2010 AMC 10A Problems/Problem 5

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)
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2010 AMC 10A Problems/Problem 5

Contents

Problem 5

Solution

Solution 2

Video Solution

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