Question

In an equilateral triangle, 3 coins of radii 1 unit each are kept
so that they touch each other and also the sides of the triangle.
Area of the triangle is

• A. $4+2\sqrt{3}$
• B. $6+4\sqrt{3}$
• C. $12+\frac{7\sqrt{3}}{4}$
• D. $3+\frac{7\sqrt{3}}{4}$

Answer 1

Answer: (B)

$6+4\sqrt{3}$

The sides of the equilateral triangles are tangents to the circle.

Hence the length of the sides of the triangles will be

$a=(r+\frac{r}{tan{30}^0})$ where r is the radius of the circle

$=2(1+\frac{1}{tan{30}^0})$

$=2(1+\sqrt{3})$

Hence the the area of the triangle

$=\frac{\sqrt{3}}{4}{a}^2$

$=\frac{\sqrt{3}}{4}\left(2\right(1+\sqrt{3}){)}^2$

$=\frac{\sqrt{3}}{4}\times 4(1+\sqrt{3}{)}^2$

$=\sqrt{3}(1+\sqrt{3}{)}^2$

$=\sqrt{3}(1+3+2\sqrt{3})$

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