Question

What is the perimeter of a polygon with vertices at (-3, 1), (5, 1), (-3, 4), (5, 4)?

Answer 1

The perimeter of the polygon is 22.

Explanation:

Let
`A(x,y) =(-3,1)`,
`B(x,y) = (5,1)`,
`C(x,y) = (-3, 4)` and
`D(x,y) =(5,4)` the vertices of the polygon. From Geometry we know that number of side of polygons equals their number of vertices, then we notice that a quadrilateral is here. In addition, we conclude that such quadrilateral is a rectangle:

1)
`x_(A) = x_(C)`

2)
`x_(B) = x_(D)`

3)
`y_(A) = y_(B)`

4)
`y_(C) = y_(D)`

Hence, we find that
`AB = CD` and
`AC = BD`.

Then we find all lengths of the rectangle by Pythagorean Theorem:

`AB = \sqrt{[5-(-3)]^(2)+(1-1)^(2)}`

`AB =8`

`AC = \sqrt{[(-3)-(-3)]^(2)+(4-1)^(2)}`

`AC = 3`

By Geometry, we conclude that
`CD = 8` and
`BD = 3`.

Then, the perimeter of the rectangle (
`p`), dimensionless, is defined by the following formula:

`p = 2\cdot (s+l)` (Eq. 1)

Where:

`s` - Shortest length, dimensionless.

`l` - Longest length, dimensionless.

If we know that
`s = 3` and
`l = 8`, then the perimeter of the rectangle is:

`p = 2\cdot (3+8)`

`p = 22`

The perimeter of the polygon is 22.