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>. |
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| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/ib6LCvSLo7M | ||
| + | |||
| + | ~Education, the Study of Everything | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2012|ab=A|num-b=1|num-a=3}} | {{AMC10 box|year=2012|ab=A|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:57, 1 July 2023
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 
.
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
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. 
