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

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== Solution ==
 
== Solution ==
<math>\fbox{}</math>
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<asy>
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pair A,B,C,D,E;
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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);
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label("$A$",(0,-0.5),S); label("$B$",B,SE); label("$C$",C,SW); label("$D$",D,SE); label("$E$",E,SW);
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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));
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</asy>
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<math>\fbox{B}</math>
  
 
== See also ==
 
== See also ==

Revision as of 21:42, 13 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; 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); label("$A$",(0,-0.5),S); label("$B$",B,SE); label("$C$",C,SW); label("$D$",D,SE); label("$E$",E,SW); 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] $\fbox{B}$

See also

1952 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 36
Followed by
Problem 38
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All AHSME Problems and Solutions

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