Question

A contractor is building a new house. She creates a map with the coordinates A(6,12) , B(6,−6) , C(−4,−6) , and D(−4,12) . Each unit on the coordinate plane represents 1 yard. What is the perimeter of the outside of the house?(1 point)

Answer 1

To find the perimeter of the house, we need to calculate the lengths of all the sides of the quadrilateral formed by the points A, B, C, and D.

The coordinates of the points are:
- A(6, 12)
- B(6, -6)
- C(-4, -6)
- D(-4, 12)

Let's calculate the lengths of each side:

1. Length of AB:
\[
AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} = \sqrt{(6 - 6)^2 + (-6 - 12)^2} = \sqrt{0 + (-18)^2} = \sqrt{324} = 18 \text{ yards}
\]

2. Length of BC:
\[
BC = \sqrt{(x_C - x_B)^2 + (y_C - y_B)^2} = \sqrt{(-4 - 6)^2 + (-6 - (-6))^2} = \sqrt{(-10)^2 + 0^2} = \sqrt{100} = 10 \text{ yards}
\]

3. Length of CD:
\[
CD = \sqrt{(x_D - x_C)^2 + (y_D - y_C)^2} = \sqrt{(-4 - (-4))^2 + (12 - (-6))^2} = \sqrt{0 + (18)^2} = \sqrt{324} = 18 \text{ yards}
\]

4. Length of DA:
\[
DA = \sqrt{(x_A - x_D)^2 + (y_A - y_D)^2} = \sqrt{(6 - (-4))^2 + (12 - 12)^2} = \sqrt{(10)^2 + 0^2} = \sqrt{100} = 10 \text{ yards}
\]

Now, we can calculate the perimeter by summing up the lengths of all sides:
\[
\text{Perimeter} = AB + BC + CD + DA = 18 + 10 + 18 + 10 = 56 \text{ yards}
\]

Thus, the perimeter of the outside of the house is **56 yards**.