Difference between revisions of "2005 AMC 12B Problems/Problem 5"

(Problem)
(Solution)
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== Solution ==
 
== Solution ==
 +
There are 80 tiles.  Each tile has <math>[\mbox{square} - 4 \cdot (\mbox{quarter circle})]</math> shaded.  Thus:
 +
 +
<cmath>
 +
\begin{align*}
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\mbox{shaded area} &= 80 ( 1 - 4 \cdot (1/4) \cdot \pi \cdot (1/2)^2) \\
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&= 80(1-(1/4)\pi) \\
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&= \boxed{80-20\pi}.
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\end{align*}
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</cmath>
  
 
== See also ==
 
== See also ==
 
* [[2005 AMC 12B Problems]]
 
* [[2005 AMC 12B Problems]]

Revision as of 22:34, 17 April 2009

Problem

An $8$-foot by $10$-foot floor is tiles 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]

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

Solution

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

\begin{align*} \mbox{shaded area} &= 80 ( 1 - 4 \cdot (1/4) \cdot \pi \cdot (1/2)^2) \\ &= 80(1-(1/4)\pi) \\ &= \boxed{80-20\pi}. \end{align*}

See also