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

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We count <math>3 \cdot 3=9</math> unit squares in the middle, and <math>4</math> small triangles each with an area of <math>1</math>. Thus, the answer is <math>9+4=\boxed{\textbf{(C) } 13}</math>
 
We count <math>3 \cdot 3=9</math> unit squares in the middle, and <math>4</math> small triangles 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==
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==Solution 2==
  
 
We count the half squares and count 8 of them, then count the middle 9 squares and you will count 6 of them, thus, <math>8+6=\boxed{\textbf{(C) } 13}</math>
 
We count the half squares and count 8 of them, then count the middle 9 squares and you will count 6 of them, thus, <math>8+6=\boxed{\textbf{(C) } 13}</math>

Revision as of 20:24, 7 January 2020

Problem 4

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

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

Solution 2

We count the half squares and count 8 of them, then count the middle 9 squares and you will count 6 of them, thus, $8+6=\boxed{\textbf{(C) } 13}$

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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