Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2001 AMC 10 Problems/Problem 15

Revision as of 16:00, 16 March 2011 by Pidigits125 (talk | contribs) (→‎Solution)

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 $600$.

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

[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)--(7,4),dashed) [/asy]

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems/Problem_15&oldid=37526"