Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
(→‎Solution 2)
(5 intermediate revisions by 5 users not shown)
Line 1: Line 1:
−
==Problem 4==
+
==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>?
  
Line 14: Line 14:
 
==Solution==
 
==Solution==
  
−
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>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>
  
−
==Solution==
+
==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 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
  
 
==See Also==
 
==See Also==

Revision as of 15:14, 19 January 2021

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

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 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 $9+4=\boxed{\textbf{(C) } 13}$ That is how I did it ~avamarora

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_4&oldid=142801"