Difference between revisions of "2007 AMC 10A Problems/Problem 15"
Mariekitty (talk | contribs) m (→Solution 1) |
Mariekitty (talk | contribs) m (→Alternate Solution) |
||
Line 14: | Line 14: | ||
<cmath>A=(2+3\sqrt{2})^2=4+6\sqrt{2}+6\sqrt{2}+18=22+12\sqrt{2} \Rightarrow \text{(B)}</cmath> | <cmath>A=(2+3\sqrt{2})^2=4+6\sqrt{2}+6\sqrt{2}+18=22+12\sqrt{2} \Rightarrow \text{(B)}</cmath> | ||
− | == | + | === Solution 2 === |
+ | |||
+ | ---- | ||
+ | |||
Extend two radii from the larger circle to the centers of the two smaller circles above. This forms a right triangle of sides <math>3, 3, 3\sqrt{2}</math>. The length of the hypotenuse of the right triangle plus twice the radius of the smaller circle is equal to the side of the square. It follows, then <cmath> A = (2+3\sqrt{2})^2 = 22 + 12\sqrt{2} \Rightarrow \text{(B)}</cmath> | Extend two radii from the larger circle to the centers of the two smaller circles above. This forms a right triangle of sides <math>3, 3, 3\sqrt{2}</math>. The length of the hypotenuse of the right triangle plus twice the radius of the smaller circle is equal to the side of the square. It follows, then <cmath> A = (2+3\sqrt{2})^2 = 22 + 12\sqrt{2} \Rightarrow \text{(B)}</cmath> |
Revision as of 11:33, 31 December 2011
Contents
[hide]Problem
Four circles of radius are each tangent to two sides of a square and externally tangent to a circle of radius , as shown. What is the area of the square?
Solution 1
The diagonal has length . Therefore the sides have length , and the area is
Solution 2
Extend two radii from the larger circle to the centers of the two smaller circles above. This forms a right triangle of sides . The length of the hypotenuse of the right triangle plus twice the radius of the smaller circle is equal to the side of the square. It follows, then
See Also
2007 AMC 10A (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 10 Problems and Solutions |