Difference between revisions of "2004 AMC 12B Problems/Problem 7"
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== Problem == | == Problem == | ||
− | A square has sides of length 10, and a circle centered at one of its vertices has radius 10. What is the area of the union of the regions enclosed by the square and the circle? | + | A square has sides of length <math>10</math>, and a circle centered at one of its vertices has radius <math>10</math>. What is the area of the union of the regions enclosed by the square and the circle? |
− | <math> | + | <math>\mathrm{(A)\ }200+25\pi\quad\mathrm{(B)\ }100+75\pi\quad\mathrm{(C)\ }75+100\pi\quad\mathrm{(D)\ }100+100\pi\quad\mathrm{(E)\ }100+125\pi</math> |
== Solution == | == Solution == |
Revision as of 22:50, 23 July 2014
- The following problem is from both the 2004 AMC 12B #7 and 2004 AMC 10B #9, so both problems redirect to this page.
Problem
A square has sides of length , and a circle centered at one of its vertices has radius . What is the area of the union of the regions enclosed by the square and the circle?
Solution
The area of the circle is , the area of the square is .
Exactly of the circle lies inside the square. Thus the total area is .
See Also
2004 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
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 |
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.