarrow x=\frac{2}{\sqrt{15}}\]You need to login to perform this action.
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