Difference between revisions of "2018 AMC 8 Problems/Problem 4"
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draw((1,3)--(2,4)--(2,5)--(3,6)--(4,5)--(5,5)--(6,4)--(5,3)--(5,2)--(4,1)--(3,2)--(2,2)--cycle,black+2bp); | draw((1,3)--(2,4)--(2,5)--(3,6)--(4,5)--(5,5)--(6,4)--(5,3)--(5,2)--(4,1)--(3,2)--(2,2)--cycle,black+2bp); | ||
</asy> | </asy> | ||
| − | + | Solution | |
| + | We count <math>3*3=9</math> unit squares in the middle, and <math>4</math> unit squares on the side after combining the triangles. Thus, the answer is <math>9+4=\boxed{13}, \textbf{(A)}</math> | ||
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 12.5 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13.5 \qquad \textbf{(E) } 14</math> | <math>\textbf{(A) } 12 \qquad \textbf{(B) } 12.5 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13.5 \qquad \textbf{(E) } 14</math> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2018|num-b=3|num-a=5}} | {{AMC8 box|year=2018|num-b=3|num-a=5}} | ||
Revision as of 13:24, 21 November 2018
Problem 4
The twelve-sided figure shown has been drawn on 
graph paper. What is the area of the figure in 
?
![[asy] unitsize(8mm); for (int i=0; i<7; ++i) { draw((i,0)--(i,7),gray); draw((0,i+1)--(7,i+1),gray); } draw((1,3)--(2,4)--(2,5)--(3,6)--(4,5)--(5,5)--(6,4)--(5,3)--(5,2)--(4,1)--(3,2)--(2,2)--cycle,black+2bp); [/asy]](http://latex.artofproblemsolving.com/f/4/d/f4d92072ddf7f5314e710dc9641eb84c3ea32879.png)
Solution
We count 
unit squares in the middle, and 
unit squares on the side after combining the triangles. Thus, the answer is 
See Also
| 2018 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 | ||
