Difference between revisions of "2006 AMC 10A Problems/Problem 16"
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− | Alternatively, using the fact that Area = radius <math>\cdot</math> semiperimeter, and knowing that <math>AC = \sqrt{FC^2 + 64}</math>, we can set up the equation <cmath>2FC + 2 \sqrt{FC^2 + 64} = 8FC.</cmath> Grouping and getting rid of the square root gives us <math>FC^2 + 64 = 9FC^2</math>, meaning that <math>FC = 2 \sqrt{2}</math>. | + | Alternatively, using the fact that Area = radius <math>\cdot</math> semiperimeter, and knowing that <math>AC = \sqrt{FC^2 + 64}</math>, we can set up the equation <cmath>2FC + 2 \sqrt{FC^2 + 64} = 8FC.</cmath> Grouping and getting rid of the square root gives us <math>FC^2 + 64 = 9FC^2</math>, meaning that <math>FC = 2 \sqrt{2}</math>. Thus, the area of the triangle is <math>\boxed{16 \sqrt{2}, D}</math> ~ booking(Please feel free to edit this, since this is my first time writing a solution!) |
== See also == | == See also == |
Latest revision as of 10:44, 30 November 2024
Contents
[hide]Problem
A circle of radius is tangent to a circle of radius
. The sides of
are tangent to the circles as shown, and the sides
and
are congruent. What is the area of
?
Solution 1
Let the centers of the smaller and larger circles be and
, respectively.
Let their tangent points to
be
and
, respectively.
We can then draw the following diagram:
We see that . Using the first pair of similar triangles, we write the proportion:

By the Pythagorean Theorem, we have .
Now using ,

Hence, the area of the triangle is
Solution 2
Since we have that
.
Since we know that the total length of
We also know that , so
Also, since we have that
Since we know that and
we have that
This equation simplified gets us
Let
By the Pythagorean Theorem on we have that
We know that ,
and
so we have
Simplifying, we have that
Recall that .
Therefore,
Since the height is we have the area equal to
Thus our answer is .
~mathboy282
Alternatively, using the fact that Area = radius semiperimeter, and knowing that
, we can set up the equation
Grouping and getting rid of the square root gives us
, meaning that
. Thus, the area of the triangle is
~ booking(Please feel free to edit this, since this is my first time writing a solution!)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.