question_answer

The area  of the equilateral triangle which containing three coins of unity radius is [IIT Screening 2005]

A) \[6+4\sqrt{3}\ sq.\ units\]

B)  \[8+\sqrt{3}\] sq. units

C) \[4+\frac{7\sqrt{3}}{2}\] sq. units

D) \[12+2\sqrt{3}\]sq. units

Correct Answer: A

Solution :

In \[\Delta B{{C}_{1}}M\]; \[BM=({{C}_{1}}M).\,\,\cot 30\] Þ \[BM=\sqrt{3}\] Þ Similarly, \[CN=\sqrt{3}\]and \[MN={{C}_{1}}{{C}_{2}}=1+1=2\] Hence, side \[BC=\sqrt{3}+\sqrt{3}+2=2(1+\sqrt{3})\] Þ Area of equilateral triangle \[=\frac{\sqrt{3}}{4}{{[2(1+\sqrt{3})]}^{2}}\]=\[AB=OA-OB=h(\cot {{15}^{o}}-\cot {{30}^{o}})\]sq units.