Difference between revisions of "1951 AHSME Problems/Problem 2"

Latest revision as of 11:19, 5 July 2013

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:

$(\mathrm{A})\ \frac{x^2}2 \qquad (\mathrm{B})\ 2x^2 \qquad (\mathrm{C})\ \frac{2x^2}9 \qquad (\mathrm{D})\ \frac{x^2}{18} \qquad (\mathrm{E})\ \frac{x^2}{72}$

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 $\mathrm{D}$ .

1951 AHSC (Problems • Answer Key • Resources)
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Difference between revisions of "1951 AHSME Problems/Problem 2"

Latest revision as of 11:19, 5 July 2013

Problem

Solution

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