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Difference between revisions of "2006 AMC 10B Problems/Problem 24"

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Therefore, the area of hexagon <math>AOBCPD</math> is <math>2\cdot12\sqrt{2}= 24\sqrt{2} \Rightarrow B </math>
 
Therefore, the area of hexagon <math>AOBCPD</math> is <math>2\cdot12\sqrt{2}= 24\sqrt{2} \Rightarrow B </math>
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== See Also ==
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== See also ==
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*[[2006 AMC 10B Problems]]
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{{AMC10 box|year=2006|ab=B|num-b=23|num-a=25}}
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*[[2006 AMC 10B Problems/Problem 23|Previous Problem]]
 
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*[[2006 AMC 10B Problems/Problem 25|Next Problem]]
 
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 19:51, 10 May 2011

Problem

Circles with centers $O$ and $P$ have radii $2$ and $4$, respectively, and are externally tangent. Points $A$ and $B$ on the circle with center $O$ and points $C$ and $D$ on the circle with center $P$ are such that $AD$ and $BC$ are common external tangents to the circles. What is the area of the concave hexagon $AOBCPD$?

2006amc10b24.gif

$\mathrm{(A) \ } 18\sqrt{3}\qquad \mathrm{(B) \ } 24\sqrt{2}\qquad \mathrm{(C) \ } 36\qquad \mathrm{(D) \ } 24\sqrt{3}\qquad \mathrm{(E) \ } 32\sqrt{2}$

Solution

Since a tangent line is perpendicular to the radius containing the point of tangency, $\angle OAD = \angle PDA = 90^\circ$.

Construct a perpendicular to $DP$ that goes through point $O$. Label the point of intersection $X$.

Clearly $OADX$ is a rectangle.

Therefore $DX=2$ and $PX=2$.

By the Pythagorean Theorem: $OX = \sqrt{6^2 - 2^2} = 4\sqrt{2}$.

The area of $OADX$ is $2\cdot4\sqrt{2}=8\sqrt{2}$.

The area of $OXP$ is $\frac{1}{2}\cdot2\cdot4\sqrt{2}=4\sqrt{2}$.

So the area of quadrilateral $OADP$ is $8\sqrt{2}+4\sqrt{2}=12\sqrt{2}$.

Using similar steps, the area of quadrilateral $OBCP$ is also $12\sqrt{2}$

Therefore, the area of hexagon $AOBCPD$ is $2\cdot12\sqrt{2}= 24\sqrt{2} \Rightarrow B$

See also

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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