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Difference between revisions of "1958 AHSME Problems/Problem 21"

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== Solution ==
 
== Solution ==
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Draw <math>OE</math>. Since triangles <math>AOB</math>, <math>COE</math>, and <math>DOE</math> are congruent, you can fit triangle <math>AOE</math> twice in <math>CED</math>. Thus, our answer is <math>2:1</math>, or <math>(E)</math>.
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Draw <math>OE</math>. Since triangles <math>AOB</math>, <math>COE</math>, and <math>DOE</math> are congruent, you can fit triangle <math>AOE</math> twice in <math>CED</math>. Thus, our answer is <math>(E)2:1</math>.
 
<math>\fbox{}</math>
 
<math>\fbox{}</math>
  

Latest revision as of 00:06, 1 January 2024

Problem

In the accompanying figure $\overline{CE}$ and $\overline{DE}$ are equal chords of a circle with center $O$. Arc $AB$ is a quarter-circle. Then the ratio of the area of triangle $CED$ to the area of triangle $AOB$ is:

[asy] draw(circle((0,0),10),black+linewidth(.75)); draw((-10,0)--(0,0)--(10,0)--(0,10)--cycle,dot); MP("O",(0,0),N);MP("C",(-10,0),W);MP("D",(10,0),E);;MP("E",(0,10),N); draw((-sqrt(70),-sqrt(30))--(sqrt(30),-sqrt(70))--(0,0)--cycle,dot); MP("A",(-sqrt(70),-sqrt(30)),SW);MP("B",(sqrt(30),-sqrt(70)),SE); [/asy]

$\textbf{(A)}\ \sqrt {2} : 1\qquad \textbf{(B)}\ \sqrt {3} : 1\qquad \textbf{(C)}\ 4 : 1\qquad \textbf{(D)}\ 3 : 1\qquad \textbf{(E)}\ 2 : 1$

Solution

Draw $OE$. Since triangles $AOB$, $COE$, and $DOE$ are congruent, you can fit triangle $AOE$ twice in $CED$. Thus, our answer is $(E)2:1$. $\fbox{}$

See Also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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All AHSME Problems and Solutions

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