Difference between revisions of "2004 AMC 8 Problems/Problem 14"
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<math>\textbf{(A)}\ 15\qquad \textbf{(B)}\ 18\frac12 \qquad \textbf{(C)}\ 22\frac12 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 41</math> | <math>\textbf{(A)}\ 15\qquad \textbf{(B)}\ 18\frac12 \qquad \textbf{(C)}\ 22\frac12 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 41</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | + | ||
| − | + | <asy> | |
| + | unitsize(5mm); | ||
| + | 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); | ||
| + | draw((0,0)--(10,0)--(10,10)--(3,4)--(0,5)--cycle); | ||
| + | draw((10,4)--(0,4)--cycle); | ||
| + | |||
| + | dot("$A$", (0,5), W); | ||
| + | dot("$B$", (3,4), N); | ||
| + | dot("$C$", (10,10), NE); | ||
| + | dot("$D$", (0,4), W); | ||
| + | dot("$E$", (10,4), E); | ||
| + | dot("$F$", (0,0), SW); | ||
| + | dot("$G$", (10,0), SE); | ||
| + | dot("$H$", (4,0), S); | ||
| + | </asy> | ||
| + | |||
| + | Divide the shape up as above. | ||
| + | <cmath>Area = [DEGF] + [ABD] + [BCE] - [AFH] - [CGH] = 4 \cdot 10 + \frac12 \cdot 1 \cdot 3 + \frac12 \cdot 7 \cdot 6 - \frac12 \cdot 5 \cdot 4 - \frac12 \cdot 6 \cdot 10 = 40 + \frac32 + 21 - 10 - 30 = \boxed{\textbf{(C)}\ 22\frac12}</cmath> | ||
| + | |||
| + | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | 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]], | ||
<cmath>\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}</cmath> | <cmath>\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}</cmath> | ||
| + | |||
| + | ==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}} | ||
Latest revision as of 09:13, 19 January 2024
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 1
![[asy] unitsize(5mm); 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); draw((0,0)--(10,0)--(10,10)--(3,4)--(0,5)--cycle); draw((10,4)--(0,4)--cycle); dot("$A$", (0,5), W); dot("$B$", (3,4), N); dot("$C$", (10,10), NE); dot("$D$", (0,4), W); dot("$E$", (10,4), E); dot("$F$", (0,0), SW); dot("$G$", (10,0), SE); dot("$H$", (4,0), S); [/asy]](http://latex.artofproblemsolving.com/0/0/5/005f9f59d2c18f343506c5ccec951cf72f6aa5aa.png)
Divide the shape up as above.
![\[Area = [DEGF] + [ABD] + [BCE] - [AFH] - [CGH] = 4 \cdot 10 + \frac12 \cdot 1 \cdot 3 + \frac12 \cdot 7 \cdot 6 - \frac12 \cdot 5 \cdot 4 - \frac12 \cdot 6 \cdot 10 = 40 + \frac32 + 21 - 10 - 30 = \boxed{\textbf{(C)}\ 22\frac12}\]](http://latex.artofproblemsolving.com/4/5/1/451699e68db782227a57cee6b41a225de3111d0b.png)
Solution 2
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 3
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. 
