| If the length of each side of an equilateral triangle is increased by 2 units, the area is found to be increased by \[3+\sqrt{3}\]sq units. The length of each side of the triangle is |
A) \[\sqrt{3}\,\,\text{units}\]
B) \[3\sqrt{3}\,\,\text{units}\]
C) \[3\,\,\text{units}\]
D) \[1+3\sqrt{3}\,\,\text{units}\]
Correct Answer: A
Solution :
| Let the original side of equilateral triangle be x units |
| New length of each side \[=(x+2)\] |
| According to the question, |
| \[\frac{\sqrt{3}}{4}{{(x+2)}^{2}}-\frac{\sqrt{3}}{4}({{x}^{2}})=3+\sqrt{3}\] |
| \[\Rightarrow \] \[\frac{\sqrt{3}}{4}[{{(x+2)}^{2}}-{{x}^{2}}]=3+\sqrt{3}\] |
| \[\Rightarrow \] \[\frac{\sqrt{3}}{4}[4x+4]=3+\sqrt{3}\] |
| \[\Rightarrow \] \[4\sqrt{3}x+4\sqrt{3}=12+4\sqrt{3}\] |
| \[\Rightarrow \] \[4\sqrt{3}x=12\] |
| \[x=\frac{12}{4\sqrt{3}}=\sqrt{3}\] |
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