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iogann1982 [59]
3 years ago
10

Lupe’s Construction is building a rectangular house that is 16 feet longer than it is wide. They will be putting a 3-foot wide s

idewalk all the way around the house. There will be a protective metal strip around the outside edge of the sidewalk. If the house will have a perimeter of 136 feet, how will the four pieces of the protective strip need to be
Mathematics
1 answer:
mr Goodwill [35]3 years ago
5 0

Answer:

The length of the four pieces of protective strip is 96 feet

Step-by-step explanation:

Let the w be the width of the rectangular house and L is length. Since L = w + 16, its perimeter P = 2(w + L)

= 2(w + w + 16)

= 2(2w + 16)

Since the perimeter of the house P = 136 feet,

136 = 2(2w + 16)

dividing through by 2, we have

136/2 = (2w + 16)

68 = (2w + 16)

collecting like terms, we have

68 - 16 = 2w

52 = 2w

dividing through by 2, we have

w = 52/2

w = 26 feet.

Now, since the side walk is 3 foot wide all around, the width of the house plus sidewalk = w + 3 and the length of the house plus sidewalk = L + 3 = w + 16 + 3 = w + 19.

So, the perimeter of the house plus sidewalk which is the perimeter of the four pieces of protective strip is P' = 2(w + 3 + w + 19)

= 2(w + 22)

= 2(26 + 22)

= 2(48)

= 96 feet

So, the length of the four pieces of protective strip is 96 feet.

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