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.

FOR EDGE

Answer 1

Answer:

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

Explanation:

If you are confused on how to find areas of parallelogram that don't have any parallel sides to the x/y axes, then create a rectangle around the parallelogram and subtract the areas of the triangle from the total area of the rectangle. Example is shown in desmos below:

(Luckily, the first parallelogram happens to be a square so the method above is not necessary)

Area of first parallelogram:

$(\sqrt{2^2 + 4^2} )^2$ = 20

Area of second parallelogram:

(4*8) - (2*2 + 6*2) = 16

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

Answer 2

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.

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