Difference between revisions of "2010 AMC 10A Problems/Problem 5"
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==Solution== | ==Solution== | ||
− | 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>. | + | 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== | ||
+ | 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== | ||
+ | https://youtu.be/C1VCk_9A2KE?t=290 | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AMC10 box|year=2010|ab=A|num-b=4|num-a=6}} | ||
+ | {{MAA Notice}} |
Latest revision as of 06:12, 29 June 2022
Problem 5
The area of a circle whose circumference is is
. What is the value of
?
Solution
If the circumference of a circle is , the radius would be
. Since the area of a circle is
, the area is
. The answer is
.
Solution 2
By definition, is the ratio of the circumference to the diameter. Since the circumference is
, the diameter must be
and the radius is
. Therefore, by the area of circle formula
the area is
and
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.