A) \[1:\sqrt{3}-\sqrt{2}:\sqrt{3}+\sqrt{2}\]
B) \[1:\sqrt{2}-1:\sqrt{3}-\sqrt{2}\]
C) \[1:\sqrt{2}-1:\sqrt{3}\]
D) \[1:\sqrt{2}:\sqrt{3}-1\]
Correct Answer: B
Solution :
\[\text{S=AB=}\frac{\text{1}}{\text{2}}\text{g}{{\text{t}}_{\text{1}}}^{\text{2}}\] \[\Rightarrow \text{2S=AC=}\frac{\text{1}}{\text{2}}\text{g}{{\left( {{\text{t}}_{\text{1}}}\text{+}{{\text{t}}_{\text{2}}} \right)}^{\text{2}}}\] \[\text{and 3S = AD = }\frac{1}{2}G{{\left( {{\text{t}}_{\text{1}}}\text{+}{{\text{t}}_{\text{2}}}+{{t}_{3}} \right)}^{\text{2}}}\] \[{{t}_{1}}=\sqrt{\frac{2S}{g}}\] \[{{\text{t}}_{\text{1}}}\text{+}{{\text{t}}_{\text{2}}}=\sqrt{\frac{4S}{g}},{{t}_{2}}=\sqrt{\frac{4S}{g}}-\sqrt{\frac{2S}{g}}\] \[{{\text{t}}_{\text{1}}}\text{+}{{\text{t}}_{\text{2}}}+{{t}_{3}}=\sqrt{\frac{6S}{g}},{{t}_{3}}=\sqrt{\frac{6S}{g}}-\sqrt{\frac{4S}{g}}\] \[{{\text{t}}_{\text{1}}}\text{:}{{\text{t}}_{\text{2}}}\text{:}{{\text{t}}_{\text{3}}}::1:\left( \sqrt{2}-\sqrt{1} \right):\left( \sqrt{3}-\sqrt{2} \right)\]You need to login to perform this action.
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