Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2011 AMC 10B Problems/Problem 14: Difference between revisions

← Older editNewer edit →
Gina (talk | contribs)
783 edits
mNo edit summary
Nathan wailes (talk | contribs)
2,960 edits
No edit summary
Line 18: Line 18:


{{AMC10 box|year=2011|ab=B|num-b=13|num-a=15}}
{{AMC10 box|year=2011|ab=B|num-b=13|num-a=15}}
{{MAA Notice}}

Revision as of 12:12, 4 July 2013

Problem

A rectangular parking lot has a diagonal of $25$ meters and an area of $168$ square meters. In meters, what is the perimeter of the parking lot?

$\textbf{(A)}\ 52 \qquad\textbf{(B)}\ 58 \qquad\textbf{(C)}\ 62 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70$

Solution

Let the sides of the rectangular parking lot be $a$ and $b$. Then $a^2 + b^2 = 625$ and $ab = 168$. Add the two equations together, then factor. \begin{align*} a^2 + 2ab + b^2 &= 625 + 168 \times 2\\ (a + b)^2 &= 961\\ a + b &= 31 \end{align*} The perimeter of a rectangle is $2 (a + b) = 2 (31) = \boxed{\textbf{(C)} 62}$

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_14&oldid=54662"