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1952 AHSME Problems/Problem 37

Revision as of 19:21, 17 December 2024 by Clever14710owl (talk | contribs) (→‎Solution)
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Problem

Two equal parallel chords are drawn $8$ inches apart in a circle of radius $8$ inches. The area of that part of the circle that lies between the chords is:

$\textbf{(A)}\ 21\frac{1}{3}\pi-32\sqrt{3}\qquad \textbf{(B)}\ 32\sqrt{3}+21\frac{1}{3}\pi\qquad \textbf{(C)}\ 32\sqrt{3}+42\frac{2}{3}\pi \qquad\\ \textbf{(D)}\ 16\sqrt {3} + 42\frac {2}{3}\pi \qquad \textbf{(E)}\ 42\frac {2}{3}\pi$

Solution 1

[asy] pair A,B,C,D,E,F,G; A=(0,0); B=(-5,4sqrt(3)+2); C=(5,4sqrt(3)+2); D=(-5,-4sqrt(3)-0.5); E=(5,-4sqrt(3)-0.5); F=(-4,0); G=(4,-sqrt(2)/2); label("$A$",(0,-0.5),S); label("$B$",B,SE); label("$C$",C,SW); label("$D$",D,SE); label("$E$",E,SW); label("$F$",F,W); label("$G$",G,NE); draw(circle(A,8)); draw((-4,-4sqrt(3))--(-4,4sqrt(3))); draw((4,-4sqrt(3))--(4,4sqrt(3))); draw((-4,-4sqrt(3))--(4,4sqrt(3))); draw((4,-4sqrt(3))--(-4,4sqrt(3))); draw((4,0)--(-4,0)); 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] 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=$4\sqrt{3}$, as are DF, EG, and GC.

It then follows that the area of triangles BAD and CAE are $16\sqrt{3}$.

Since AFB and AFD are 30-60-90 triangles, the area of sector BAD is $\frac{64\pi}{3}$, as is sector CAE.

Thus, the area outside of the two chords is $\frac{128\pi}{3}-32\sqrt{3}$. Since we want the area inside the two chords, you can subtract the outside from the whole circle, which is $64\pi-(\frac{128\pi}{3}-32\sqrt{3})$ $\Rightarrow \frac{64\pi}{3}+32\sqrt{3}$, or $\fbox{B}$.


Solution 2

The area inside the chords are the arcs plus the triangles. We know that AFB and AFD are 30-60-90 triangles, which means that angle BAC is angle DAE which is 60 degrees. This means that the sum of the sectors is $\frac{64 \pi} {3}$. The triangles are both equal to $\frac{8\sqrt{3} \cdot 4} {2}$, so the sum of the two is $32\sqrt{3}$. Finally, we add the two together. This yields the answer, $\frac{64\pi}{3}+32\sqrt{3}$, or $\fbox{B}$.

~clever14710owl

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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