Difference between revisions of "2010 AMC 10A Problems/Problem 5"

(Solution)
m
(One intermediate revision by one other user not shown)
Line 18: Line 18:
 
If the circumference of a circle is <math>24\pi</math>, the radius would be <math>12</math>. Since the area of a circle is <math>\pi r^2</math>, the area is <math>144\pi</math>. The answer is <math>\boxed{E}</math>.
 
If the circumference of a circle is <math>24\pi</math>, the radius would be <math>12</math>. Since the area of a circle is <math>\pi r^2</math>, the area is <math>144\pi</math>. The answer is <math>\boxed{E}</math>.
  
 +
==Video Solution==
 +
https://youtu.be/C1VCk_9A2KE?t=290
  
 +
~IceMatrix
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2010|ab=A|num-b=4|num-a=6}}
 
{{AMC10 box|year=2010|ab=A|num-b=4|num-a=6}}
 +
{{MAA Notice}}

Revision as of 06:19, 26 May 2020

Problem 5

The area of a circle whose circumference is $24\pi$ is $k\pi$. What is the value of $k$?

$\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 24 \qquad \mathrm{(D)}\ 36 \qquad \mathrm{(E)}\ 144$

Solution

If the circumference of a circle is $24\pi$, the radius would be $12$. Since the area of a circle is $\pi r^2$, the area is $144\pi$. The answer is $\boxed{E}$.

Video Solution

https://youtu.be/C1VCk_9A2KE?t=290

~IceMatrix

See Also

2010 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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 AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png