Difference between revisions of "2022 AMC 8 Problems/Problem 1"
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==Solution== | ==Solution== | ||
− | Note that the given symbol can be split into <math>5</math> diamonds. Each of the diamonds contain <math>4</math> triangles, so each of the diamonds has an area of <math>0.5 \cdot 4 = 2</math>. Hence, our answer is <math>2 \cdot 5 = \boxed{\textbf{(A) }10}</math>. | + | Note that the given symbol can be split into <math>5</math> diamonds. Each of the diamonds contain <math>4</math> triangles, so each of the diamonds has an area of <math>0.5 \cdot 4 = 2</math>. Hence, our answer is <math>2 \cdot 5 = \boxed{\textbf{(A) } 10}</math>. |
<i>pog</i> | <i>pog</i> |
Revision as of 10:12, 28 January 2022
Problem
The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the area of the logo in square inches?
usepackage("mathptmx"); defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); filldraw((1,2)--(2,1)--(3,2)--(4,1)--(5,2)--(4,3)--(5,4)--(4,5)--(3,4)--(2,5)--(1,4)--(2,3)--(1,2)--cycle, mediumgray*0.5 + lightgray*0.5); draw((0,0)--(6,0), gray); draw((0,1)--(6,1), gray); draw((0,2)--(6,2), gray); draw((0,3)--(6,3), gray); draw((0,4)--(6,4), gray); draw((0,5)--(6,5), gray); draw((0,6)--(6,6), gray); draw((0,0)--(0,6), gray); draw((1,0)--(1,6), gray); draw((2,0)--(2,6), gray); draw((3,0)--(3,6), gray); draw((4,0)--(4,6), gray); draw((5,0)--(5,6), gray); draw((6,0)--(6,6), gray); (Error making remote request. Unexpected URL sent back)
Solution
Note that the given symbol can be split into diamonds. Each of the diamonds contain triangles, so each of the diamonds has an area of . Hence, our answer is .
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See Also
2022 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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.