Difference between revisions of "2010 AMC 8 Problems/Problem 6"
| (11 intermediate revisions by 6 users not shown) | |||
| Line 13: | Line 13: | ||
==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}} | {{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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
