Cho AB = 1/3 CD; SΔABT = 1cm^2. Tính SABCD

To find the area \( S_{ABCD} \) of the quadrilateral \( ABCD \) given \( S_{\Delta ABT} = 1 \, \text{cm}^2 \) and \( AB = \frac{1}{3} CD \), we can follow these steps:

1. **Identify Relationship Between Areas**: Since \( AB \) is \( \frac{1}{3} CD \), we can express the relationship of the triangles based on their bases and heights.

2. **Area of \( \Delta CDT \)**: The triangle \( CDT \) will have three times the length of \( AB \) if they share the same height from point \( T \). Thus, we can conclude that:
\[
S_{\Delta CDT} = 3 \times S_{\Delta ABT} = 3 \times 1 \, \text{cm}^2 = 3 \, \text{cm}^2
\]

3. **Calculate the Area of Quadrilateral \( ABCD \)**:
\[
S_{ABCD} = S_{\Delta ABT} + S_{\Delta CDT}
\]
\[
S_{ABCD} = 1 \, \text{cm}^2 + 3 \, \text{cm}^2 = 4 \, \text{cm}^2
\]

Therefore, the area \( S_{ABCD} = 4 \, \text{cm}^2 \).