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 \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) | + | <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)24\sqrt{2}}</math> |
== Solution 2 == | == Solution 2 == | ||
− | <math> | + | <math>ADPO</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, <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\sqrt{2}</math>, and the total area is two trapezoids, or <math>\boxed{24\sqrt{2}}</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\sqrt{2}</math>, and the total area is two trapezoids, or <math>\boxed{24\sqrt{2}}</math>. |
== See also == | == See also == |
Latest revision as of 16:40, 6 October 2019
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 |
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