Question

Three circular coins each of radii $1$cm are kept in an equilateral triangle so that all the three coins touch each other and also the sides of the triangle.Area of the triangle is

• A. $(4+2\sqrt{3}){cm}^2$
• B. $\frac{1}{4}(12+7\sqrt{3}){cm}^2$
• C. $\frac{1}{4}(48+7\sqrt{3}){cm}^2$
• D. $(6+4\sqrt{3}){cm}^2$

Answer 1

The correct option is D $(6+4\sqrt{3}){cm}^2$

Let $r=1$cm
$a=BC=BD+DE+EC$
$=BD+C_1{C}_2+BD$
$=2.BD+C_1{C}_2$
$=2r\cot {30}^0+2r$
$=2r(\sqrt{3}+1)$
$=2(\sqrt{3}+1)$ since $r=1$cm
$\therefore$Area of $△ABC=\frac{\sqrt{3}}{4}{a}^2$
$=\frac{\sqrt{3}}{4}{\left(2(\sqrt{3}+1)\right)}^2$
$=\frac{\sqrt{3}}{4}\times 4(4+2\sqrt{3})$
$=\sqrt{3}(4+2\sqrt{3})=6+4\sqrt{3}$sq.cm