Question 467675
First you did not include any given lengths of the sides to determine area so I will give the solution in general terms.
I hope I depicted the figure accurately from your description.
Assuming the bottom half of square is cut into 2 Equal triangles.
Then area of 1 of those triangles is one-fourth area of outer square.
Also notice this triangle is a special 45-45-90 right triangle.
*Diagonal line across square cuts corners into two 45 degree angles* 
Thus smaller triangle above is also a special 45-45-90 right triangle.
The sides of such a triangle are (x,x, and {{{sqrt(2)}}}x)
The area of such a triangle is:
A = {{{x^2/2}}}
In terms of the hypotenuse, h where h = {{{sqrt(2)x}}} then x = {{{h/sqrt(2)}}} 
A = {{{((h/sqrt(2))^2)/2}}} = {{{h^2/4}}}
Now given the side length of the outer square (s) and the hypotenuse of smaller triangle (c).
**Notice that s is the hypotenuse of the larger shaded triangle** 
Use the above formula to obtain the area of the two shaded triangles
Area = {{{(s^2/4) + (c^2/4)}}} = {{{(s^2 + c^2)/4}}}