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Difference between revisions of "1953 AHSME Problems/Problem 11"

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== Solution ==
 
== Solution ==
  
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We notice that since the running track is simply the area of the outer circle that is outside of the inner circle, the radius of the larger circle must be precisely <math>10</math> feet larger than the radius of the smaller circle.
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Call the radius of the outer circle <math>r_1</math> and that of the inner circle <math>r_2</math>. The width of the track is <math>r_1-r_2</math>. The circumference of a circle is <math>2\pi</math> times the radius, so the difference in circumferences is <math>2\pi r_1-2\pi r_2=10\pi</math> feet. If we divide each side by <math>2\pi</math>, we get <math>r_1-r_2=\boxed{5}</math> feet.
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Since the circumference of a circle is calculated as <math>2\pi{r}</math> where <math>r</math> is the radius, we know that the circumference of the smaller circle is <math>2\pi{r}</math> and the circumference of the larger circle is <math>2\pi(r+10)=2\pi{r}+20\pi</math>.  
 
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The difference between the circumferences is <math>2\pi{r}+20\pi-2\pi{r}=20\pi\approx20\cdot3=\boxed{\textbf{(C) } 5\text{ feet}}</math>.
 
  
 
==See Also==
 
==See Also==

Revision as of 00:28, 9 April 2020

A running track is the ring formed by two concentric circles. If the circumferences of the two circles differ by $10\pi$ feet, how wide is the track in feet?

Solution

Call the radius of the outer circle $r_1$ and that of the inner circle $r_2$. The width of the track is $r_1-r_2$. The circumference of a circle is $2\pi$ times the radius, so the difference in circumferences is $2\pi r_1-2\pi r_2=10\pi$ feet. If we divide each side by $2\pi$, we get $r_1-r_2=\boxed{5}$ feet.

See Also

1953 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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All AHSME Problems and Solutions

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