Difference between revisions of "2006 AMC 10A Problems/Problem 16"
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<div style="text-align:center;"><math>\frac{AD}{AF} = \frac{DO_1}{CF} \Longrightarrow \frac{2\sqrt{2}}{8} = \frac{1}{CF} \Longrightarrow CF = 2\sqrt{2}</math></div> | <div style="text-align:center;"><math>\frac{AD}{AF} = \frac{DO_1}{CF} \Longrightarrow \frac{2\sqrt{2}}{8} = \frac{1}{CF} \Longrightarrow CF = 2\sqrt{2}</math></div> | ||
| − | The area of the triangle is <math>\frac{1}{2} | + | The area of the triangle is <math>\frac{1}{2}\cdot AF \cdot BC = \frac{1}{2}\cdot AF \cdot (2\cdot CF) = AF \cdot CF = 8\left(2\sqrt{2}\right) = 16\sqrt{2}\ \mathrm{(D)}</math>. |
== See also == | == See also == | ||
Revision as of 11:23, 6 June 2009
Problem
A circle of radius 1 is tangent to a circle of radius 2. The sides of 
are tangent to the circles as shown, and the sides 
and 
are congruent. What is the area of ?
Solution
Note that 
. Using the first pair of similar triangles, we write the proportion:
By the Pythagorean Theorem we have that 
.
Now using 
,
The area of the triangle is 
.
See also
| 2006 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 | ||
