Difference between revisions of "2001 AMC 10 Problems/Problem 15"
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<math> \textbf{(A)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 25 </math> | <math> \textbf{(A)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 25 </math> | ||
| − | == | + | == Solution 1 == |
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| + | Drawing the problem out, we see we get a parallelogram with a height of <math> 40 </math> and a base of <math> 15 </math>, giving an area of <math> 600 </math>. | ||
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If we look at it the other way, we see the distance between the stripes is the height and the base is <math> 50 </math>. | If we look at it the other way, we see the distance between the stripes is the height and the base is <math> 50 </math>. | ||
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The area is still the same, so the distance between the stripes is <math> 600/50 = \boxed{\textbf{(C)}\ 12} </math>. | The area is still the same, so the distance between the stripes is <math> 600/50 = \boxed{\textbf{(C)}\ 12} </math>. | ||
| − | + | == Solution 2 == | |
Alternatively, we could use similar triangles--the <math> 30-40-50 </math> triangle (created by the length of the bordering stripe and the difference between the two curbs) is similar to the <math> x-y-15 </math> triangle, where we are trying to find <math> y </math> (the shortest distance between the two stripes). Therefore, <math> y </math> would have to be <math> \boxed{\textbf{(C)}\ 12} </math>. | Alternatively, we could use similar triangles--the <math> 30-40-50 </math> triangle (created by the length of the bordering stripe and the difference between the two curbs) is similar to the <math> x-y-15 </math> triangle, where we are trying to find <math> y </math> (the shortest distance between the two stripes). Therefore, <math> y </math> would have to be <math> \boxed{\textbf{(C)}\ 12} </math>. | ||
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| + | ==Video Solution by Daily Dose of Math== | ||
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| + | https://youtu.be/1jRoajU-d5E?si=GzO1MEfu_5lqa8pA | ||
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| + | ~Thesmartgreekmathdude | ||
== See Also == | == See Also == | ||
Latest revision as of 20:44, 15 July 2024
Problem
A street has parallel curbs 
feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The length of the curb between the stripes is 
feet and each stripe is 
feet long. Find the distance, in feet, between the stripes.
Solution 1
Drawing the problem out, we see we get a parallelogram with a height of and a base of , giving an area of 
.
If we look at it the other way, we see the distance between the stripes is the height and the base is .
![[asy] draw((0,0)--(5,0)); draw((2.5,5)--(7.5,5)); draw((0,0)--(2.5,5)); draw((5,0)--(7.5,5),linewidth(2)); draw((2,4)--(6,2),dashed);[/asy]](http://latex.artofproblemsolving.com/2/2/c/22cc747ff3d38f7c11bc0bb8fa93ecba438b001b.png)
The area is still the same, so the distance between the stripes is 
.
Solution 2
Alternatively, we could use similar triangles--the 
triangle (created by the length of the bordering stripe and the difference between the two curbs) is similar to the 
triangle, where we are trying to find 
(the shortest distance between the two stripes). Therefore, would have to be 
.
Video Solution by Daily Dose of Math
https://youtu.be/1jRoajU-d5E?si=GzO1MEfu_5lqa8pA
~Thesmartgreekmathdude
See Also
| 2001 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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. 
