Difference between revisions of "2001 AMC 10 Problems/Problem 15"

Latest revision as of 20:44, 15 July 2024

Problem

A street has parallel curbs $40$ feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The length of the curb between the stripes is $15$ feet and each stripe is $50$ feet long. Find the distance, in feet, between the stripes.

$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 25$

Solution 1

Drawing the problem out, we see we get a parallelogram with a height of $40$ and a base of $15$ , giving an area of $600$ .

If we look at it the other way, we see the distance between the stripes is the height and the base is $50$ .

$[asy] draw((0,0)--(5,0)); draw((2.5,5)--(7.5,5)); draw((0,0)--(2.5,5)); draw((5,0)--(7.5,5),linewidth(2)); draw((2,4)--(6,2),dashed);[/asy]$

The area is still the same, so the distance between the stripes is $600/50 = \boxed{\textbf{(C)}\ 12}$ .

Solution 2

Alternatively, we could use similar triangles--the $30-40-50$ triangle (created by the length of the bordering stripe and the difference between the two curbs) is similar to the $x-y-15$ triangle, where we are trying to find $y$ (the shortest distance between the two stripes). Therefore, $y$ would have to be $\boxed{\textbf{(C)}\ 12}$ .

Video Solution by Daily Dose of Math

https://youtu.be/1jRoajU-d5E?si=GzO1MEfu_5lqa8pA

~Thesmartgreekmathdude

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.

@@ Line 9: / Line 9: @@
 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>.
-<asy>
-draw((0,0)--(5,0),linewidth(2));
-draw((2.5,5)--(7.5,5));
-draw((0,0)--(2.5,5));
-draw((5,0)--(7.5,5));
-draw((2.5,5)--(2.5,0),dashed);</asy>
 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>.
@@ Line 30: / Line 26: @@
 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>.
+==Video Solution by Daily Dose of Math==
+https://youtu.be/1jRoajU-d5E?si=GzO1MEfu_5lqa8pA
+~Thesmartgreekmathdude
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