Answer:
![\large\boxed{\large\boxed{47unit^2}}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B%5Clarge%5Cboxed%7B47unit%5E2%7D%7D)
Explanation:
The complete question is:
<em>A hole the size of a photograph is cut from a red piece of paper to use in a picture frame.</em>
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<em>On a coordinate plane, 2 squares are shown. The photograph has points (-3, -2), (- 2, 2), (2, 1), and (1, -3). The red paper has points (- 4, 4), (4, 4), (4, -4), and (-4, -4).</em>
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<em>What is the area of the piece of red paper after the hole for the photograph has been cut?</em>
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<h2><em>Solution</em></h2>
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The area of the piece of redpaper after the hole has been cut is equal to the area of the big square less the area of the small rectangle.
<u>1. Area of the big rectangle</u>
Vertices:
- <em>(- 4, 4) </em>
- <em>(4, 4)</em>
- <em>(4, -4)</em>
- <em>(-4, -4)</em>
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The side length is the distance between two consecutive vertices:
The two points (-4,4) and (4,4) has the same y-coordinate, thus the length is just the absolute value of the difference of the x-coordinates:
The area of this square is ![(8unit)^2=64unit^2](https://tex.z-dn.net/?f=%288unit%29%5E2%3D64unit%5E2)
<u>2. Area of the small square</u>
Vertices:
- <em>(-3, -2)</em>
- <em>(- 2, 2) </em>
- <em>(2, 1)</em>
- <em>(1, -3)</em>
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To find the side length use the distance formula for two consecutive points, for instance (2,1) and (-2,2):
![d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\ \\d=\sqrt{(2-(-2))^2+(2-1)^2}=\sqrt{4^2+1^2}=\sqrt{17}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28x_1-x_2%29%5E2%2B%28y_1-y_2%29%5E2%7D%5C%5C%20%5C%5Cd%3D%5Csqrt%7B%282-%28-2%29%29%5E2%2B%282-1%29%5E2%7D%3D%5Csqrt%7B4%5E2%2B1%5E2%7D%3D%5Csqrt%7B17%7D)
The area is:
![(\sqrt{17}unit)^2}=17unit^2](https://tex.z-dn.net/?f=%28%5Csqrt%7B17%7Dunit%29%5E2%7D%3D17unit%5E2)
<u>3. Area of the piece of red paper after the holw for the photograph has been cut:</u>
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Find the difference:
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![64unit^2-17unit^2=47unit^2](https://tex.z-dn.net/?f=64unit%5E2-17unit%5E2%3D47unit%5E2)
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