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>.
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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 11:12, 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);
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$\textbf{(A) } 10 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$

Solution

Note that the given symbol can be split into $5$ diamonds. Each of the diamonds contain $4$ triangles, so each of the diamonds has an area of $0.5 \cdot 4 = 2$. Hence, our answer is $2 \cdot 5 = \boxed{\textbf{(A) } 10}$.

pog

See Also

2022 AMC 8 (ProblemsAnswer KeyResources)
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

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