Question

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

Answer 1

Let radius of the coin `O_2D=r=1`

Side of the equilateral triangle

`BC=BD+DE+EC`

`=>BC=2BD+2r` [since `DE=2r`]

`=>BC=2O_2Dcot30^@+2r`

`=>BC=2rcot30^@+2r`

`=>BC=2*1*sqrt3+2*1=2+2sqrt3` unit

Area of `DeltaABC`

`=sqrt3/4BC^2`

`=sqrt3/4(2+2sqrt3)^2=sqrt3(4+2sqrt3)` squnit

`=6+4sqrt3` squnit

Answer 2

`6 + 4 sqrt{3}`

Explanation:

Without drawing the picture I see the centers make a small equilateral triangle out of two radii, so sides `2` Outside that, each radius to the side (two per circle) makes a rectangle sides `1` and `2`, 3 rectangles total. What remains is six 30/60/90 right triangles, short side `1` so equivalent to three more equilateral triangles of side `2`. The area of an equilateral triangle is `sqrt{3}/4 s^2` so the total area of the triangle is:

`4 sqrt{3}/4(2^2) + 3(1 times 2) = 6 + 4 sqrt{3}`

[Forgot to hit Post on this one before.]