Difference between revisions of "1968 AHSME Problems/Problem 11"
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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 == |
Latest revision as of 20:16, 17 July 2024
Problem
If an arc of on circle
has the same length as an arc of
on circle
, the ratio of the area of circle
to that of circle
is:
Solution
Let the length of the two arcs be denoted . 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
measures
radians, and the central angle of circle
measures
radians. Let the radius of circle
be
and the radius of circle
be
. Then, we know that the arc length
, and so
. From this, we see that
. Thus, our answer is
.
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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