Difference between revisions of "2018 AMC 8 Problems/Problem 4"
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The twelve-sided figure shown has been drawn on <math>1 \text{ cm}\times 1 \text{ cm}</math> graph paper. What is the area of the figure in <math>\text{cm}^2</math>? | The twelve-sided figure shown has been drawn on <math>1 \text{ cm}\times 1 \text{ cm}</math> graph paper. What is the area of the figure in <math>\text{cm}^2</math>? | ||
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<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> | ||
| + | |||
| + | ==Solution 1== | ||
| + | |||
| + | We count <math>3 \cdot 3=9</math> unit squares in the middle, and <math>8</math> small triangles, which gives 4 rectangles each with an area of <math>1</math>. Thus, the answer is <math>9+4=\boxed{\textbf{(C) } 13}</math>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | We can see here that there are <math>9</math> total 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 <math>4</math> corners so we add that to the original 9 squares to get <math>9+4=\boxed{\textbf{(C) } 13}</math>. That is how I did it. | ||
| + | |||
| + | ~avamarora | ||
| + | |||
| + | ==Solution 3== | ||
| + | |||
| + | We can apply Pick's Theorem here. There are <math>8</math> lattice points, and <math>12</math> lattice points on the boundary. Then, | ||
| + | |||
| + | <cmath> 8 + 12 \div 2 - 1 = \boxed {\textbf{(C) }13}.</cmath> | ||
| + | |||
| + | ==Solution 4== | ||
| + | |||
| + | Use the Shoelace Theorem. | ||
| + | |||
| + | ~SaxStreak | ||
| + | |||
| + | == Video Solution (CRITICAL THINKING!!!)== | ||
| + | https://youtu.be/7qY99daRZUA | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/huLjsiLQS90 | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/51K3uCzntWs?t=1338 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2018|num-b=3|num-a=5}} | {{AMC8 box|year=2018|num-b=3|num-a=5}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 20:18, 29 March 2023
Contents
Problem
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 1
We count 
unit squares in the middle, and 
small triangles, which gives 4 rectangles each with an area of 
. Thus, the answer is 
.
Solution 2
We can see here that there are 
total 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 
corners so we add that to the original 9 squares to get . That is how I did it.
~avamarora
Solution 3
We can apply Pick's Theorem here. There are lattice points, and 
lattice points on the boundary. Then,
![\[8 + 12 \div 2 - 1 = \boxed {\textbf{(C) }13}.\]](http://latex.artofproblemsolving.com/8/d/4/8d4221bda95dc3240fa56afce546af5959eeed19.png)
Solution 4
Use the Shoelace Theorem.
~SaxStreak
Video Solution (CRITICAL THINKING!!!)
~Education, the Study of Everything
Video Solution
~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/51K3uCzntWs?t=1338
~ pi_is_3.14
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. 
