Given:
The vertices of a rectangle are (6, 3), (−6, 3), (−6, −1), and (6, −1).
To find:
The perimeter of the rectangle.
Solution:
Distance formula:
![d=\sqrt{(x_2-x_1)^2-(y_2-y_1)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28x_2-x_1%29%5E2-%28y_2-y_1%29%5E2%7D)
Let the vertices of a rectangle are A(6, 3), B(−6, 3), C(−6, −1), and D(6, −1).
Using the distance formula, we get
![AB=\sqrt{(-6-6)^2-(3-3)^2}](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B%28-6-6%29%5E2-%283-3%29%5E2%7D)
![AB=\sqrt{(-12)^2-(0)^2}](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B%28-12%29%5E2-%280%29%5E2%7D)
![AB=\sqrt{144}](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B144%7D)
![AB=12](https://tex.z-dn.net/?f=AB%3D12)
Similarly,
![BC=\sqrt{\left(-6-\left(-6\right)\right)^2+\left(-1-3\right)^2}=4](https://tex.z-dn.net/?f=BC%3D%5Csqrt%7B%5Cleft%28-6-%5Cleft%28-6%5Cright%29%5Cright%29%5E2%2B%5Cleft%28-1-3%5Cright%29%5E2%7D%3D4)
![CD=\sqrt{\left(6-\left(-6\right)\right)^2+\left(-1-\left(-1\right)\right)^2}=12](https://tex.z-dn.net/?f=CD%3D%5Csqrt%7B%5Cleft%286-%5Cleft%28-6%5Cright%29%5Cright%29%5E2%2B%5Cleft%28-1-%5Cleft%28-1%5Cright%29%5Cright%29%5E2%7D%3D12)
![AD\sqrt{\left(6-6\right)^2+\left(-1-3\right)^2}=4](https://tex.z-dn.net/?f=AD%5Csqrt%7B%5Cleft%286-6%5Cright%29%5E2%2B%5Cleft%28-1-3%5Cright%29%5E2%7D%3D4)
Now, the perimeter of the rectangle is:
![Perimeter=AB+BC+CD+AD](https://tex.z-dn.net/?f=Perimeter%3DAB%2BBC%2BCD%2BAD)
![Perimeter=12+4+12+4](https://tex.z-dn.net/?f=Perimeter%3D12%2B4%2B12%2B4)
![Perimeter=32](https://tex.z-dn.net/?f=Perimeter%3D32)
Therefore, the perimeter of the rectangle is 32 units.