Difference between revisions of "2012 AMC 10A Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | Cutting the square in half will bisect one pair of sides while the other side will remain unchanged. Thus, the new square is <math>\frac{8}{2} | + | Cutting the square in half will bisect one pair of sides while the other side will remain unchanged. Thus, the new square is <math>\frac{8}{2}</math> by <math>8</math>, or <math>\boxed{\textbf{(E)}\ 4\ \text{by}\ 8}</math>. |
== See Also == | == See Also == | ||
Revision as of 13:33, 13 July 2021
Problem
A square with side length 8 is cut in half, creating two congruent rectangles. What are the dimensions of one of these rectangles?
Solution
Cutting the square in half will bisect one pair of sides while the other side will remain unchanged. Thus, the new square is 
by 
, or 
.
See Also
| 2012 AMC 10A (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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
