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

Revision as of 16:17, 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 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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

@@ Line 18: / Line 18: @@
 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 <math> \boxed{\textbf{(C)}\ 12} </math>.
+== See Also ==
+{{AMC10 box|year=2001|num-b=14|num-a=16}}

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