For Part A, in creating the square, you did it correctly.
Now for Part B, you can create a right triangle by adding a point at -4, -3, and we'll call it, say, T.
Then you can use the Pythagorean's Theorem to find the length of one side of the square.
The Pythagorean's Theorem is
![a^{2} + b^{2} = c^{2}](https://tex.z-dn.net/?f=%20a%5E%7B2%7D%20%2B%20%20b%5E%7B2%7D%20%3D%20c%5E%7B2%7D%20)
A would be the distance between P and T, and B would be the distance between Q and T. So then, a would be 7, and B would be 2.
To fill it in, it would be
![7^{2} + 2^{2} = c^{2}](https://tex.z-dn.net/?f=7%5E%7B2%7D%20%2B%20%202%5E%7B2%7D%20%3D%20c%5E%7B2%7D%20)
or 49 + 4 =
![c^{2}](https://tex.z-dn.net/?f=%20c%5E%7B2%7D%20)
or 53 =
![c^{2}](https://tex.z-dn.net/?f=c%5E%7B2%7D%20)
or
![\sqrt{53}](https://tex.z-dn.net/?f=%20%5Csqrt%7B53%7D%20)
= x
Which is equal to about 7.28
Now multiply that by 4 to get the perimeter, or 29.12
So the perimeter would be about 29.12