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1952 AHSME Problems/Problem 37: Difference between revisions

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\textbf{(E)}\ 42\frac {2}{3}\pi </math>
\textbf{(E)}\ 42\frac {2}{3}\pi </math>


== Solution ==
== Solution 1 ==
<asy>
<asy>
pair A,B,C,D,E,F,G;
pair A,B,C,D,E,F,G;
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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.
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=<math>4\sqrt{3}</math>, as are DF, EG, and GC.
By pythagorean theorem, BF=<math>4\sqrt{3}</math>, as are DF, EG, and GC.
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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  
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>.
<math>\fbox{B}</math>.
==Solution 2==
The area inside the chords are the arcs plus the triangles. We know that <math>\triangle AFB</math> and <math>\triangle AFD</math> are 30-60-90 triangles, which means that <math>\angle BAC = \angle DAE = 60^\circ</math>. This means that the sum of the sectors is <math>\frac{64 \pi} {3}</math>. The triangles are both equal to <math>\frac{8\sqrt{3} \cdot 4} {2}</math>, so the sum of the two is <math>32\sqrt{3}</math>. Finally, we add the two together. This yields the answer, <math>\frac{64\pi}{3}+32\sqrt{3}</math>, or <math>\fbox{B}</math>.
~clever14710owl


== See also ==
== See also ==

Latest revision as of 18:19, 13 January 2025

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 $\triangle AFB$ and $\triangle AFD$ are 30-60-90 triangles, which means that $\angle BAC = \angle DAE = 60^\circ$. 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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