Difference between revisions of "2006 AMC 12B Problems/Problem 15"
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== Solution 2 == | == Solution 2 == | ||
− | ADOP and OPBC are congruent right trapezoids with legs 2 and 4 and with OP equal to 6. Draw an altitude from O to either DP or CP, creating a rectangle with width 2 and base x, and a right triangle with one leg 2, the hypotenuse 6, and the other x. Using | + | <math>ADOP</math> and <math>OPBC</math> are congruent right trapezoids with legs <math>2</math> and <math>4</math> and with <math>OP</math> equal to <math>6</math>. Draw an altitude from <math>O</math> to either <math>DP</math> or <math>CP</math>, creating a rectangle with width <math>2</math> and base <math>x</math>, and a right triangle with one leg <math>2</math>, the hypotenuse <math>6</math>, and the other <math>x</math>. Using |
− | the Pythagorean theorem, x is equal to <math>4\sqrt{2}</math>, and x is also equal to the height of the trapezoid. The area of the trapezoid is thus <math>\frac{1}{2}\cdot(4+2)\cdot4\sqrt{2} = 12</math> | + | the Pythagorean theorem, <math>x</math> is equal to <math>4\sqrt{2}</math>, and <math>x</math> is also equal to the height of the trapezoid. The area of the trapezoid is thus <math>\frac{1}{2}\cdot(4+2)\cdot4\sqrt{2} = 12</math> |
− | + | and the total area is two trapezoids, or <math>\boxed{24\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}} |
Revision as of 20:47, 30 January 2012
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 |