In an equilateral triangle , 3 coins of radii 1 unit each are kept in such a way that they touch each other and also the sides of the triangle.
Then the area of the triangle (in sq.units) is

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$3 + 8 \sqrt{3}$

$4 + 5 \sqrt{2}$

$6 + 4 \sqrt{3}$

$4 + 6 \sqrt{3}$

answer is B.

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Detailed Solution

It can be seen that the side of the triangle is

$\begin{array}{l} \tan 30^{\circ} = \frac{E G}{G C} \Rightarrow G G = B H = \sqrt{3} \\ ∴ B C = 2 + 2 \sqrt{3} \end{array}$

$∴$ Area of an equilateral triangle

$= \frac{\sqrt{3}}{4} a^{2} = 6 + 4 \sqrt{3}$

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