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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]]}}
 
{{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?
+
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
+
\textbf{(A) }\ 80-20\pi      \qquad
−
\mathrm{(B)}\ 60-10\pi      \qquad
+
\textbf{(B) }\ 60-10\pi      \qquad
−
\mathrm{(C)}\ 80-10\pi      \qquad
+
\textbf{(C) }\ 80-10\pi      \qquad
−
\mathrm{(D)}\ 60+10\pi      \qquad
+
\textbf{(D) }\ 60+10\pi      \qquad
−
\mathrm{(E)}\ 80+10\pi
+
\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:
+
There are <math>80</math> tiles.  Each tile has <math>[\mbox{square} - 4 \cdot (\mbox{quarter circle})]</math> shaded.  Thus:
  
 
<cmath>
 
<cmath>
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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{(A) 80-20\pi}.
+
&= \boxed{\textbf{(A) }80-20\pi}.
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
−
4 quarters of a circle is a circle so that may save you 0.5 seconds :)
+
4 quarters of a circle is a circle so that may save you 0.5 seconds (very much lots) :)
  
 
== See also ==
 
== See also ==

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 (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. AMC logo.png

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