Difference between revisions of "2005 AMC 12B Problems/Problem 5"
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| + | {{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #5]] and [[2005 AMC 10B Problems|2005 AMC 10B #8]]}} | ||
== Problem == | == Problem == | ||
| + | An <math>8 </math>-foot by <math>10</math>-foot bathroom floor is tiled with square tiles of size <math>1</math> foot by <math>1</math> foot. Each tile has a pattern consisting of four white quarter circles of radius <math>1/2</math> 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> | ||
| + | |||
| + | <math> | ||
| + | \textbf{(A) }\ 80-20\pi \qquad | ||
| + | \textbf{(B) }\ 60-10\pi \qquad | ||
| + | \textbf{(C) }\ 80-10\pi \qquad | ||
| + | \textbf{(D) }\ 60+10\pi \qquad | ||
| + | \textbf{(E) }\ 80+10\pi | ||
| + | </math> | ||
== Solution == | == Solution == | ||
| + | There are <math>80</math> tiles. Each tile has <math>[\mbox{square} - 4 \cdot (\mbox{quarter circle})]</math> shaded. Thus: | ||
| + | |||
| + | <cmath> | ||
| + | \begin{align*} | ||
| + | \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) \\ | ||
| + | &= \boxed{\textbf{(A) }80-20\pi}. | ||
| + | \end{align*} | ||
| + | </cmath> | ||
| + | 4 quarters of a circle is a circle so that may save you 0.5 seconds (very much lots) :) | ||
== See also == | == See also == | ||
| − | + | {{AMC10 box|year=2005|ab=B|num-b=7|num-a=9}} | |
| + | {{AMC12 box|year=2005|ab=B|num-b=4|num-a=6}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
| + | [[Category:Area Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 14:09, 23 June 2024
- 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 bathroom floor is tiled 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 
tiles. Each tile has ![$[\mbox{square} - 4 \cdot (\mbox{quarter circle})]$](http://latex.artofproblemsolving.com/6/1/7/617c9a81577193c333f0b80c64514805a1e55857.png)
shaded. Thus:
4 quarters of a circle is a circle so that may save you 0.5 seconds (very much lots) :)
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. 
