2005 AMC 12B Problems/Problem 7: Difference between revisions
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== Solution 2== | == Solution 2== | ||
You can also assign <math>x</math> and <math>y</math> to be <math>0</math>. Then you can easily see that the diagonals are <math>6</math> and <math>8</math>. Multiply and divide by <math>2</math> to get D. <math>24</math> | You can also assign <math>x</math> and <math>y</math> to be <math>0</math>. Then you can easily see that the diagonals are <math>6</math> and <math>8</math>. Multiply and divide by <math>2</math> to get D. <math>24</math> | ||
==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 <math>3x + 4y = 12.</math> 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("<math>x</math>",(5,0),S); | |||
label("<math>y</math>",(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 == | == See also == | ||
{{AMC12 box|year=2005|ab=B|num-b=6|num-a=8}} | {{AMC12 box|year=2005|ab=B|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:15, 27 December 2020
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 | |
These problems are copyrighted © by the Mathematical Association of America. 
