Difference between revisions of "2018 AMC 8 Problems/Problem 4"
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==Solution 2== | ==Solution 2== | ||
| − | We | + | We can see here that there are 9 full squares in the middle. We also see that the triangles that make the corners of the shape have an area half the squares' area. Then we can easily find that each corner has an area of one square and there are 4 corners so we add that to the original 9 squares to get <math>8+5=\boxed{\textbf{(C) } 13}</math> That is how I did it ~avamarora |
==See Also== | ==See Also== | ||
Revision as of 13:33, 1 November 2020
Contents
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 
small triangles each with an area of 
. Thus, the answer is 
Solution 2
We can see here that there are 9 full squares in the middle. We also see that the triangles that make the corners of the shape have an area half the squares' area. Then we can easily find that each corner has an area of one square and there are 4 corners so we add that to the original 9 squares to get 
That is how I did it ~avamarora
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
