Find the center of gravity of a square lamina of side 4a from which a rectangular section of side 2a has been removed from one corner, with one vertex coinciding with a vertex of the square.

To find the center of gravity of the resulting lamina, we will use the principle of moments by considering the original square and subtracting the removed square section.

Step 1: Define the coordinate system and properties of the original square.
Let the vertex A of the square ABCD be at the origin $(0,0)$.
Given AB = 4a and BC = 4a, the vertices are:
A = $(0,0)$
B = $(4a,0)$
C = $(4a,4a)$
D = $(0,4a)$

The area of the original square ABCD, $A_1$, is $(4a)^2 = 16a^2$.
The center of gravity of the original square, $G_1$, is at its geometric center:
$