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 "2004 AMC 12B Problems/Problem 6"

m (→‎Problem)
m (→‎Solution)
Line 8: Line 8:
 
== Solution ==
 
== Solution ==
  
−
The directions "southwest" and "southeast" are orthogonal. Thus the described situation is a right triangle with legs 8 miles and 10 miles long. The hypotenuse length is <math>\sqrt{8^2 + 10^2} \sim 12.8</math>, and thus the answer is <math>\mathrm{(A)}</math>.
+
The directions "southwest" and "southeast" are orthogonal. Thus the described situation is a right triangle with legs <math>8</math> miles and <math>10</math> miles long. The hypotenuse length is <math>\sqrt{8^2 + 10^2}\approx12.8</math>, and thus the answer is <math>\boxed{\mathrm{(A)}\ 13}</math>.
  
−
Without a calculator one can note that <math>8^2 + 10^2 = 164 < 169 = 13^2\rightarrow\boxed{A}</math>.
+
Without a calculator one can note that <math>8^2+10^2=164<169=13^2\Rightarrow\boxed{\mathrm{(A)}\ 13}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 21:31, 22 July 2014

The following problem is from both the 2004 AMC 12B #6 and 2004 AMC 10B #8, so both problems redirect to this page.

Problem

Minneapolis-St. Paul International Airport is $8$ miles southwest of downtown St. Paul and $10$ miles southeast of downtown Minneapolis. Which of the following is closest to the number of miles between downtown St. Paul and downtown Minneapolis?

$\mathrm{(A)\ }13\qquad\mathrm{(B)\ }14\qquad\mathrm{(C)\ }15\qquad\mathrm{(D)\ }16\qquad\mathrm{(E)\ }17$

Solution

The directions "southwest" and "southeast" are orthogonal. Thus the described situation is a right triangle with legs $8$ miles and $10$ miles long. The hypotenuse length is $\sqrt{8^2 + 10^2}\approx12.8$, and thus the answer is $\boxed{\mathrm{(A)}\ 13}$.

Without a calculator one can note that $8^2+10^2=164<169=13^2\Rightarrow\boxed{\mathrm{(A)}\ 13}$.

See Also

2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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 12 Problems and Solutions
2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_6&oldid=62643"