Difference between revisions of "2008 AMC 8 Problems/Problem 18"
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Part 6: Same as part 2: <math>10</math> | Part 6: Same as part 2: <math>10</math> | ||
Total: <math>10\pi + 10 + 5\pi + 20 + 5\pi + 10 = \boxed{E = 20\pi + 40}</math> | Total: <math>10\pi + 10 + 5\pi + 20 + 5\pi + 10 = \boxed{E = 20\pi + 40}</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=NJs4rFyXFwQ | ||
+ | —DSA_Catachu | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2008|num-b=17|num-a=19}} | {{AMC8 box|year=2008|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:13, 3 May 2020
Contents
Problem
Two circles that share the same center have radii meters and meters. An aardvark runs along the path shown, starting at and ending at . How many meters does the aardvark run?
Solution
We will deal with this part by part: Part 1: 1/4 circumference of big circle= Part 2: Big radius minus small radius= Part 3: 1/4 circumference of small circle= Part 4: Diameter of small circle: Part 5: Same as part 3: Part 6: Same as part 2: Total:
Video Solution
https://www.youtube.com/watch?v=NJs4rFyXFwQ —DSA_Catachu
See Also
2008 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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.