Difference between revisions of "2006 AMC 10A Problems/Problem 16"
m (Fixed a minor terminology error.) |
m (→Solution) |
||
Line 19: | Line 19: | ||
<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 12:23, 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 |