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> | ||
− | ==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 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 21:24, 7 January 2020
Contents
Problem 4
The twelve-sided figure shown has been drawn on graph paper. What is the area of the figure in
?
Solution
We count unit squares in the middle, and
small triangles each with an area of
. Thus, the answer is
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,
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.