Max Area of Rectangle within Isosceles Triangle

[Izzhov]

The Problem:
To find the maximum area of a rectangle within the boundaries of an isosceles triangle with a base of 10 and congruent sides' lengths of 13.

How Far I Was Able to Get:
I was able to prove that the max area was 30 if one of the sides of the rectangle coincides with one of the sides of the isosceles triangle. However, I was not able to prove that the area of the rectangle could not be greater if none of the sides of the rectangle coincide with any of the sides of the isosceles triangle (meaning that one or more of the vertices of the rectangle is not touching any of the edges of the triangle).

 

[AlephZero]

The area of a rectangle is the same as the area of a parallelogram with the same base and height.

Think about the maximum area of parallelogram you can fit into the triangle.

 

[Izzhov]

Hmm... I think I get it. Since a triangle can be made out of four congruent triangles, and a parallelogram can be made out of two congruent triangles, the maximum area has to be when the triangles of the parallelogram line up with two of the triangles comprising the larger one, which is one half the area of the larger triangle. Is this correct?