Difference between revisions of "1951 AHSME Problems/Problem 2"
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Let <math>w</math> be the width. Then <math>l = 2w</math>, and the perimeter is <math>x = 2(2w)+2w = 6w \implies w = \frac{x}6</math>. The area is <math>wl = w(2w) = 2w^2 = 2\left(\frac{x^2}{36}\right) = \frac{x^2}{18}</math>, so the answer is <math>\mathrm{D}</math>. | Let <math>w</math> be the width. Then <math>l = 2w</math>, and the perimeter is <math>x = 2(2w)+2w = 6w \implies w = \frac{x}6</math>. The area is <math>wl = w(2w) = 2w^2 = 2\left(\frac{x^2}{36}\right) = \frac{x^2}{18}</math>, so the answer is <math>\mathrm{D}</math>. | ||
| − | == See | + | == See Also == |
| − | {{AHSME box|year=1951|num-b=1|num-a=3}} | + | {{AHSME 50p box|year=1951|num-b=1|num-a=3}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
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 
yards of fencing. The area in terms of is:
Solution
Let 
be the width. Then 
, and the perimeter is 
. The area is 
, so the answer is 
.
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
