Difference between revisions of "2006 AMC 12B Problems/Problem 15"
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Draw the altitude from <math>O</math> onto <math>DP</math> and call the point <math>H</math>. Because <math>\angle OAD</math> and <math>\angle ADP</math> are right angles due to being tangent to the circles, and the altitude creates <math>\angle OHD</math> as a right angle. <math>ADHO</math> is a rectangle with <math>OH</math> bisecting <math>DP</math>. The length <math>OP</math> is <math>4+2</math> and <math>HP</math> has a length of <math>2</math>, so by pythagorean's, <math>OH</math> is <math>\sqrt 32</math>. | Draw the altitude from <math>O</math> onto <math>DP</math> and call the point <math>H</math>. Because <math>\angle OAD</math> and <math>\angle ADP</math> are right angles due to being tangent to the circles, and the altitude creates <math>\angle OHD</math> as a right angle. <math>ADHO</math> is a rectangle with <math>OH</math> bisecting <math>DP</math>. The length <math>OP</math> is <math>4+2</math> and <math>HP</math> has a length of <math>2</math>, so by pythagorean's, <math>OH</math> is <math>\sqrt 32</math>. | ||
− | <math>2 \cdot \sqrt 32 + \frac{1}{2}\cdot2\cdot \ | + | <math>2 \cdot \sqrt{32} + \frac{1}{2}\cdot2\cdot \sqrt{32} = 3\sqrt{32} = 12\sqrt{2}</math>, which is half the area of the hexagon, so the area of the entire hexagon is <math>2\cdot12\sqrt{2} = \boxed{(B)} \qquad24\sqrt{2}</math> |
== See also == | == See also == | ||
{{AMC12 box|year=2006|ab=B|num-b=14|num-a=16}} | {{AMC12 box|year=2006|ab=B|num-b=14|num-a=16}} |
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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
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 |