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Difference between revisions of "2004 AMC 10A Problems/Problem 19"

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==Solution 1==
 
==Solution 1==
 
The cylinder can be "unwrapped" into a rectangle, and we see that there are two stripes which is a parallelogram with base <math>3</math> and height <math>40</math>, each. Thus, we get <math>3\times40\times2=240\Rightarrow\boxed{\mathrm{(C)}\ 240}</math>
 
The cylinder can be "unwrapped" into a rectangle, and we see that there are two stripes which is a parallelogram with base <math>3</math> and height <math>40</math>, each. Thus, we get <math>3\times40\times2=240\Rightarrow\boxed{\mathrm{(C)}\ 240}</math>
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==Solution 2 (Complement Counting)== Like in Solution 1 "unwrap" the lateral surface of the cylinder into a rectangle, the width of the rectangle is the circumference of the circular base which is <math>30\times\pi</math>. And the length is just the height of the cylinder, <math>80</math>. Now if we don't see that the stripe is just a parallelogram we can calculate the area of the two right triangles with legs <math>80</math> and <math>30\pi-3</math>, and subtract our result from the total area of the rectangle to get the area of the stripe. Thus the area of the stripe is <math>80\times30\pi</math> <math>-(2(</math>80\times(30\pi-3))/2)<math>. And simplifying this our answer is </math>3\times40\times2=240\Rightarrow\boxed{\mathrm{(C)}\ 240}$ ~blankbox
 
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 20:21, 13 August 2025

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]

$\mathrm{(A) \ } 120 \qquad \mathrm{(B) \ } 180 \qquad \mathrm{(C) \ } 240 \qquad \mathrm{(D) \ } 360 \qquad \mathrm{(E) \ } 480$

Solution 1

The cylinder can be "unwrapped" into a rectangle, and we see that there are two stripes which is a parallelogram with base $3$ and height $40$, each. Thus, we get $3\times40\times2=240\Rightarrow\boxed{\mathrm{(C)}\ 240}$

Video Solution

https://youtu.be/PYTi6qZUAPw

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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

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