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Difference between revisions of "1952 AHSME Problems/Problem 37"

(Created page with "== Problem == Two equal parallel chords are drawn <math>8</math> inches apart in a circle of radius <math>8</math> inches. The area of that part of the circle that lies between ...")
 
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== Solution ==
 
== Solution ==
−
<math>\fbox{}</math>
+
<asy>
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pair A,B,C,D,E,F,G;
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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));
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draw((-4,-4sqrt(3))--(-4,4sqrt(3)));
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draw((4,-4sqrt(3))--(4,4sqrt(3)));
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draw((-4,-4sqrt(3))--(4,4sqrt(3)));
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draw((4,-4sqrt(3))--(-4,4sqrt(3)));
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draw((4,0)--(-4,0));
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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));
 +
</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=<math>4\sqrt{3}</math>, as are DF, EG, and GC.
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It then follows that the area of triangles BAD and CAE are <math>16\sqrt{3}</math>.
 +
 
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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.
 +
 
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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 19:23, 15 April 2020

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

[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 $\fbox{B}$.

See also

1952 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 36 		Followed by
Problem 38
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 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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