Difference between revisions of "2002 AMC 8 Problems/Problem 15"
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label("$E$", (4*6+2, 0), S);</asy> | label("$E$", (4*6+2, 0), S);</asy> | ||
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<math> \textbf{(A)}\text{A}\qquad\textbf{(B)}\ \text{B}\qquad\textbf{(C)}\ \text{C}\qquad\textbf{(D)}\ \text{D}\qquad\textbf{(E)}\ \text{E} </math> | <math> \textbf{(A)}\text{A}\qquad\textbf{(B)}\ \text{B}\qquad\textbf{(C)}\ \text{C}\qquad\textbf{(D)}\ \text{D}\qquad\textbf{(E)}\ \text{E} </math> | ||
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2002|num-b=14|num-a=16}} | {{AMC8 box|year=2002|num-b=14|num-a=16}} | ||
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Latest revision as of 03:47, 25 November 2019
Problem
Which of the following polygons has the largest area?
![[asy] size(330); int i,j,k; for(i=0;i<5; i=i+1) { for(j=0;j<5;j=j+1) { for(k=0;k<5;k=k+1) { dot((6i+j, k)); }}} draw((0,0)--(4,0)--(3,1)--(3,3)--(2,3)--(2,1)--(1,1)--cycle); draw(shift(6,0)*((0,0)--(4,0)--(4,1)--(3,1)--(3,2)--(2,1)--(1,1)--(0,2)--cycle)); draw(shift(12,0)*((0,1)--(1,0)--(3,2)--(3,3)--(1,1)--(1,3)--(0,4)--cycle)); draw(shift(18,0)*((0,1)--(2,1)--(3,0)--(3,3)--(2,2)--(1,3)--(1,2)--(0,2)--cycle)); draw(shift(24,0)*((1,0)--(2,1)--(2,3)--(3,2)--(3,4)--(0,4)--(1,3)--cycle)); label("$A$", (0*6+2, 0), S); label("$B$", (1*6+2, 0), S); label("$C$", (2*6+2, 0), S); label("$D$", (3*6+2, 0), S); label("$E$", (4*6+2, 0), S);[/asy]](http://latex.artofproblemsolving.com/6/b/8/6b83ac33b3e3e61ab2ca2c1c80dcbd8766f52066.png)
Solution
Each polygon can be partitioned into unit squares and right triangles with sidelength 
. Count the number of boxes enclosed by each polygon, with the unit square being , and the triangle being being 
. A has 5, B has 5, C has 5, D has 4.5, and E has 5.5. Therefore, the polygon with the largest area is 
.
See Also
| 2002 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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. 
