Let us take two copies of $ Y $-pieces [ or pair of pants ] with each boundary geodesic of length $ l $, and glue them together without any twisting to obtain a genus 2 closed orientable hyperbolic surface $M$. I want to find out the

a) length of the shortest simple closed geodesic and b) Cheeger's isoperimetric constant for this surface. FYI : Cheeger's constant for a closed surface $M $ is defined as

infimum of $ \frac{l(A)}{minimum ( \area of B and \area of B')} $ where B and B' are the two components of $ M \backslash A $, where infimum is taken over all 1 dimensional geodesic submanifolds A, i.e. union of simple closed geodesics A separating $M$ and boundary of A is a part of both boundary of B and that of B'. Any help ?

I believe length of the shortest geodesic should be minimum over some multiples of l, coming from measuring the length of geodesics in $M$ which "we first intuitively see" while drawing a diagram of a genus 2 surface, i.e.the ones surrounding just one "hole", and the ones surrounding "both holes" in $M$, the ones surrounding the "handles".I also beleive the geodesics in the isotopy classe of curves "winding around" a collar of $M$ should have more length.But I want to make it rigorous.

The reason I asked this question is if we can show that Cheeger's constant is say $ \geq \frac{l}{100} $, then it gives us a concrete way to construct a cosed hyperbolic surface with arbitrarily large eigenvalue, by Cheeger's inequality.