Trigonometric Equations Question 252

Question: The area of the equilateral triangle which containing three coins of unity radius is

[IIT Screening 2005]

Options:

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

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Answer:

Correct Answer: A

Solution:

  • In $ \Delta BC_1M $ ; $ BM=(C_1M).\cot 30 $
    Þ $ BM=\sqrt{3} $
    Þ Similarly, $ CN=\sqrt{3} $ and $ MN=C_1C_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.