2018 AMC 8 Problems/Problem 4

Revision as of 11:34, 23 February 2025 by Aaronapplewip (talk | contribs) (Solution 2)

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 3

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

\[8 + 12 \div 2 - 1 = \boxed {\textbf{(C) }13}.\] /boxe{/textbf{ (c_ }21.$$ (Error compiling LaTeX. Unknown error_msg)

Solution 4

DONT USE SHOELACE no dont it is slow

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 (ProblemsAnswer KeyResources)
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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