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The figure below represents a play space that Logan fenced in for his dog.

10 ft
- 12 ft2
Logan is getting a second dog and wants to increase the length of the play
space by 3 feet and the width by 3 feet. What will be the difference in the area,
in square feet, between the original play space and the new play space?
Show your work.

The figure below represents a play space that Logan fenced in for his dog. 10 ft - 12 ft-example-1

1 Answer

7 votes

Answer:

75 square feet

Explanation:

The original dimensions are 10 ft by 12 ft, so the area is ...

(10 f)(12 ft) = 120 ft^2

The new dimensions are 13 ft by 15 ft, so the area is ...

(13 ft)(15 ft) = 195 ft^2

The difference in area between the original and new play spaces is ...

new area - original area = 195 ft^2 - 120 ft^2 = 75 ft^2

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There are some other ways you can figure this.

1) 3 ft added to the 10 ft side adds an area of (3 ft)(10 ft) = 30 ft^2

3 ft added to the 12 ft side adds an area of (3 ft)(12 ft) = 36 ft^2

The corner area added is a square 3 ft on a side, so is (3 ft)^2 = 9 ft^2

The total of these area additions is ...

(30 +36 +9) ft^2 = 75 ft^2

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2) Adding a 3-ft wide area along both sides and the corner makes a "path" whose width is 3 ft and whose centerline length is 11.5 ft + 13.5 ft = 25 ft. Then the area of that "path" is (3 ft)(25 ft) = 75 ft^2.

The figure below represents a play space that Logan fenced in for his dog. 10 ft - 12 ft-example-1
User Josh Hull
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