1958 AHSME Problems/Problem 21

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 $(E)2:1$ . $\fbox{}$

1958 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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1958 AHSME Problems/Problem 21

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