Question

The area of the equilateral triangle, in which three coins of radius $1cm$ are placed, as shown in the figure is

• A. $(6+4\sqrt{3})sqcm$
• B. $(4\sqrt{3},-6)sqcm$
• C. $(7+4\sqrt{3})sqcm$
• D. $4\sqrt{3}sqcm$

Answer 1

Answer: (A)

$(6+4\sqrt{3})sqcm$

$\begin{array}{c}\tan {30}^{\circ }=\frac{AC}{BC}\\ \frac{1}{\sqrt{3}}=\frac{1}{BC}\\ BC=\sqrt{3}\\ \Rightarrow sideof\Delta is2+2\sqrt{3}\\ Areaofequal\Delta =\frac{\sqrt{3}}{4}\left(2+2\sqrt{3}\right)\\ =\frac{\sqrt{3}}{4}\times \left\{4+4\times 3+8\sqrt{3}\right\}\\ =\sqrt{3}\left(4+2\sqrt{3}\right)\\ =\left(4\sqrt{3}+6\right)c{m}^2\end{array}$