Difference between revisions of "2016 AMC 10A Problems/Problem 21"
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<math>\frac{1}{2}|(2\sqrt{2}+4\sqrt{2}+4\sqrt{6})-(6\sqrt{2}+2\sqrt{2}+2\sqrt{6})|</math> | <math>\frac{1}{2}|(2\sqrt{2}+4\sqrt{2}+4\sqrt{6})-(6\sqrt{2}+2\sqrt{2}+2\sqrt{6})|</math> | ||
− | <math>= \sqrt{6}-\sqrt{2} \fbox{D}</math>. | + | <math>= \sqrt{6}-\sqrt{2}</math> <math>\fbox{D}</math>. |
==See Also== | ==See Also== |
Revision as of 21:44, 15 February 2016
Circles with centers and , having radii and , respectively, lie on the same side of line and are tangent to at and , respectively, with between and . The circle with center is externally tangent to each of the other two circles. What is the area of triangle ?
Solution
Notice that we can find in two different ways: and , so
. Additionally, . Therefore, . Similarly, . We can calculate easily because . .
Plugging into first equation, the two sums of areas, .
.
Solution 2
Use the Shoelace Thm!
Let the center of the first circle of radius 1 be at (0, 1).
Draw the trapezoid and using Pythagorean Thm., we get that so the center of the second circle of radius 2 is at .
Draw the trapezoid and using Pythagorean Thm., we get that so the center of the third circle of radius 3 is at .
Now, we may use the Shoelace Thm!
.
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 |
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 12 Problems and Solutions |
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