| ABCD is a rhombus in which\[\angle C=60{}^\circ ,\] Then,\[AC:BD\] is equal to |
 |
A) \[\sqrt{3}:1\]
B) 3 : 1
C) \[\sqrt{3}:\sqrt{2}\]
D) 3 : 2
Correct Answer: A
Solution :
| ABCD is a rhombus. So, its all the sides are equal. |
| \[BC=DC\] |
| \[\Rightarrow \] \[\angle BDC=\angle DBC=x{}^\circ \] |
| Also, \[\angle BCD=60{}^\circ \] |
| In \[\Delta BCD,\] \[2x+60{}^\circ =180{}^\circ \] |
| \[\Rightarrow \] \[2x=120{}^\circ \]\[\Rightarrow \,\]\[x=60{}^\circ \] |
| So, \[\Delta BCD\] is an equilateral triangle. |
| \[BD=BC=a\] |
| Now, \[A{{B}^{2}}=O{{A}^{2}}+O{{B}^{2}}\] |
| \[O{{A}^{2}}=A{{B}^{2}}-O{{B}^{2}}={{a}^{2}}-{{\left( \frac{a}{2} \right)}^{2}}={{a}^{2}}-\frac{{{a}^{2}}}{4}=\frac{3{{a}^{2}}}{4}\] |
| \[OA=\frac{\sqrt{3}a}{2}\] |
| \[\therefore \] \[AC=2\times OA=\sqrt{3}\,a\] |
| Thus, \[AC:BD=\frac{\sqrt{3}a}{a}=\frac{\sqrt{3}}{1}\] |
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