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1951 AHSME Problems/Problem 2

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Problem

A rectangular field is half as wide as it is long and is completely enclosed by x yards of fencing. The area in terms of x is:

(\mathrm{A})\ \frac{x^2}2 \qquad (\mathrm{B})\ 2x^2 \qquad (\mathrm{C})\ \frac{2x^2}9 \qquad (\mathrm{D})\ \frac{x^2}{18} \qq...

Solution

Let w be the width. Then l = 2w, and the perimeter is x = 2(2w)+2w = 6w \implies w = \frac{x}6. The area is wl = w(2w) = 2w^2 = 2\left(\frac{x^2}{36}\right) = \frac{x^2}{18}, so the answer is \mathrm{D}.

See also

1951 AHSME (Problems)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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