Question

Maya is using a coordinate plane to create a map of her yard drawn to scale. Each unit

on the coordinate plane represents 1 meter. Her yard forms a quadrilateral with vertices
A(5, -3), B(10, -3), C(12, -11), and D(7, -11). Maya would like to fence in her entire
yard. Fencing costs $200 per meter. How much will it cost Maya to fence in her yard?

Answer 1

$5,298.48

Explanation:

Given vertices:

• A = (5, -3)
• B = (10, -3)
• C = (12, -11)
• D = (7, -11)

As points A and B, and points C and D, have the same y-coordinates, AB is parallel to CD.

As there is a difference of 5 units between points A and B, and C and D, AB = CD.

As AB is parallel to CD and AB = CD ⇒ AD = BD and AD is parallel to BD.

Therefore, the quadrilateral ABCD is a parallelogram (see attachment).

To find the perimeter of ABCD, calculate the side lengths of AB and AD using the Distance Formula.

`\boxed{\begin{minipage}{4.7 cm}\underline{Distance Formula}\\\\$d=√((x_2-x_1)^2+(y_2-y_1)^2)$\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are the two endpoints.\end{minipage}}`

`\begin{aligned}AB & =√((x_B-x_A)^2+(y_B-y_A)^2)\\& =√((10-5)^2+(-3-(-3))^2)\\& =√((5)^2+(0)^2)\\& =√(25)\\& =5\:\:\sf units\end{aligned}`

`\begin{aligned}AD & =√((x_D-x_A)^2+(y_D-y_A)^2)\\& =√((7-5)^2+(-11-(-3))^2)\\& =√((2)^2+(-8)^2)\\& =√(4+64)\\& =√(68)\:\:\sf units\end{aligned}`

The perimeter of a two-dimensional shape is the distance all the way around the outside.

`\begin{aligned}\textsf{Perimeter of $ABCD$} & = AB + CD + AD + BC\\& = 2\:AB + 2\:AD\\& = 2 \cdot 5+2 \cdot √(68)\\& = 10+2 √(68)\\& = 10+2√(4 \cdot 17)\\& = 10+2√(4)√(17)\\& = 10+2 \cdot2√(17)\\& = 10+4√(17)\:\:\sf units\end{aligned}`

As 1 unit = 1 meter, the perimeter of the fence is (10 + 4√17) m.

If fencing costs $200 per meter:

`\begin{aligned}\sf \implies Cost \:of\:fencing& = 200(10 + 4 √(17))\\& = 2000+800 √(17)\\& = 5298.4845\\ & = \$5298.48\end{aligned}`

Therefore, it will cost Maya $5,298.48 to fence her yard.