Calculus of Variations and Lagrange Multipliers

Yes. For the general theorem, see this Wikipedia page. In the particular case which you consider, you can actually consider the minimization problem

$$ \int_{-1}^1 f(x) + \lambda (\sqrt{ 1 + f'(x)^2} - \pi) dx $$

which leads to the Euler-Lagrange equation

$$ \left( \frac{f'}{\sqrt{1 + (f')^2}}\right)' = \lambda^{-1} $$

which leads to a somewhat unappealing looking Lagrangian with $f''$ in the denominator when you plug $\lambda$ back in.

(This is, of course, not completely unexpected, as your functional $A$ is not bounded below on the set $L[f] = \pi$. To see this, it suffice to note that $L[f + c] = L[f]$ by definition, while $A[f + c] = A[f] + c$. So for a decreasing sequence of $c\searrow -\infty$, $A[f + c]$ decreases without bound.)