1995 AHSME Problems/Problem 10
Revision as of 13:59, 5 July 2013 by Nathan wailes (talk | contribs)
Problem
The area of the triangle bounded by the lines and
is
Solution
![[asy] defaultpen(fontsize(8)); draw((0,0)--(6,6)--(-6,6)--(0,0)); draw((0,-1)--(0,8)); draw((-7,0)--(7,0)); label("$(6,6)$",(6,6), (1,1));label("$(-6,6)$",(-6,6),(-1,1)); label("$y=x$",(3,3),(1,-1));label("$y=-x$",(-3,3),(-1,-0.5)); label("$6$",(-3,6),(0,1));label("$6$",(3,6),(0,1)); label("$6$",(0,3),(-1,0)); [/asy]](http://latex.artofproblemsolving.com/6/8/6/6863401960a6ad7e98ced6f1e4b5c85e86470a46.png)
The height of the triangle is , and the width of the triangle is
. Thus the area of the triangle is
.
See also
1995 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.