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

Latest revision as of 05:12, 29 June 2022

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 $\boxed{E}$ .

Solution 2

By definition, $\pi$ is the ratio of the circumference to the diameter. Since the circumference is $24\pi$ , the diameter must be $24$ and the radius is $12$ . Therefore, by the area of circle formula $A=\pi r^{2}$ the area is $12^{2}\pi=144\pi$ and $k=144 \Longrightarrow \boxed{\textbf{(E)} 144}$ .

Video Solution

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

~IceMatrix

2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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All AMC 10 Problems and Solutions

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

@@ Line 19: / Line 19: @@
+==Solution 2==
+By definition, <math>\pi</math> is the ratio of the circumference to the diameter. Since the circumference is <math>24\pi</math>, the diameter must be <math>24</math> and the radius is <math>12</math>. Therefore, by the area of circle formula <math>A=\pi r^{2}</math> the area is <math>12^{2}\pi=144\pi</math> and <math>k=144 \Longrightarrow \boxed{\textbf{(E)} 144}</math>.
+==Video Solution==
+https://youtu.be/C1VCk_9A2KE?t=290
+~IceMatrix
 == See Also ==
 {{AMC10 box|year=2010|ab=A|num-b=4|num-a=6}}
+{{MAA Notice}}

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