Difference between revisions of "2005 AMC 12B Problems/Problem 5"
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\mbox{shaded area} &= 80 \left( 1 - 4 \cdot \dfrac{1}{4} \cdot \pi \cdot \left(\dfrac{1}{2}\right)^2\right) \\ | \mbox{shaded area} &= 80 \left( 1 - 4 \cdot \dfrac{1}{4} \cdot \pi \cdot \left(\dfrac{1}{2}\right)^2\right) \\ | ||
&= 80\left(1-\dfrac{1}{4}\pi\right) \\ | &= 80\left(1-\dfrac{1}{4}\pi\right) \\ | ||
| − | &= \boxed{80-20\pi}. | + | &= \boxed{(A) 80-20\pi}. |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
Revision as of 19:47, 25 February 2018
- The following problem is from both the 2005 AMC 12B #5 and 2005 AMC 10B #8, so both problems redirect to this page.
Problem
An 
-foot by 
-foot floor is tiles with square tiles of size 
foot by foot. Each tile has a pattern consisting of four white quarter circles of radius 
foot centered at each corner of the tile. The remaining portion of the tile is shaded. How many square feet of the floor are shaded?
![[asy] unitsize(2cm); defaultpen(linewidth(.8pt)); fill(unitsquare,gray); filldraw(Arc((0,0),.5,0,90)--(0,0)--cycle,white,black); filldraw(Arc((1,0),.5,90,180)--(1,0)--cycle,white,black); filldraw(Arc((1,1),.5,180,270)--(1,1)--cycle,white,black); filldraw(Arc((0,1),.5,270,360)--(0,1)--cycle,white,black); [/asy]](http://latex.artofproblemsolving.com/c/b/e/cbe2bb4966beb1c1cbc889938c8e120b7d80639b.png)
Solution
There are 80 tiles. Each tile has ![$[\mbox{square} - 4 \cdot (\mbox{quarter circle})]$](http://latex.artofproblemsolving.com/6/1/7/617c9a81577193c333f0b80c64514805a1e55857.png)
shaded. Thus:
See also
| 2005 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 | ||
| 2005 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
