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Difference between revisions of "2018 AMC 8 Problems/Problem 4"

(Created page with "==Problem 4== 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</ma...")
 
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==Problem 4==
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==Problem==
 
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>
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==Solution 1==
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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>.
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==Solution 2==
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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.
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~avamarora
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==Solution 3==
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We can apply Pick's Theorem here. There are <math>8</math> lattice points, and <math>12</math> lattice points on the boundary. Then,
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<cmath> 8 + 12 \div 2 - 1 = \boxed {\textbf{(C) }13}.</cmath>
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==Solution 4==
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Use the Shoelace Theorem.
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~SaxStreak
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== Video Solution (CRITICAL THINKING!!!)==
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https://youtu.be/7qY99daRZUA
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/huLjsiLQS90
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~savannahsolver
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==Video Solution by OmegaLearn==
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https://youtu.be/51K3uCzntWs?t=1338
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~ pi_is_3.14
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==See Also==
 
{{AMC8 box|year=2018|num-b=3|num-a=5}}
 
{{AMC8 box|year=2018|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 20:18, 29 March 2023

Problem

The twelve-sided figure shown has been drawn on $1 \text{ cm}\times 1 \text{ cm}$ graph paper. What is the area of the figure in $\text{cm}^2$?

[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]

$\textbf{(A) } 12 \qquad \textbf{(B) } 12.5 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13.5 \qquad \textbf{(E) } 14$

Solution 1

We count $3 \cdot 3=9$ unit squares in the middle, and $8$ small triangles, which gives 4 rectangles each with an area of $1$. Thus, the answer is $9+4=\boxed{\textbf{(C) } 13}$.

Solution 2

We can see here that there are $9$ 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 $4$ corners so we add that to the original 9 squares to get $9+4=\boxed{\textbf{(C) } 13}$. That is how I did it.

~avamarora

Solution 3

We can apply Pick's Theorem here. There are $8$ lattice points, and $12$ lattice points on the boundary. Then,

\[8 + 12 \div 2 - 1 = \boxed {\textbf{(C) }13}.\]

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

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

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