Difference between revisions of "2011 AMC 10B Problems/Problem 14"
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== Solution 2 (Quick) == | == Solution 2 (Quick) == | ||
− | We see the answer choices or the perimeter are integers. Therefore, the sides of the rectangle are most likely integers that satisfy <math>a^2+b^2=25^2</math>. In other words, <math>(a,b,25)</math> is a set of Pythagorean triples. | + | We see the answer choices or the perimeter are integers. Therefore, the sides of the rectangle are most likely integers that satisfy <math>a^2+b^2=25^2</math>. In other words, <math>(a,b,25)</math> is a set of Pythagorean triples. Guessing and checking, we have <math>(7,24,25)</math> as the triplet, as the area is <math>7 \cdot 24 = 168</math> as requested. Therefore, the perimeter is <math>2(7+24)=\boxed{\textbf{(C)} 62}</math>. |
== See Also== | == See Also== |
Latest revision as of 10:05, 13 July 2021
Problem
A rectangular parking lot has a diagonal of meters and an area of square meters. In meters, what is the perimeter of the parking lot?
Solution
Let the sides of the rectangular parking lot be and . Then and . Add the two equations together, then factor. The perimeter of a rectangle is
Solution 2 (Quick)
We see the answer choices or the perimeter are integers. Therefore, the sides of the rectangle are most likely integers that satisfy . In other words, is a set of Pythagorean triples. Guessing and checking, we have as the triplet, as the area is as requested. Therefore, the perimeter is .
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 | ||
All AMC 10 Problems and Solutions |
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