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

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

$draw((0,0)--(5,0),linewidth(2)); draw((2.5,5)--(7.5,5)); draw((0,0)--(2.5,5)); draw((5,0)--(7.5,5)); draw((2.5,5)--(2.5,0),dashed)$

If we look at it the other way, we see the distance between the stripes is the height and the base is $50$ . The area is obviously still the same, so the distance between the stripes is $600 \div 50 = \boxed{\textbf{(C)}\ 12}$ .

$draw((0,0)--(5,0));draw((2.5,5)--(7.5,5));draw((0,0)--(2.5,5));draw((5,0)--(7.5,5),linewidth(2));draw((2,4)--(7,4),dashed)$

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2001 AMC 10 Problems/Problem 15

Problem

Solution