question_answer

Let D be the middle point of the side BC of a triangle ABC. If the triangle ADC is equilateral, then \[{{a}^{2}}:{{b}^{2}}:{{c}^{2}}\]is equal to  [Pb. CET 2004]

A) \[1:4:3\]

B) \[4:1:3\]

C) \[4:3:1\]

D) \[3:4:1\]

Correct Answer: B

Solution :

\[\cos 120{}^\circ =\frac{{{x}^{2}}+{{x}^{2}}-A{{B}^{2}}}{2{{x}^{2}}}\] Þ \[\frac{2{{x}^{2}}-A{{B}^{2}}}{2{{x}^{2}}}=\frac{-1}{2}\] Þ \[4{{x}^{2}}-2A{{B}^{2}}=-2{{x}^{2}}\] Þ  \[3{{x}^{2}}=A{{B}^{2}}\]Þ\[AB=x\sqrt{3}\] Þ  \[{{a}^{2}}:{{b}^{2}}:{{c}^{2}}={{(2x)}^{2}}:{{x}^{2}}:{{(x\sqrt{3})}^{2}}\] =  \[4{{x}^{2}}:{{x}^{2}}:3{{x}^{2}}\] = \[4:1:3\].