2001 AMC 10 Problems/Problem 15

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 $600$ .

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

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

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

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 $\boxed{\textbf{(C)}\ 12}$ .

2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2001 AMC 10 Problems/Problem 15

Contents

Problem

Solution 1

Solution 2

See Also