Difference between revisions of "2004 AMC 12B Problems/Problem 7"
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Exactly <math>1/4</math> of the circle lies inside the square. Thus the total area is <math>\dfrac34 S_{\bigcirc} + S_{\square} = \boxed{100+75\pi} \Longrightarrow \mathrm{(B)}</math>. | Exactly <math>1/4</math> of the circle lies inside the square. Thus the total area is <math>\dfrac34 S_{\bigcirc} + S_{\square} = \boxed{100+75\pi} \Longrightarrow \mathrm{(B)}</math>. | ||
| + | |||
| + | <asy> | ||
| + | Draw(Circle((0,0),10)); | ||
| + | Draw((0,0)--(10,0)--(10,10)--(0,10)--(0,0)); | ||
| + | label("$10$",(5,0),S); | ||
| + | label("$10$",(0,5),W); | ||
| + | dot((0,0)); | ||
| + | </asy> | ||
== See Also == | == See Also == | ||
Revision as of 14:05, 4 July 2012
- The following problem is from both the 2004 AMC 12B #7 and 2004 AMC 10B #9, so both problems redirect to this page.
Problem 7
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?
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 
.
![[asy] Draw(Circle((0,0),10)); Draw((0,0)--(10,0)--(10,10)--(0,10)--(0,0)); label("$10$",(5,0),S); label("$10$",(0,5),W); dot((0,0)); [/asy]](http://latex.artofproblemsolving.com/9/f/c/9fcea2965eceab15b23e988726a4b3446b01b59f.png)
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 12B (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 12 Problems and Solutions | |
