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== |
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]], | 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]], | ||
Line 26: | Line 26: | ||
==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+\frac{5}{2}-1=\boxed{\textbf{(C)}\ 22\frac12}</cmath> | 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> | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | <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 = [DGEF] + [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] | ||
==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 09:10, 19 January 2024
Contents
[hide]Problem
What is the area enclosed by the geoboard quadrilateral below?
Solution 1
Let the bottom left corner be . The points would then be and . Applying the Shoelace Theorem,
Solution 2
The figure contains interior points and boundary points. Using Pick's Theorem, the area is
Solution 3
Divide the shape up as above.
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.