Three Circles touching each other

Place the circles on a coordinate plane so that the center between the three circles is at (0, 0). Then the left circle has an equation of $(x + R)^2 + (y - \frac{\sqrt{3}}{3}R)^2 = R^2$ , the right circle $(x - R)^2 + (y - \frac{\sqrt{3}}{3}R)^2 = R^2$ , and the bottom circle $x^2 + (y + \frac{2\sqrt{3}}{3}R)^2 = R^2$ .

The $y$ -coordinate of $T$ can be found by solving $0^2 + (T_y + \frac{2\sqrt{3}}{3}R)^2 = R^2$ , which is $T_y = (1 - \frac{2\sqrt{3}}{3})R$ .

The $x$ -coordinate of $Q$ can be found by solving $(Q_x - R)^2 + (T_y - \frac{\sqrt{3}}{3}R)^2 = R^2$ , which is $Q_x = (\sqrt{2\sqrt{3}-3}+1)R$

The $x$ -coordinate of $R$ is $R_x = R$ .

The $y$ -coordinate of $R$ can be found by solving $R_x^2 + (R_y + \frac{2\sqrt{3}}{3}R)^2 = R^2$ , which is $R_y = (\frac{\sqrt{3}}{3} - 1)R$ .

The height of the trapezoid is then $h = T_y - R_y = (2 - \sqrt{3})R$ .

The median of the trapezoid is then $m = \frac{1}{2}(2Q_x + 2R_x) =(\sqrt{2\sqrt{3}-3}+2)R$

The area of the trapezoid is $A = (2 - \sqrt{3})(\sqrt{2\sqrt{3}-3}+2)R^2 = 10$ , which solves to $R^2 \approx 13.919 \approx \boxed{14}$ .