Question

The pentagon ABCDE is graphed on a coordinate plane with vertices A(0, 0), B(3, 5), C(3, 8), D(8, 5), and E(8, 0). What is the pentagon's perimeter?

Answer 1

Answer:

AB=3-0,5-0

AB=3^2+5^2

AB=9+25 square root

AB=5.83

BC=3-3,8-5

BC=0,3

BC=3^2 BC=9 square root

BC=3

CD=8-3,5-8

CD=5,-3

CD=5^2+-3^2

CD=25+9square root

CD=√34

CD=5.83

DE=8-8,0-5

DE=0+-5^2

DE=√25

DE=5

EA=8-0,0-0

EA=8^2

EA=√64

EA=8

perimeter=8+5+5.83+5.83+3= 27.66 units

=27.66 units

Answer 2

The perimeter of the pentagon is $\sqrt{73}+21\,\,\,{\rm units}$.

Given:

A pentagon ABCDE is graphed on a coordinate plane with vertices A(0, 0), B(3, 5), C(3, 8), D(8, 5), and E(8, 0).

To Find:

The pentagon's perimeter.

Solution:

The perimeter of the pentagon is the sum of all of its sides.

To find the length of all of its sides, the distance between two vertices is needed to be computed which will be computed using the distance formula given by $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.

The length of "AC" is

$AC=\sqrt{(3-0)^2+(8-0)^2}\\=\sqrt{9+64}\\ =\sqrt{73}\,\,{\rm units}$

The length of "BC" is

$BC=\sqrt{(3-3)^2+(5-8)^2}\\=\sqrt{0+9}\\ =\sqrt{9}\\=3\,\,{\rm units}$

The length of "BD" is

$BC=\sqrt{(8-3)^2+(5-5)^2}\\=\sqrt{5^2+0}\\=5\,\,\,{\rm units}$

The length of "ED" is

$ED=\sqrt{(8-8)^2+(0-5)^2}\\=\sqrt{0+(5)^2}\\=\sqrt{(5)^2}\\=5\,\,\,{\rm units}$

The length of "AE" is

$AE=\sqrt{(8-0)^2+(0-0)^2}\\=\sqrt{(8)^2}\\=8\,\,{\rm units}$

Add the length of AC, BC, BD, DE, and AE.

$\sqrt{73}+3+5+5+8=\sqrt{73}+21\,\,\,{\rm units}$

Thus, the perimeter of the pentagon is $\sqrt{73}+21\,\,\,{\rm units}$.