A rectangular fish pond is 7ft longer than it is wide. A wooden walk is 1ft wide and is placed around the pond. The area covered by the pond and the walk is 58ft (squared) greater than the area covered by the pond alone. What are the dimensions of the pond?
2007-11-19 14:57:59 · 2 answers · asked by daniella26 2 in Science & Mathematics ➔ Mathematics
Thanks so much for the help!!
2007-11-19 15:29:52 · update #1
Draw two rectangles. On the smaller one, label the width "w" and the length (w+7) from the information provided in the first sentence of the problem.
On the larger rectangle, add the width of the walk, 1 foot on each side of the pond. Therefore, the width is (w+2) and the length is (w+9).
Area = Length * width
For the pond alone: A = (w+7) * w = w^2 + 7w
For pond & walk: A+58 = (w+9) * (w+2) = w^2 + 11w + 18
We have a system of two equations and two unknowns. I will solve by subtracting the first equation from the second equation.
A + 58 = w^2 + 11w + 18
A = w^2 + 7w
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58 = 4w + 18
w = 10 feet
length = w+7 = 17 feet
2007-11-19 15:18:55 · answer #1 · answered by Katy D 4 · 1⤊ 0⤋
Since the walk is 1 foot wide and occupies 58 square feet, the perimeter of the pond is 54 feet (deduct 4 corner squares).
l = 7+w
2l + 2w = 54
14 +4w = 54
w = 10
l = 17
2007-11-19 15:10:21 · answer #2 · answered by Computer Guy 7 · 1⤊ 0⤋