Difference between revisions of "2004 AMC 10A Problems/Problem 19"
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==Solution== | ==Solution== | ||
| − | The cylinder can be "unwrapped" into a rectangle, and we see that the stripe is a parallelogram with base <math>3</math> and height <math>80</math>. <math>3\times80=240\Rightarrow\boxed{\mathrm{(C)}\ 240}</math> | + | The cylinder can be "unwrapped" into a rectangle, and we see that the stripe is a parallelogram with base <math>3</math> and height <math>80</math>. Thus, we get <math>3\times80=240\Rightarrow\boxed{\mathrm{(C)}\ 240}</math> |
==See also== | ==See also== | ||
Revision as of 14:57, 12 February 2017
Problem
A white cylindrical silo has a diameter of 30 feet and a height of 80 feet. A red stripe with a horizontal width of 3 feet is painted on the silo, as shown, making two complete revolutions around it. What is the area of the stripe in square feet?
![[asy] size(250);defaultpen(linewidth(0.8)); draw(ellipse(origin, 3, 1)); fill((3,0)--(3,2)--(-3,2)--(-3,0)--cycle, white); draw((3,0)--(3,16)^^(-3,0)--(-3,16)); draw((0, 15)--(3, 12)^^(0, 16)--(3, 13)); filldraw(ellipse((0, 16), 3, 1), white, black); draw((-3,11)--(3, 5)^^(-3,10)--(3, 4)); draw((-3,2)--(0,-1)^^(-3,1)--(-1,-0.89)); draw((0,-1)--(0,15), dashed); draw((3,-2)--(3,-4)^^(-3,-2)--(-3,-4)); draw((-7,0)--(-5,0)^^(-7,16)--(-5,16)); draw((3,-3)--(-3,-3), Arrows(6)); draw((-6,0)--(-6,16), Arrows(6)); draw((-2,9)--(-1,9), Arrows(3)); label("$3$", (-1.375,9.05), dir(260), fontsize(7)); label("$A$", (0,15), N); label("$B$", (0,-1), NE); label("$30$", (0, -3), S); label("$80$", (-6, 8), W);[/asy]](http://latex.artofproblemsolving.com/d/b/0/db0ee67d222e80fda19ce7927dc5efb178a81aff.png)
Solution
The cylinder can be "unwrapped" into a rectangle, and we see that the stripe is a parallelogram with base 
and height 
. Thus, we get 
See also
| 2004 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
