7. In the plane of the regular dodecagon $A_{1} A_{2} \cdots A_{12}$, there are $\qquad$ squares that have at least two vertices in $\left\{A_{1}, A_{2}, \cdots, A_{12}\right\}$ (answer with a number).
7. 183 .

From the 12 vertices of a regular dodecagon, there are $\mathrm{C}_{12}^{2}$ different ways to choose any two vertices. For each pair of vertices forming a segment $A_{i} A_{j}$, there are two squares with $A_{i} A_{j}$ as a side, and one square with $A_{i} A_{j}$ as a diagonal. Thus, each segment $A_{i} A_{j}$ corresponds to 3 squares, but the three squares $A_{1} A_{4} A_{7} A_{10}, A_{2} A_{5} A_{8} A_{11}, A_{3} A_{6} A_{9} A_{12}$ are each counted 6 times. Therefore, the total number of squares is $3 \mathrm{C}_{12}^{2}-3 \times(6-1)=3 \times 66-15=183$.