Difference between revisions of "2006 AMC 8 Problems/Problem 7"
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== Solution == | == Solution == | ||
| − | Using the formulas of circles, <math> C=2 \pi r </math> and <math> A= \pi r^2 </math>, we find that circle <math> Y </math> has a radius of <math> 4 </math> and circle <math> Z </math> has a radius of <math> 3 </math>. Also, circle X has a radius of \pi. Thus, the order from smallest to largest radius is <math> \boxed{\textbf{(E)}\ X, Z, Y} </math>. | + | Using the formulas of circles, <math> C=2 \pi r </math> and <math> A= \pi r^2 </math>, we find that circle <math> Y </math> has a radius of <math> 4 </math> and circle <math> Z </math> has a radius of <math> 3 </math>. Also, circle X has a radius of \pi . Thus, the order from smallest to largest radius is <math> \boxed{\textbf{(E)}\ X, Z, Y} </math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2006|num-b=6|num-a=8}} | {{AMC8 box|year=2006|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:17, 13 January 2024
Problem 7
Circle 
has a radius of 
. Circle 
has a circumference of 
. Circle 
has an area of 
. List the circles in order from smallest to the largest radius.
Solution
Using the formulas of circles, 
and 
, we find that circle has a radius of 
and circle has a radius of 
. Also, circle X has a radius of \pi . Thus, the order from smallest to largest radius is 
.
See Also
| 2006 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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. 
