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

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

(→‎Solution)
Line 5: Line 5:
 
<math> \textbf{(A)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 25 </math>
 
<math> \textbf{(A)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 25 </math>
  
−
== Solution ==
+
== Solutions ==
 +
 
 +
=== Solution 1 ===
  
 
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>.
 
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));
+
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>. The area is obviously still the same, so the distance between the stripes is <math> 600 \div 50 = \boxed{\textbf{(C)}\ 12} </math>.
−
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>. The area is obviously still the same, so the distance between the stripes is <math> 600 \div 50 = \boxed{\textbf{(C)}\ 12} </math>.
+
=== Solution 2 ===
  
−
[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]
+
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 $ \boxed{\textbf{(C)}\ 12}.

Revision as of 16:02, 16 March 2011

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.

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

Solutions

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

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}.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems/Problem_15&oldid=37527"