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Difference between revisions of "2010 AMC 10A Problems/Problem 5"
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==Solution==
 
==Solution==
  
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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>.
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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>.
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==Solution 2==
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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>.
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==Video Solution==
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https://youtu.be/C1VCk_9A2KE?t=290
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~IceMatrix
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== See Also ==
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{{AMC10 box|year=2010|ab=A|num-b=4|num-a=6}}
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{{MAA Notice}}

Latest revision as of 06:12, 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

See Also

2010 AMC 10A (Problems • Answer Key • Resources)
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

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