In the figure, the area of square $WXYZ$ is $25 \text{ cm}^2$. The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In $\triangle ABC$, $AB = AC$, and when $\triangle ABC$ is folded over side $\overline{BC}$, point $A$ coincides with $O$, the center of square $WXYZ$. What is the area of $\triangle ABC$, in square centimeters?

$\textbf{(A)}\ \frac{15}4\qquad\textbf{(B)}\ \frac{21}4\qquad\textbf{(C)}\ \frac{27}4\qquad\textbf{(D)}\ \frac{21}2\qquad\textbf{(E)}\ \frac{27}2$
The side lengths of square $\text{WXYZ}$ must be 5 cm, since the area is $25\ {cm}^2$. First, you should determine the height of $\triangle{ABC}$. The distance from $\text{O}$ to line $\text{WZ}$ must be 2.5 cm, since line $\text{WX}$ = 5 cm, and the distance from $\text{O}$ to $\text{Z}$ is half of that. The distance from line $\text{WZ}$ to line $\text{BC}$ must be 2, since the side lengths of the small squares are 1, and there are two squares from line $\text{WZ}$ to line $\text{BC}$. So, the height of $\triangle{ABC}$ must be 4.5, which is 2.5 + 2. The length of $\text{BC}$ can be determined by subtracting 2 from 5, since the length of $\text{WZ}$ is 5, and the two squares in the corners give us 2 together. This gives us the base for $\triangle{ABC}$, which is 3. Then, we multiply 4.5 by 3 and divide by 2, to get an answer of $\boxed{\textbf{(C)}\ \frac{27}{4}}$.