1952 AHSME Problems/Problem 37
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. Label the intersection points of the diameter with the chords F and G. 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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