The area of the equilateral triangle which co | vedclass

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(A) Let the radius of each coin be $r = 1$. The centers of the three coins form an equilateral triangle with side length $2r = 2$.Let the vertices of

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$6 + 4\sqrt{3} \; 	ext{sq. units}$

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(A) Let the radius of each coin be $r = 1$. The centers of the three coins form an equilateral triangle with side length $2r = 2$.Let the vertices of the large equilateral triangle be $A, B, C$ and the centers of the coins be $C_1, C_2, C_3$.The distance from a vertex (e.g.,$B$) to the projection of the center of the nearest coin $(M)$ on the side $BC$ is given by $BM = r \cot(30^\circ) = 1 	imes \sqrt{3} = \sqrt{3}$.The distance between the centers of the two base coins is $MN = 2r = 2$.By symmetry,the side length $s$ of the large equilateral triangle is $s = BM + MN + NC = \sqrt{3} + 2 + \sqrt{3} = 2 + 2\sqrt{3} = 2(1 + \sqrt{3})$.The area of an equilateral triangle with side $s$ is given by $\frac{\sqrt{3}}{4} s^2$.Area $= \frac{\sqrt{3}}{4} [2(1 + \sqrt{3})]^2 = \frac{\sqrt{3}}{4} 	imes 4(1 + \sqrt{3})^2 = \sqrt{3}(1 + 3 + 2\sqrt{3}) = \sqrt{3}(4 + 2\sqrt{3}) = 4\sqrt{3} + 6 \; 	ext{sq. units}$.

The area of the equilateral triangle which contains three coins of unit radius is

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