Let $X$ be a compact smooth 2-dimensional Riemannian manifold with boundary. Assume that the Gauss curvature of $X$ is at least $\kappa$, the diameter is at most $D$, and the second fundamental form of the boundary is at least $\lambda$.


**Does there exist an upper bound on the area of $X$ in terms of $\kappa, D,\lambda$ only?**


Remarks. (1) In the case of the empty boundary the answer is known to be positive (e.g. it follows from the Bishop inequality).


(2) If the boundary is non-empty and $\lambda\geq 0$, then the answer is also positive. In this case $X$ is locally geodesically convex and hence it is an Alexandrov space of curvature at most $\kappa$ in the sense of Alexandrov, and for such spaces there is a generalization of the Bishop inequality due to Burago-Gromov-Perelman. (Although there is a little bit more direct but longer explanation.)



Estimate of area of 2-dimensional surface
Yes, there is a bound; your problem can be reduced to the case of convex boundary using the following trick.


Without loss of generality, we can assume that $\lambda=-\tfrac1{10}$ and $\kappa=-1$.


In this case you can attach a collar to your surface locally isometric to the tubular neighborhood in of line in the Lobachevsky plane which has curvature of boundary $\tfrac1{10}$. In the obtained surface boundary is convex and it has curvature $\ge -1$ in the sense of Alexandrov (by Alexandrov's gluing theorem). 
The diameter, area and perimeter of the obtained surface can be bounded in terms of the diameter area and perimeter of the original surface and the other way around. 


