Difference between revisions of "2003 AMC 12A Problems/Problem 14"
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Either way, our answer is <math>\boxed{D}</math>. | Either way, our answer is <math>\boxed{D}</math>. | ||
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| + | == See Also == | ||
| + | {{AMC12 box|year=2003|ab=A|num-b=13|num-a=15}} | ||
Revision as of 23:38, 27 April 2011
Since the area of square ABCD is 16, the side length must be 4. Thus, the side length of triangle AKB is 4, and the height of AKB, and thus DMC, is 
.
The diagonal of the square KNMC will then be 
. From here there are 2 ways to proceed:
First: Since the diagonal is , the side length is 
, and the area is thus 
.
Second: Since a square is a rhombus, the area of the square is 
, where 
and 
are the diagonals of the rhombus. Since the diagonal is , the area is 
.
Either way, our answer is 
.
See Also
| 2003 AMC 12A (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 12 Problems and Solutions | |
