A hole the size of a photograph is cut from a red piece of paper to use in a picture frame. On a coordinate plane, 2 squares are shown. The photograph has points (negative 3, negative 2), (negative 2, 2), (2, 1), and (1, negative 3). The red paper has points (negative 4, 4), (4, 4), (4, negative 4), and (negative 4, negative 4). What is the area of the piece of red paper after the hole for the photograph has been cut? 17 square units 25 square units 39 square units 47 square units

D.

Step-by-step explanation:

First, find the area of the two squares:

The photograph: (-3, -2), (-2, 2), (2, 1), and (1, -3).

The square is diagonally aligned, so we need to find the length between two points in order to find the area.

I will use (1, -3) and (2, 1).

There is a 1 unit length and a 4 unit height. We can use the Pythagorean theorem to find the hypotenuse of the triangle, which is the length of the square's side:

1^2 + 4^2 = x^2

1 + 16 = x^2

x = square root of 17

The side length for the square of the photograph is the square root of 17, so the area of the photograph is 17 units^2.

The red paper has side lengths of 8. The distance between (-4, 4) and (4, 4) is 8 units wide, so we do not need to use the Pythagorean theorem.

Now that we know the area of the red paper and photograph, you can subtract the area to find the red paper with the hole:

64 - 17 = 47 square units.

D

Step-by-step explanation:

Let's plot these points (see attachment).

We need to find the area of the paper and then the area of the photograph and subtract the photograph area from the red paper area.

The area of the big paper is easy; the side lengths of the square are 8 by 8, so the area is 8 * 8 = 64 square units.

Now, to find the area of the photograph, we just need the length of one side because it's a square. We use the distance formula between any two adjacent points; let's do (-2, 2) and (2, 1):

d =

Then the area is (√17)² = 17.

Subtract 17 from 64:

64 - 17 = 47

The area is 47 square units, or D.

answer for #1 is c

answer for #2 is d

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