2005 AMC 12B Problems/Problem 7
Problem
What is the area enclosed by the graph of 
?
Solution 1
If we get rid of the absolute values, we are left with the following 4 equations (using the logic that if 
, then 
is either 
or 
):
We can then put these equations in slope-intercept form in order to graph them.
Now you can graph the lines to find the shape of the graph:
![[asy] Label f; f.p=fontsize(6); xaxis(-8,8,Ticks(f, 4.0)); yaxis(-6,6,Ticks(f, 3.0)); fill((0,-3)--(4,0)--(0,3)--(-4,0)--cycle,grey); draw((-4,-6)--(8,3), Arrows(4)); draw((4,-6)--(-8,3), Arrows(4)); draw((-4,6)--(8,-3), Arrows(4)); draw((4,6)--(-8,-3), Arrows(4));[/asy]](http://latex.artofproblemsolving.com/a/1/9/a19def037694543a1771b3897d68d519dd066fcf.png)
We can easily see that it is a rhombus with diagonals of 
and 
. The area is 
, or 
Solution 2
You can also assign 
and 
to be 
. Then you can easily see that the diagonals are and . Multiply and divide by 
to get D. 
Solution 3
The graph is symmetric with respect to both coordinate axes, and in the first quadrant it coincides with the graph of the line 
Therefore the region is a rhombus, and the area is\[
\text{Area} = 4\left(\frac{1}{2}(4\cdot 3)\right) = 24 \rightarrow \boxed{D}.
\][asy]
draw((-5,0)--(5,0),Arrow);
draw((0,-4)--(0,4),Arrow);
label("",(5,0),S);
label("",(0,4),E);
label("4",(4,0),S);
label("-4",(-4,0),S);
label("3",(0,3),NW);
label("-3",(0,-3),SW);
draw((4,0)--(0,3)--(-4,0)--(0,-3)--cycle,linewidth(0.7));
[/asy]
~Alcumus
See also
| 2005 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 6 |
Followed by Problem 8 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
