Difference between revisions of "2004 AMC 12B Problems/Problem 7"
m (→Problem) |
m (→Solution) |
||
| Line 8: | Line 8: | ||
== Solution == | == Solution == | ||
| − | The area of the circle is <math>S_{\bigcirc}=100\pi</math> | + | The area of the circle is <math>S_{\bigcirc}=100\pi</math>; the area of the square is <math>S_{\square}=100</math>. |
| − | Exactly <math>1 | + | Exactly <math>\frac{1}{4}</math> of the circle lies inside the square. Thus the total area is <math>\dfrac34 S_{\bigcirc}+S_{\square}=\boxed{\mathrm{(B)\ }100+75\pi}</math>. |
<asy> | <asy> | ||
Revision as of 23:53, 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 
.
![[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 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. 
