calculus - help: isosceles triangle circumscribing a circle of radius r

Please help me show that: the equilateral triangle of altitute $3r$ is the isosceles triangle of least area circumscribing a circle of radius $r$.

Iassumed the following:

base = $2a$

height = $h$

radius of circle = $r$

Area = $\frac{1}{2}(2a)h = ah$

$tan(2\theta) = \frac{h}{a}$ and $tan\theta = \frac{r}{a} $

Ii tried using double angle identities to represent h in terms of a and \theta but I have failed miserably.

This is what I ended up with:
$\displaystyle{%
h={2r\cos^{2}\left(\theta\right) \over 2\cos^{2}\left(\theta\right) - 1}}$

I know i need to differentiate the area and set to zero, but I can't get it. please direct me how to solve this problem. Thank you;