Difference between revisions of "2004 AMC 8 Problems/Problem 14"
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==Solution 2== | ==Solution 2== | ||
| − | The figure contains <math>21</math> interior points and <math>5</math> boundary points. Using [[Pick's Theorem]], the area is <cmath>21+5 | + | 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 12:42, 29 January 2023
Contents
Problem
What is the area enclosed by the geoboard quadrilateral below?
![[asy] unitsize(3mm); defaultpen(linewidth(.8pt)); dotfactor=2; for(int a=0; a<=10; ++a) for(int b=0; b<=10; ++b) { dot((a,b)); }; draw((4,0)--(0,5)--(3,4)--(10,10)--cycle); [/asy]](http://latex.artofproblemsolving.com/7/6/0/76034624e12f53431e5b0c0d4dd5abb475dc1fbb.png)
Solution
Let the bottom left corner be 
. The points would then be 
and 
. Applying the Shoelace Theorem,
![\[\text{Area} = \frac12 \begin{vmatrix} 4 & 0 \\ 0 & 5 \\ 3 & 4 \\ 10 & 10 \end{vmatrix} = \frac12 |(20+30)-(15+40+40)| = \frac12 |50-95| = \boxed{\textbf{(C)}\ 22\frac12}\]](http://latex.artofproblemsolving.com/2/f/f/2ff044d57d10f3e743aa129b0d6e4b0450bc5d95.png)
Solution 2
The figure contains 
interior points and 
boundary points. Using Pick's Theorem, the area is ![\[21+\frac{5}{2}-1=\boxed{\textbf{(C)}\ 22\frac12}\]](http://latex.artofproblemsolving.com/f/5/e/f5e1586abbf61db3a45123dd45f1f2b1c1d8c6bd.png)
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. 
