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This problem is from the 2001 AMC 10. It is Problem 15.

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 $

What is the difference between the distance between the stripes and the length of the curb between the stripes? My diagram: enter image description here

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    I take it as the perpendicular distance from one stripe to the other. – 2017-01-07

2 Answers 2

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enter image description here

The distance between the strips is the distance between those parallel lines and the length of the curb is the distance which is measured through the given line i.e. the road.

$\triangle ABE$ is similar to $\triangle CDE$

$AB$ is the distance between strips.

Therefore,

$\displaystyle \ \ \ \ \ \frac{CD}{AB}=\frac{CE}{AE}$

$\displaystyle \Rightarrow\frac{40}{AB}=\frac{50}{15}$

$\displaystyle \Rightarrow AB=12$

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    What did you label as 10? – 2017-01-07
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    Are you referring the $10$ near $A$ or $D$, Actually that is $\angle\theta$ – 2017-01-07
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    Oh ok. How do you know the triangles are similar? – 2017-01-07
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    $\angle CEA$ is common to both right triangles $\Rightarrow$ $\angle BAE=\angle EAD$. Hence by AAA similarity condition triangles are similar. – 2017-01-07
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The crosswalk is a parallelogram with sides $15$ and $50$. The height corresponding to the small side is $40$. Let $x$ be the height corresponding to the longer side. The area formula gives $15 \cdot 40 = 50 \cdot x$.