Question

The area of the equilateral triangle, in which three coins of radius 1 cm 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

The correct option is A $(6+4\sqrt{3})sqcm$

Let radius of the coin $O_{2}D=r=1$

Side of the equilateral triangle

$BC=BD+DE+EC$

$\Rightarrow BC=2BD+2r$ [since DE=2r]

$\Rightarrow BC=2O_{2}D\cot 30+2r$

$\Rightarrow BC=2r\cot 30+2r$

$\Rightarrow BC=2\cdot 1\cdot \sqrt{3}+2\cdot 1=2+2\sqrt{3}$

Area of $△ABC$

$=\frac{\sqrt{3}}{4}B{C}^2$

$=\frac{\sqrt{3}}{4}(2+2\sqrt{3}{)}^2$

$=\sqrt{3}(4+2\sqrt{3})$

$=6+4\sqrt{3}$