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Nastasia [14]
3 years ago
6

What is the length of line segment "BC"?​

Mathematics
1 answer:
melisa1 [442]3 years ago
8 0
3x-5=2x
2x-3x=-x
-5=-x
-5/-1=5
-x/-1=x
5=x
2×5=10
BC is 10 units
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What is the area of the piece of red paper after the hole for
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Answer:

     \large\boxed{\large\boxed{47unit^2}}

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>

<em />

<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>

<em />

<em>What is the area of the piece of red paper after the hole for the photograph has been cut?</em>

<em />

<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>
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<em />

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:

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The area of this square is    (8unit)^2=64unit^2

<u>2. Area of the small square</u>

Vertices:

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  • <em>(- 2, 2) </em>
  • <em>(2, 1)</em>
  • <em>(1, -3)</em>

<em />

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}

The area is:

          (\sqrt{17}unit)^2}=17unit^2

<u>3. Area of the piece of red paper after the holw for the photograph has been cut:</u>

<u />

Find the difference:

<u />

         64unit^2-17unit^2=47unit^2

<em />

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