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Difference between revisions of "1968 AHSME Problems/Problem 11"

(Created page with "== Problem == If an arc of <math>60^{\circ}</math> on circle <math>I</math> has the same length as an arc of <math>45^{\circ}</math> on circle <math>II</math>, the ratio of the ...")
 
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== Solution ==
 
== Solution ==
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Let the length of the two arcs be denoted <math>x</math>. First, we acknowledge that the length of an [[arc]] of a circle is given by the measure of that arc's [[central angle]] in [[radians]] times the radius of that circle. Note that the central angle of circle <math>I</math> measures <math>\frac{\pi}{3}</math> radians, and the central angle of circle <math>II</math> measures <math>\frac{\pi}{4}</math> radians. Let the radius of circle <math>I</math> be <math>r_1</math> and the radius of circle <math>II</math> be <math>r_2</math>. Then, we know that the arc length <math>x=\frac{\pi}{3}r_1=\frac{\pi}{4}r_2</math>, and so <math>\frac{r_1}{r_2}=\frac{3}{4}</math>. From this, we see that <math>\frac{r_1^2}{r_2^2}=\frac{\pi r_1^2}{\pi r_2^2}=\frac{[I]}{[II]}=\frac{9}{16}</math>. Thus, our answer is <math>\fbox{(B) 9:16}</math>.
  
 
== See also ==
 
== See also ==
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{{AHSME box|year=1968|num-b=10|num-a=12}}   
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{{AHSME 35p box|year=1968|num-b=10|num-a=12}}   
  
 
[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:16, 17 July 2024

Problem

If an arc of $60^{\circ}$ on circle $I$ has the same length as an arc of $45^{\circ}$ on circle $II$, the ratio of the area of circle $I$ to that of circle $II$ is:

$\text{(A) } 16:9\quad \text{(B) } 9:16\quad \text{(C) } 4:3\quad \text{(D) } 3:4\quad \text{(E) } \text{none of these}$

Solution

Let the length of the two arcs be denoted $x$. First, we acknowledge that the length of an arc of a circle is given by the measure of that arc's central angle in radians times the radius of that circle. Note that the central angle of circle $I$ measures $\frac{\pi}{3}$ radians, and the central angle of circle $II$ measures $\frac{\pi}{4}$ radians. Let the radius of circle $I$ be $r_1$ and the radius of circle $II$ be $r_2$. Then, we know that the arc length $x=\frac{\pi}{3}r_1=\frac{\pi}{4}r_2$, and so $\frac{r_1}{r_2}=\frac{3}{4}$. From this, we see that $\frac{r_1^2}{r_2^2}=\frac{\pi r_1^2}{\pi r_2^2}=\frac{[I]}{[II]}=\frac{9}{16}$. Thus, our answer is $\fbox{(B) 9:16}$.

See also

1968 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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