12. Let $P$ be a point on the diagonal $A C$ of the square $A B C D$. If $A P$ is one-fourth of the length of one side of the square and the area of the quadrilateral $A B P D$ is 1 square unit, find the area of $A B C D$.
12. $4 \sqrt{2}$ square units
Let $s$ be the length of one side of the square $A B C D$. Let $(A B P)$ and $(A B P D)$ denote the areas of $\triangle A B P$ and quadrilateral $A B P D$, respectively. Then $(A B P)=\frac{1}{2}(A B P D)=\frac{1}{2}$.
Since the diagonals of a square are perpendicular to each other and the length of each diagonal of $A B C D$ is equal to $\frac{1}{2} \sqrt{2} s$, we have
$$
(A B P)=\frac{1}{2}(\text { base } \times \text { height })=\frac{1}{2}\left(\frac{s}{4}\right)\left(\frac{\sqrt{2} s}{2}\right)=\frac{\sqrt{2} s^{2}}{16} .
$$

Since $(A B P)$ is also equal to $\frac{1}{2}$, we get $s^{2}=4 \sqrt{2}$ square units as the area of the square $A B C D$.