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>. Thus, the order from smallest to largest radius is <math> \boxed{\textbf{(B)}\ Z, X, Y} </math>.
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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 <math> \pi </math>. Thus, the order from smallest to largest radius is <math> \boxed{\textbf{(B)}\ Z, X, Y} </math>.
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==Video Solution by WhyMath==
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https://youtu.be/J5-hDWd28tM
  
 
==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}}

Latest revision as of 13:21, 29 October 2024

Problem 7

Circle $X$ has a radius of $\pi$. Circle $Y$ has a circumference of $8 \pi$. Circle $Z$ has an area of $9 \pi$. List the circles in order from smallest to the largest radius.

$\textbf{(A)}\ X, Y, Z\qquad\textbf{(B)}\ Z, X, Y\qquad\textbf{(C)}\ Y, X, Z\qquad\textbf{(D)}\ Z, Y, X\qquad\textbf{(E)}\ X, Z, Y$

Solution

Using the formulas of circles, $C=2 \pi r$ and $A= \pi r^2$, we find that circle $Y$ has a radius of $4$ and circle $Z$ has a radius of $3$. Also, circle X has a radius of $\pi$. Thus, the order from smallest to largest radius is $\boxed{\textbf{(B)}\ Z, X, Y}$.

Video Solution by WhyMath

https://youtu.be/J5-hDWd28tM

See Also

2006 AMC 8 (ProblemsAnswer KeyResources)
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

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