SOLUTION: The length of the Serumore Hotel's lobby exceeds its width by 4 feet. A rug covers the floor of the lobby except for a border 2 feet wide all around. If the area of this border o

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Question 317123: The length of the Serumore Hotel's lobby exceeds its width by 4 feet. A rug covers the floor of the lobby except for a border 2 feet wide all around. If the area of this border of exposed floor is 68 square feet, what is the area of the rug?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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The length of the Serumore Hotel's lobby exceeds its width by 4 feet.
A rug covers the floor of the lobby except for a border 2 feet wide all around.
If the area of this border of exposed floor is 68 square feet, what is the area of the rug?
:
Let x = the width of the rug
then
(x+4) = the length of the rug
and
x(x+4) = x^2 + 4x = the area of the rug
:
(x+4) = overall width
(x+4) + 4 = (x+8) = overall length
and
(x+4)*(x+8) = x^2 + 12x + 32 =overall area
:
Overall area - rug area = 68
x^2 + 12x + 32 - (x^2 + 4x) = 68
x^2 + 12x + 32 - x^2 - 4x = 68
x^2 - x^2 + 12x - 4x = 68 - 32
8x = 36
x = 36%2F8
x = 4.5
Find the area of the rug
A = x(x+4)
A = 4.5(4.5+4)
A = 4.5(8.5)
A = 38.25 sq/ft
:
:
Check:
Overall area = 4.5^2 + 12(4.5) + 32 = 106.25
106.25 - 38.25 = 68 confirms our solution