Question

Three coins of unit radius are placed in an equilateral triangle as shown in the figure. The area of the equilateral triangle is

• A. $6+4\sqrt{3}$
• B. $20+8\sqrt{3}$
• C. $7+4\sqrt{3}$
• D. $2+\sqrt{3}$

Answer 1

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

In $\Delta ABC,acos{60}^{\circ }=1$
$a=2$
So, $BC=a\cos {30}^{\circ }$
$=\sqrt{3}$
So, side of equilateral $\Delta$ is $s=2+2\sqrt{3}$
So, are a of $\Delta =\frac{\sqrt{3}}{4}{s}^2$
$=\frac{\sqrt{3}}{4}(2+2\sqrt{3}{)}^2$
$=\sqrt{3}(4+2\sqrt{3})$
$=6+4\sqrt{3}$