2018 AMC 8 Problems/Problem 4

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, $8+6=\boxed{\textbf{(C) } 13}$

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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All AJHSME/AMC 8 Problems and Solutions

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2018 AMC 8 Problems/Problem 4

Contents

Problem 4

Solution

Solution 2

See Also