Difference between revisions of "1986 AHSME Problems/Problem 27"
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==Solution== | ==Solution== | ||
+ | <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 15:54, 2 August 2016
Problem
In the adjoining figure, is a diameter of the circle,
is a chord parallel to
,
and
intersects
at
, with
. The ratio of the area of
to that of
is
Solution
and
are similar isosceles triangles. It remains to find the square of the ratio of their sides. Draw in
. Because
is a diameter,
. Thus,
So
The answer is thus
.
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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