Difference between revisions of "2006 AMC 10A Problems/Problem 16"
Dairyqueenxd (talk | contribs) (→Solution 1) |
Dairyqueenxd (talk | contribs) (→Solution 2) |
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Let <math>FC = a</math> | Let <math>FC = a</math> | ||
− | By the Pythagorean Theorem on <math>\triangle{AFC},</math> we have that <math>AF^2 + FC^2 = AC^2.</math> | + | By the [[Pythagorean Theorem]] on <math>\triangle{AFC},</math> we have that <math>AF^2 + FC^2 = AC^2.</math> |
We know that <math>AF = 8</math>, <math>FC = a</math> and <math>AC = 3a</math> so we have <math>8^2 + a^2 = (3a)^2.</math> | We know that <math>AF = 8</math>, <math>FC = a</math> and <math>AC = 3a</math> so we have <math>8^2 + a^2 = (3a)^2.</math> | ||
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Since the height is <math>AF = 8,</math> we have the area equal to <math>\frac{4 \sqrt{2} \cdot 8}{2}=16\sqrt{2}.</math> | Since the height is <math>AF = 8,</math> we have the area equal to <math>\frac{4 \sqrt{2} \cdot 8}{2}=16\sqrt{2}.</math> | ||
− | Thus our answer is <math>\boxed{\ | + | Thus our answer is <math>\boxed{\textbf{(D) }16 \sqrt{2}}</math>. |
~mathboy282 | ~mathboy282 |
Revision as of 11:06, 17 December 2021
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
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.