An 8-inch by 8-inch square is folded along a diagonal creating a triangular region. This resulting triangular region is then folded so that the right angle vertex just meets the midpoint of the hypotenuse. What is the area of the resulting trapezoidal figure in square inches?

[asy]
fill((0,0)--(10,0)--(10,10)--(0,10)--cycle,gray(.7));
draw((0,0)--(10,0)--(10,10)--(0,10)--cycle,linewidth(1));
fill((20,0)--(20,10)--(30,10)--cycle,gray(0.7));
draw((20,0)--(20,10)--(30,10)--cycle,linewidth(1));
draw((20,0)--(30,0)--(30,10),dashed);
fill((40,0)--(40,5)--(45,10)--(50,10)--cycle,gray(0.7));
draw((40,0)--(40,5)--(45,10)--(50,10)--cycle,linewidth(1));
draw((40,5)--(45,5)--(45,10));
draw((40,5)--(40,10)--(45,10),dashed);
draw((40,0)--(50,0)--(50,10),dashed);
draw((12,5)--(18,5),Arrow);
draw((32,5)--(38,5),Arrow);

[/asy]
Break the figure up into smaller $4\times4$ squares by making two cuts, one vertical cut down the center and one horizontal cut across the center. In the top left small square, half is occupied by part of the trapezoid (since a diagonal of a square splits the square into two equal areas). Similarly, in the top right small square and bottom left small square, half is occupied by part of the trapezoid. In the bottom right small square, none is occupied by the trapezoid. In all, there are three half-squares that make up the trapezoid. Since each small square has area $4\cdot4=16$ square inches, the area of the trapezoid is $\frac{3}{2}\cdot16=24$ square inches.