1. Since we're talking about a square, we'll have 4 sides that measure the same. This way, the perimeter is:
![\begin{gathered} 4s=4AB=4\cdot9 \\ \rightarrow36 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/tj90c4hjbcnsdq1nm7mxk4jtiyavru4g7h.png)
2. Notice that the transformation contracts distance by a factor of 3 (all distances are reduce to a third part).
Consequently, sides will be reduce to their third part too. This way, the new perimeter would be:
![\begin{gathered} 4\cdot((1)/(3)\cdot9) \\ \rightarrow12 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/rwo8w9sz2n8d0tifuoudvpato7x11iw5r0.png)
3. The area of the square is the lenght of its sides squared. This way, the area of ABCD is:
![9^2=81](https://img.qammunity.org/2023/formulas/mathematics/high-school/ekxlq48b355q812tf2vhmwnd5vlmi3s8zg.png)
4. Knowing that the new sides measure 3, the new area would be:
![3^2=9](https://img.qammunity.org/2023/formulas/mathematics/high-school/4sid5au9qtxmz9me94k68s2539uyhme2gp.png)