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 floor is tiles 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?
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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>
 
<asy>
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<math>
 
<math>
\mathrm{(A)}\ 80-20\pi      \qquad
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\textbf{(A) }\ 80-20\pi      \qquad
\mathrm{(B)}\ 60-10\pi      \qquad
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\textbf{(B) }\ 60-10\pi      \qquad
\mathrm{(C)}\ 80-10\pi      \qquad
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\textbf{(C) }\ 80-10\pi      \qquad
\mathrm{(D)}\ 60+10\pi      \qquad
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\textbf{(D) }\ 60+10\pi      \qquad
\mathrm{(E)}\ 80+10\pi
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\textbf{(E) }\ 80+10\pi
 
</math>
 
</math>
  
 
== Solution ==
 
== Solution ==
There are 80 tiles.  Each tile has <math>[\mbox{square} - 4 \cdot (\mbox{quarter circle})]</math> shaded.  Thus:
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There are <math>80</math> tiles.  Each tile has <math>[\mbox{square} - 4 \cdot (\mbox{quarter circle})]</math> shaded.  Thus:
  
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
\mbox{shaded area} &= 80 ( 1 - 4 \cdot (1/4) \cdot \pi \cdot (1/2)^2) \\
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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) \\
&= 80(1-(1/4)\pi) \\
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&= 80\left(1-\dfrac{1}{4}\pi\right) \\
&= \boxed{80-20\pi}.
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&= \boxed{\textbf{(A) }80-20\pi}.
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
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4 quarters of a circle is a circle so that may save you 0.5 seconds (very much lots) :)
  
 
== See also ==
 
== See also ==
* [[2005 AMC 12B Problems]]
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{{AMC10 box|year=2005|ab=B|num-b=7|num-a=9}}
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{{AMC12 box|year=2005|ab=B|num-b=4|num-a=6}}
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[[Category:Introductory Geometry Problems]]
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[[Category:Area Problems]]
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{{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 $8$-foot by $10$-foot bathroom floor is tiled with square tiles of size $1$ foot by $1$ foot. Each tile has a pattern consisting of four white quarter circles of radius $1/2$ 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]

$\textbf{(A) }\ 80-20\pi      \qquad \textbf{(B) }\ 60-10\pi      \qquad \textbf{(C) }\ 80-10\pi      \qquad \textbf{(D) }\ 60+10\pi      \qquad \textbf{(E) }\ 80+10\pi$

Solution

There are $80$ tiles. Each tile has $[\mbox{square} - 4 \cdot (\mbox{quarter circle})]$ shaded. Thus:

\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*} 4 quarters of a circle is a circle so that may save you 0.5 seconds (very much lots) :)

See also

2005 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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