Question

Three circles of radii 1 cm each are drawn in an equilateral triangle ABC such that all the circles touch one another. Find the area of the triangle.

Answer 1

Refer to the attached image.

Let the side of the triangle AC = 'a'

Let 'x' be the distance between the vertex A and O.

Now, consider a = 2x + 2r

In triangle, AOD,

$\tan A = \frac{r}{x}$

$\tan 30^\circ = \frac{r}{x}$

$\frac{1}{\sqrt 3} = \frac{r}{x}$

Now, a = 2x + 2r

$a = 2r(\frac{x}{r} + 1)$

$a = 2r(\sqrt 3+1)$

Since, r = 1

Therefore, a = $2(\sqrt 3+1)$

Area of equilateral triangle = $\frac{\sqrt 3}{4}a^2$

= $\frac{\sqrt 3}{4}(2(\sqrt3+1))^2$

= $\sqrt 3(\sqrt3+1)^2$

= $3\sqrt3+6$ square units.

Hence, the area of equilateral triangle is $3\sqrt3+6$ square units.