2010 AMC 8 Problems/Problem 6: Difference between revisions
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==Problem== | ==Problem== | ||
Which of the following has the greatest number of | Which of the following figures has the greatest number of lines of symmetry? | ||
<math> \textbf{(A)}\ \text{ | <math> \textbf{(A)}\ \text{equilateral triangle} </math> | ||
<math> \textbf{(B)}\ \text{ | <math> \textbf{(B)}\ \text{non-square rhombus} </math> | ||
<math> \textbf{(C)}\ \text{ | <math> \textbf{(C)}\ \text{non-square rectangle} </math> | ||
<math> \textbf{(D)}\ \text{ | <math> \textbf{(D)}\ \text{isosceles trapezoid} </math> | ||
<math> \textbf{(E)}\ \text{ | <math> \textbf{(E)}\ \text{square} </math> | ||
==Solution== | ==Solution== | ||
An equilateral triangle has 3 lines of symmetry. | An equilateral triangle has <math>3</math> lines of symmetry. | ||
A non-square rhombus has 2 lines of symmetry. | A non-square rhombus has <math>2</math> lines of symmetry. | ||
A non-square rectangle has 2 lines of symmetry. | A non-square rectangle has <math>2</math> lines of symmetry. | ||
An isosceles trapezoid has 1 line of symmetry. | An isosceles trapezoid has <math>1</math> line of symmetry. | ||
A square has | A square has <math>4</math> lines of symmetry. | ||
Therefore, the answer is <math>\boxed{ \textbf{(E)}\ \text{square} }</math>. | Therefore, the answer is <math>\boxed{ \textbf{(E)}\ \text{square} }</math>. | ||
==Video Solution by @MathTalks== | |||
https://youtu.be/RhyRqHMXvq0?si=m1R2q8UnLRD-KksT | |||
==Video Solution by WhyMath== | |||
https://youtu.be/hoZO5M0raTI | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=5|num-a=7}} | {{AMC8 box|year=2010|num-b=5|num-a=7}} | ||
{{MAA Notice}} | |||
Latest revision as of 10:25, 18 November 2024
Problem
Which of the following figures has the greatest number of lines of symmetry?
Solution
An equilateral triangle has 
lines of symmetry.
A non-square rhombus has 
lines of symmetry.
A non-square rectangle has lines of symmetry.
An isosceles trapezoid has 
line of symmetry.
A square has 
lines of symmetry.
Therefore, the answer is 
.
Video Solution by @MathTalks
https://youtu.be/RhyRqHMXvq0?si=m1R2q8UnLRD-KksT
Video Solution by WhyMath
See Also
| 2010 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
