Difference between revisions of "2004 AMC 8 Problems/Problem 14"
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==Solution 1== | ==Solution 1== | ||
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+ | ==Solution 2== | ||
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+ | Let the bottom left corner be <math>(0,0)</math>. The points would then be <math>(4,0),(0,5),(3,4),</math> and <math>(10,10)</math>. Applying the [[Shoelace Theorem]], | ||
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+ | <cmath>\text{Area} = \frac12 | ||
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+ | ==Solution 3== | ||
+ | The figure contains <math>21</math> interior points and <math>5</math> boundary points. Using [[Pick's Theorem]], the area is <cmath>21+\frac{5}{2}-1=\boxed{\textbf{(C)}\ 22\frac12}</cmath> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2004|num-b=13|num-a=15}} | {{AMC8 box|year=2004|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 09:12, 19 January 2024
Contents
[hide]Problem
What is the area enclosed by the geoboard quadrilateral below?
Solution 1
Divide the shape up as above.
Solution 2
Let the bottom left corner be . The points would then be and . Applying the Shoelace Theorem,
Solution 3
The figure contains interior points and boundary points. Using Pick's Theorem, the area is
See Also
2004 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.