Difference between revisions of "2006 AMC 12B Problems/Problem 15"
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Revision as of 10:46, 4 July 2013
Contents
Problem
Circles with centers and have radii 2 and 4, respectively, and are externally tangent. Points and are on the circle centered at , and points and are on the circle centered at , such that and are common external tangents to the circles. What is the area of hexagon ?
Solution
Draw the altitude from onto and call the point . Because and are right angles due to being tangent to the circles, and the altitude creates as a right angle. is a rectangle with bisecting . The length is and has a length of , so by pythagorean's, is .
, which is half the area of the hexagon, so the area of the entire hexagon is
Solution 2
and are congruent right trapezoids with legs and and with equal to . Draw an altitude from to either or , creating a rectangle with width and base , and a right triangle with one leg , the hypotenuse , and the other . Using the Pythagorean theorem, is equal to , and is also equal to the height of the trapezoid. The area of the trapezoid is thus , and the total area is two trapezoids, or .
See also
2006 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 | |
All AMC 12 Problems and Solutions |
2006 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.