question_answer

In a rhombus ABCD, ZDAB = 60° and diagonals AC = \[{{d}_{1}}\]and BD = \[{{d}_{2}}\]then\[{{d}_{1}}:{{d}_{2}}\]is

A) \[\sqrt{3}:1\]     

B)  \[\sqrt{3}:2\]

C) 1: 2                  

D)        \[2\sqrt{3}:1\]

Correct Answer: A

Solution :

[a] \[\angle DAB=60{}^\circ .\] Since.AB -AD.       \[\Delta \,ABD\] is an equilaterd. \[DB={{d}_{2}}\] So \[{{d}_{2}}\] is the side of this triangle.    \[AO=\frac{1}{2}\,\,AC=\frac{1}{2\,\,}{{d}_{1}}\]. \[O{{A}^{2}}=A{{B}^{2}}-O{{B}^{2}}\Rightarrow {{\left( \frac{1}{2}{{d}_{1}} \right)}^{2}}={{d}_{2}}^{2}-\frac{1}{4}\,\,{{d}_{2}}^{2}\] \[=\frac{3}{4}{{d}_{2}}^{2}\] \[\therefore \,\,{{d}_{1}}^{2}=3\,\,{{d}_{2}}^{2}\,\,or\,\,{{d}_{1}}=\sqrt{3}\,\,\,{{d}_{2}}\,\,i.e.{{d}_{1}}:{{d}_{2}}=\sqrt{3}:1\]