Problem:
The diameter AB of a circle of radius 2 is extended to a point D outside the circle so that BD=3. Point E is chosen so that ED=5 and line ED is perpendicular to line AD. Segment AE intersects the circle at a point C between A and E. What is the area of â–³ABC?
Answer Choices:
A. 37120​
B. 39140​
C. 39145​
D. 37140​
E. 31120​
Solution:
Because ∠ACB is inscribed in a semicircle, it is a right angle. Therefore △ABC is similar to △AED, so their areas are related as AB2 is to AE2. Because AB2=42=16 and, by the Pythagorean Theorem,
AE2=(4+3)2+52=74
this ratio is 7416​=378​. The area of △AED is 235​, so the area of △ABC is 235​⋅378​=(D)37140​​.
The problems on this page are the property of the MAA's American Mathematics Competitions
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