Here is an approach for the answer, that relies on a small (correct assumption) that would likely be possible to make more formal.
Step 1. We transform the problem into something more manageable. I would like to cut off the square on the top, together with the corresponding part of the rectangle, to get one rectangle with dimensions $2a \times (10-a)$ and that leaves another rectangle with dimensions $a \times 10$. Since we are into computing area, we can transform the second rectangle into an equivalent one with one side $2a$ and another side of $5$ (equivalent because halving one side and doubling another leaves the same area). Now connect the two rectangles together to get one rectangle of dimensions $2a \times (15-a)$.
Now we are asking how to geometrically prove that the maximum area of this rectangle happens at $a=15/2$.
Step 2. We know that if we need to maximize the area of a rectangle with a fixed perimeter of $30$, this results in finding the best $a$ such that rectangle with dimensions $a \times (15-a)$ contains the largest area. Geometrically, the optimal solution is a square with sides $a = 15-a = 15/2$.
This is exactly our problem, except we have one of the sides a linear scale larger (i.e. $2a$ instead of $a$). I would like to assume that such an extension (or any other simple multiplication of any of the sides by a constant) would not alter the correct solution.
Note: This assumption is correct, it is easy to verify using Calculus that since sucha transformation alters the objective function by a constant factor, it will have no effect on the location of the extrema, but proving this geometrically, as the author of the question requested, is non-trivial to me.
Hence the correct solution does indeed occur at $a=15/2$ and the maximum area will be $2a \times (15-a) = 15*15/2$ as desired.