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

A) \[1:4:3\]

B) \[4:1:3\]

C) \[4:3:1\]

D) \[3:4:1\]

Correct Answer: B

Solution :

[b] \[\cos 120{}^\circ =\frac{{{x}^{2}}+{{x}^{2}}-A{{B}^{2}}}{2{{x}^{2}}}\]

\[\Rightarrow \frac{2{{x}^{2}}-A{{B}^{2}}}{2{{x}^{2}}}=\frac{-1}{2}\]

\[\Rightarrow 4{{x}^{2}}-2A{{B}^{2}}=-2{{x}^{2}}\]

\[\Rightarrow 3{{x}^{2}}=A{{B}^{2}}\Rightarrow AB=x\sqrt{3}\]

\[\Rightarrow {{a}^{2}}:{{b}^{2}}:{{c}^{2}}={{(2x)}^{2}}:{{x}^{2}}:{{(x\sqrt{3})}^{2}}\]

\[=4{{x}^{2}}:{{x}^{2}}:3{{x}^{2}}=4:1:3\]