Difference between revisions of "1995 AHSME Problems/Problem 10"
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== Solution == | == Solution == | ||
| − | + | <center><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></center> | ||
The height of the triangle is <math>y = 6</math>, and the width of the triangle is <math>|x_1| + |x_2| = 2y = 12</math>. Thus the area of the triangle is <math>\frac 12 \cdot 6 \cdot 12 = 36\ \mathrm{(E)}</math>. | The height of the triangle is <math>y = 6</math>, and the width of the triangle is <math>|x_1| + |x_2| = 2y = 12</math>. Thus the area of the triangle is <math>\frac 12 \cdot 6 \cdot 12 = 36\ \mathrm{(E)}</math>. | ||
Revision as of 18:18, 5 July 2008
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 | ||
