dv said:

To generalise

The base b of the rectangle would be 6 cosθ
The height of the triangle would be 6 sinθ
Pythag:
r^2 = h^2 + (b-r)^2
r^2 = 36 (sinθ)^2 + (6cosθ – r)^2
r^2 = 36 (sinθ)^2 + 36 (cosθ)^2 + – 12cosθr + r^2
36 (sinθ^2 + cosθ^2) = 12cosθr
36 = 12cosθr
r = 3/cosθ

A = r b = 3/cosθ * 6 cosθ = 18

The irony is that my original assumption was correct. And even though my results for θ=45° and θ=0° didn’t actually use my incorrect formula, my incorrect formula give the correct answer anyway.