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

Revision as of 17:23, 17 June 2022

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

Note

Notice here that Pick's Theorem DOES NOT WORK.

Reason: There are $8$ lattice points, and $12$ lattice points on the boundary. Then,

$8 + 12 \div 2 = 14$

We find $14$ as our answer instead of the correct answer of $13$ .

Video Solution

https://youtu.be/huLjsiLQS90

~savannahsolver

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.

@@ Line 19: / Line 19: @@
 ==Solution 2==
-We can see here that there are 9 full 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 <math>9+4=\boxed{\textbf{(C) } 13}</math> That is how I did it ~avamarora
+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
+===Note===
+Notice here that Pick's Theorem '''DOES NOT WORK'''.
+Reason:
+There are <math>8</math> lattice points, and <math>12</math> lattice points on the boundary. Then,
+<cmath> 8 + 12 \div 2 = 14</cmath>
+We find <math>14</math> as our answer instead of the correct answer of <math>13</math>.
 ==Video Solution==

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