Question

On a coordinate plane, 2 parallelograms are shown. Parallelogram 1 has points (0, 2), (2, 6), (6, 4), and (4, 0). Parallelogram 2 has points (2, 0), (4, negative 6), (2, negative 8), and (0, negative 2). How do the areas of the parallelograms compare? The area of parallelogram 1 is 4 square units greater than the area of parallelogram 2. The area of parallelogram 1 is 2 square units greater than the area of parallelogram 2. The area of parallelogram 1 is equal to the area of parallelogram 2. The area of parallelogram 1 is 2 square units less than the area of parallelogram 2

Answer 1

The area of parallelogram 1 is 4 square units greater than the area of parallelogram 2

How to compare the areas?

The coordinates are given as:

Parallelogram 1: (0, 2), (2, 6), (6, 4), and (4, 0).

Parallelogram 2: (2, 0), (4, -6), (2, -8), and (0, -2)

Next, calculate the base and the height

For parallelogram 1, we have:

Base = (0, 2) and (2, 6)

Height = (6, 4), and (4, 0).

So, we have:

$Base = \sqrt{(0 - 2)^2 + (2 -6)^2} = \sqrt{20}$

$Height = \sqrt{(6 - 4)^2 + (4 -0)^2} = \sqrt{20}$

The area is:

Area = Base * Height

Area = √20 * √20

Area = 20

For parallelogram 2, we have:

Base = (0, -2) and (4, -6)

Height = (4, -6) and (2, -8)

So, we have:

$Base = \sqrt{(0 -4)^2 + (-2 +6)^2} = \sqrt{32}$

$Height = \sqrt{(4 - 2)^2 + (-6 +8)^2} = \sqrt{8}$

The area is:

Area = Base * Height

Area = √32 * √8

Area = 16

By comparing both areas, we have:

20 is greater than 16 by 4

This means that the area of parallelogram 1 is greater by 4 units

Read more about parallelogram at:

brainly.com/question/3050890

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