# Sum of Areas of Equilateral Triangles Inscribed in Circles

**Problem**

An equilateral triangle is inscribed within a circle whose diameter is 12 cm. In this triangle a circle is inscribed; and in this circle, another equilateral triangle is inscribed; and so on indefinitely. Find the sum of the areas of all the triangles.

**Solution**

$\dfrac{x_1}{2} = 6 \cos 30^\circ$

$x_1 = 6\sqrt{3} ~ \text{cm}$

$R_2 = 6 \sin 30^\circ$

$R_2 = 3 ~ \text{cm}$

The areas of all equilateral triangles form an IGP

$a_1 = A_1 = \frac{1}{2}{x_1}^2 \sin 60^\circ$

$r = \left( \dfrac{R_2}{R_1} \right)^2 = \left( \dfrac{3}{6} \right)^2 = 1/4$

$S = \dfrac{a_1}{1 - r} = \dfrac{\frac{1}{2}(6\sqrt{3})^2 \sin 60^\circ}{1 - 1/4}$

$S = 62.35 ~ \text{cm}^2$ ← *answer*

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