Difference between revisions of "1952 AHSME Problems/Problem 37"
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label("$4$",(-2.5,0.5)); label("$4$",(2.5,0.5)); label("$8$",(-2.5,3)); label("$8$",(2.5,3)); label("$8$",(-2.5,-3)); label("$8$",(2.5,-3)); | label("$4$",(-2.5,0.5)); label("$4$",(2.5,0.5)); label("$8$",(-2.5,3)); label("$8$",(2.5,3)); label("$8$",(-2.5,-3)); label("$8$",(2.5,-3)); | ||
</asy> | </asy> | ||
− | + | Draw in the diameter through A perpendicular to chords BD and CE. Then, AF=4, AG=4, and AFB ,AFD, AEG. and AGC are right triangles. We're going to find the area inside sectors BAD and CAE but outside triangles BAD and CAE. | |
− | <math>\fbox{B}</math> | + | |
+ | By pythagorean theorem, BF=<math>4\sqrt{3}</math>, as are DF, EG, and GC. | ||
+ | |||
+ | It then follows that the area of triangles BAD and CAE are <math>16\sqrt{3}</math>. | ||
+ | |||
+ | Since AFB and AFD are 30-60-90 triangles, the area of sector BAD is <math>\frac{64\pi}{3}</math>, as is sector CAE. | ||
+ | |||
+ | Thus, the area outside of the two chords is <math>\frac{128\pi}{3}-32\sqrt{3}</math>. | ||
+ | Since we want the area inside the two chords, you can subtract the outside from the whole circle, which is <math>64\pi-(\frac{128\pi}{3}-32\sqrt{3})</math> <math>\Rightarrow \frac{64\pi}{3}+32\sqrt{3}</math>, or | ||
+ | <math>\fbox{B}</math>. | ||
== See also == | == See also == |
Revision as of 21:59, 14 April 2020
Problem
Two equal parallel chords are drawn inches apart in a circle of radius inches. The area of that part of the circle that lies between the chords is:
Solution
Draw in the diameter through A perpendicular to chords BD and CE. Then, AF=4, AG=4, and AFB ,AFD, AEG. and AGC are right triangles. We're going to find the area inside sectors BAD and CAE but outside triangles BAD and CAE.
By pythagorean theorem, BF=, as are DF, EG, and GC.
It then follows that the area of triangles BAD and CAE are .
Since AFB and AFD are 30-60-90 triangles, the area of sector BAD is , as is sector CAE.
Thus, the area outside of the two chords is . Since we want the area inside the two chords, you can subtract the outside from the whole circle, which is , or .
See also
1952 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 36 |
Followed by Problem 38 | |
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All AHSME Problems and Solutions |
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