1951 AHSME Problems/Problem 2

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)
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

Page

Toolbox

Search

1951 AHSME Problems/Problem 2

Problem

Solution

See Also