Difference between revisions of "1952 AHSME Problems/Problem 46"
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== Solution == | == Solution == | ||
| − | <math>\fbox{}</math> | + | Let the larger side of the rectangle be y and the smaller side of the rectangle be x. Then, by pythagorean theorem, the diagonal of the rectangle has length <math>\sqrt{x^{2}+y^{2}}</math>. |
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| + | By the definition of the problem, the area of the new rectangle is <math>(\sqrt{x^{2}+y^{2}}+y)(\sqrt{x^{2}+y^{2}}-y)</math> | ||
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| + | Expanding this gives the area of the new rectangle to be <math>x^{2}</math>, or | ||
| + | <math>\fbox{C}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 17:59, 20 April 2020
Problem
The base of a new rectangle equals the sum of the diagonal and the greater side of a given rectangle, while the altitude of the new rectangle equals the difference of the diagonal and the greater side of the given rectangle. The area of the new rectangle is:
Solution
Let the larger side of the rectangle be y and the smaller side of the rectangle be x. Then, by pythagorean theorem, the diagonal of the rectangle has length 
.
By the definition of the problem, the area of the new rectangle is 
Expanding this gives the area of the new rectangle to be 
, or
.
See also
| 1952 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 45 |
Followed by Problem 47 | |
| 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. 
