Let's solve the problem step-by-step:
- Square $A B C D$ has $E$ and $F$ as midpoints of sides $C D$ and $A D$, respectively.
- We want to find the ratio of the area of triangle $B E A$ to the area of square $A B C D$.

Assume the side length of the square is $s$.

Coordinates (for easy calculation):
- $A=(0,0)$
- $B=(s, 0)$
- $C=(s, s)$
- $D=(0, s)$

Midpoints:
- $E$, midpoint of $\(C D: E=\left(\frac{s+0}{2}, \frac{s+s}{2}\right)=\left(\frac{s}{2}, s\right)\)$
- $F$, midpoint of $\(A D: F=\left(\frac{0+0}{2}, \frac{0+s}{2}\right)=\left(0, \frac{s}{2}\right)\)$

Using vertices of $\(\triangle B E A\)$ :
- $\(B=(s, 0)\)$
- $\(E=\left(\frac{s}{2}, s\right)\)$
- $\(A=(0,0)\)$

Area of $\triangle B E A$ is given by:

$$
\(\text { Area }=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|\)
$$


Substitute points $\(B\left(x_1, y_1\right), E\left(x_2, y_2\right), A\left(x_3, y_3\right)\)$ :

$$
\(=\frac{1}{2}\left|s(s-0)+\frac{s}{2}(0-0)+0(0-s)\right|=\frac{1}{2}|s \times s|=\frac{s^2}{2}\)
$$


Area of square $\(A B C D=s^2\)$.
Ratio of areas:

$$
\(\frac{\text { Area of } \triangle B E A}{\text { Area of square } A B C D}=\frac{\frac{s^2}{2}}{s^2}=\frac{1}{2}\)
$$


Thus, the ratio is $\(\frac{1}{2}\)$.
The correct answer is (C) $\(\frac{1}{2}\)$. The ratio of the area of triangle $B E A$ to the area of square $A B C D$ is $\(\frac{1}{2}\)$.

This is because triangle $B E A$ has half the area of the square when points $E$ and $F$ are midpoints of sides $C D$ and $A D$ respectively.

The correct answer is (C) $\(\frac{1}{2}\)$.