Find the center of mass of a combined lamina formed by a square ABCD and an isosceles triangle BCE with given dimensions.

To solve this problem, we first need to find the center of mass (CM) of the combined lamina, which consists of a square ABCD and an isosceles triangle BCE.

Let's set up a coordinate system with point A at the origin $(0,0)$.
The side length of the square ABCD is $6a$.
The coordinates of the vertices are:
A = $(0,0)$
B = $(6a,0)$
C = $(6a,6a)$
D = $(0,6a)$

For the isosceles triangle BCE, $BC = 6a$. The diagram shows $BE = CE = 5a$.
Let E be $(x_E, y_E)$. Since it's an isosceles triangle with base BC,