Difference between revisions of "2006 AMC 12B Problems/Problem 15"
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<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> | <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> | ||
| + | |||
| + | == 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 | ||
| + | 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 1/2(4+2)*4sqrt(2) = 12sqrt(2). and the total area is two trapezoids, or 24sqrt2. | ||
== 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 20:26, 29 January 2012
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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 
?
![[asy] // from amc10 problem series unitsize(0.4 cm); defaultpen(linewidth(0.7) + fontsize(11)); pair A, B, C, D; pair[] O; O[1] = (6,0); O[2] = (12,0); A = (32/6,8*sqrt(2)/6); B = (32/6,-8*sqrt(2)/6); C = 2*B; D = 2*A; draw(Circle(O[1],2)); draw(Circle(O[2],4)); draw((0.7*A)--(1.2*D)); draw((0.7*B)--(1.2*C)); draw(O[1]--O[2]); draw(A--O[1]); draw(B--O[1]); draw(C--O[2]); draw(D--O[2]); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SW); label("$D$", D, NW); dot("$O$", O[1], SE); dot("$P$", O[2], SE); label("$2$", (A + O[1])/2, E); label("$4$", (D + O[2])/2, E);[/asy]](http://latex.artofproblemsolving.com/9/8/5/985d18b9bd1bc04f16562ee9a3eabc5afe6efcbd.png)
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
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
the Pythagorean theorem, x is equal to 
, and x is also equal to the height of the trapezoid. The area of the trapezoid is thus 1/2(4+2)*4sqrt(2) = 12sqrt(2). and the total area is two trapezoids, or 24sqrt2.
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 | |
