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Difference between revisions of "1986 AHSME Problems/Problem 27"

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==Solution==
 
==Solution==
  
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<math>ABE</math> and <math>DCE</math> are similar isosceles triangles. It remains to find the square of the ratios of their sides. Draw in <math>AD</math>. Because <math>AB</math> is a diameter, <math>\angle ADB=\angle ADE=90^{\circ}</math>. Thus, <cmath>\frac{DE}{AE}=\cos\alpha</cmath> So <cmath>\frac{DE^2}{AE^2}=\cos^2\alpha</cmath> The answer is thus <math>\fbox{(C)}</math>.
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<math>ABE</math> and <math>DCE</math> are similar isosceles triangles. It remains to find the square of the ratio of their sides. Draw in <math>AD</math>. Because <math>AB</math> is a diameter, <math>\angle ADB=\angle ADE=90^{\circ}</math>. Thus, <cmath>\frac{DE}{AE}=\cos\alpha</cmath> So <cmath>\frac{DE^2}{AE^2}=\cos^2\alpha</cmath> The answer is thus <math>\fbox{(C)}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 14:54, 2 August 2016

Problem

In the adjoining figure, $AB$ is a diameter of the circle, $CD$ is a chord parallel to $AB$, and $AC$ intersects $BD$ at $E$, with $\angle AED = \alpha$. The ratio of the area of $\triangle CDE$ to that of $\triangle ABE$ is

[asy] defaultpen(fontsize(10pt)+linewidth(.8pt)); pair A=(-1,0), B=(1,0), E=(0,-.4), C=(.6,-.8), D=(-.6,-.8), E=(0,-.8/(1.6)); draw(unitcircle); draw(A--B--D--C--A); draw(Arc(E,.2,155,205)); label("$A$",A,W); label("$B$",B,C); label("$C$",C,C); label("$D$",D,W); label("$\alpha$",E-(.2,0),W); label("$E$",E,N); [/asy]

$\textbf{(A)}\ \cos\ \alpha\qquad \textbf{(B)}\ \sin\ \alpha\qquad \textbf{(C)}\ \cos^2\alpha\qquad \textbf{(D)}\ \sin^2\alpha\qquad \textbf{(E)}\ 1-\sin\ \alpha$


Solution

$ABE$ and $DCE$ are similar isosceles triangles. It remains to find the square of the ratio of their sides. Draw in $AD$. Because $AB$ is a diameter, $\angle ADB=\angle ADE=90^{\circ}$. Thus, \[\frac{DE}{AE}=\cos\alpha\] So \[\frac{DE^2}{AE^2}=\cos^2\alpha\] The answer is thus $\fbox{(C)}$.

See also

1986 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
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
All AHSME Problems and Solutions

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