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
Show Answer
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.