Question

What is the perimeter of a polygon with vertices at (-2, 1).(-2, 4). (2,7), (6,4), and

(6, 1)?
Enter your answer in the box. Do not round any side lengths.
units

Answer 1

24 units

Explanation:

The polygon with 5 vertices A (-2,1), B (-2, 4) , C (2, 7), D (6, 4) and E (6, 1).

where A
`(x_(1) , y_(1)) = (-2, 1)`, B
`(x_(2) , y_(2)) = (-2, 4)`

C
`(x_(3) , y_(3)) = (2, 7)` , D
`(x_(4) , y_(4)) = (6, 4)`

E
`(x_(5) , y_(5)) = (6, 1)`

Now using distance formula:

Length of AB =
`\sqrt{\left(x_(2)- x_(1)\right)^(2)+ \left (y_(2) -y_(1)\right)^(2)}`

=
`\sqrt{(-2+2)^(2) +(4 - 1)^(2) }`

=
`\sqrt{0^(2) +3^(2) }`

=
`\sqrt{3^(2) } = 3 unit`

Length of BC =
`\sqrt{\left(x_(3)- x_(2)\right)^(2)+ \left (y_(3) -y_(2)\right)^(2)}`

=
`\sqrt{(2- (-2))^(2) +(7 - 4)^(2) }`

=
`\sqrt{4^(2) +3^(2) }`

=
`√(16 + 9)`

=
`\sqrt{25^(2) } = 5 unit`

Length of CD =
`\sqrt{\left(x_(4)- x_(3)\right)^(2)+ \left (y_(4) -y_(3)\right)^(2)}`

=
`\sqrt{(6-2)^(2) +(4 - 7)^(2) }`

=
`\sqrt{4^(2) +(-3)^(2) }`

=
`√(16 + 9)`

=
`√(25) = 5 unit`

Length of DE =
`\sqrt{\left(x_(5)- x_(4)\right)^(2)+ \left (y_(5) -y_(4)\right)^(2)}`

=
`\sqrt{(6 - 6)^(2) +(1 - 4)^(2) }`

=
`\sqrt{0^(2) +(-3)^(2) }`

=
`√(0 + 9)`

= 3 unit

Length of EA =
`\sqrt{\left(x_(5)- x_(1)\right)^(2)+ \left (y_(5) -y_(1)\right)^(2)}`

=
`\sqrt{(6 - (- 2))^(2) +(1 - 1)^(2) }`

=
`\sqrt{(6 + 2)^(2) +(0)^(2) }`

=
`\sqrt{8^(2) +(0)^(2) }`

= 8 unit

Perimeter of polygon = Length of (AB + BC + CD + DE + EA)

= (3 + 5 + 5 + 3 + 8)

= 24 units