SOLUTION: In the middle of a rectangular room, is a rectangular rug, 13 feet by 15 feet. A strip of hardware floor of uniform width is visible around the rug. If the area of the strip is 60

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Question 935404: In the middle of a rectangular room, is a rectangular rug, 13 feet by 15 feet. A strip of hardware floor of uniform width is visible around the rug. If the area of the strip is 60 square feet, how wide (in feet) is the strip
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
u, size of uniform sptrip;
w, width of rug;
L, length of rug.
Let w=13, and L=15.
b, hardwood border area of uniform width u.
b=60.

Rug area, w%2AL.
Room area, %28w%2B2u%29%28L%2B2u%29.
Hardwood border area, %28w%2B2u%29%28L%2B2u%29-wL.
EQUATION USING BORDER AREA OF THE HARDWOOD: highlight_green%28highlight%28%28w%2B2u%29%28L%2B2u%29-wL=b%29%29


You CAN solve this completely in symbols, and then substitute the known and given values last, and evalualte u.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

In the middle of a rectangular room, is a rectangular rug, 13 feet by 15 feet. A strip of hardware floor of uniform width is visible around the rug. If the area of the strip is 60 square feet, how wide (in feet) is the strip

Let uniform width of hardwood floor, be W
Since shorter side of rug is 13 feet, shorter side of entire room = 13 + 2W
Since longer side of rug is 15 feet, longer side of entire room = 15 + 2W
Since area of entire room, equals area of rug, plus area of hardwood floor, it can be said that:
{13 + 2W)(15 + 2W) = 60 + 15(13)
195+%2B+56W+%2B+4W%5E2+=+60+%2B+195
195+%2B+56W+%2B+4W%5E2+=+255
4W%5E2+%2B+56W+%2B+195+-+255+=+0
4W%5E2+%2B+56W+-+60+=+0
4%28W%5E2+%2B+14W+-+15%29+=+4%280%29
W%5E2+%2B+14W+-+15+=+0
(W - 1)(W + 15) = 0
W, or width of strip = highlight_green%281%29 foot OR W = - 15 (ignore)