Difference between revisions of "2005 AMC 12B Problems/Problem 7"

m (Solution)
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If we get rid of the absolute values, we are left with the following 4 equations (using the logic that if <math>|a|=b</math>, then <math>a</math> is either <math>b</math> or <math>-b</math>):
 
If we get rid of the absolute values, we are left with the following 4 equations (using the logic that if <math>|a|=b</math>, then <math>a</math> is either <math>b</math> or <math>-b</math>):
  
<math><cmath>\begin{align*} 3x+4y=12 \\ -3x+4y=12 \\ 3x-4y=12 \\ -3x-4y=12 \end{align*}</cmath></math>
+
<cmath>\begin{align*} 3x+4y=12 \\ -3x+4y=12 \\ 3x-4y=12 \\ -3x-4y=12 \end{align*}</cmath>
  
 
We can then put these equations in slope-intercept form in order to graph them.
 
We can then put these equations in slope-intercept form in order to graph them.
  
<math><cmath>\begin{align*} 3x+4y=12 \,\implies\, y=-\dfrac{3}{4}x+3\\ -3x+4y=12\,\implies\, y=\dfrac{3}{4}x+3\\ 3x-4y=12\,\implies\, y=\dfrac{3}{4}x-3\\ -3x-4y=12\,\implies\, y=-\dfrac{3}{4}x-3\end{align*}</cmath></math>
+
<cmath>\begin{align*} 3x+4y=12 \,\implies\, y=-\dfrac{3}{4}x+3\\ -3x+4y=12\,\implies\, y=\dfrac{3}{4}x+3\\ 3x-4y=12\,\implies\, y=\dfrac{3}{4}x-3\\ -3x-4y=12\,\implies\, y=-\dfrac{3}{4}x-3\end{align*}</cmath>
  
 
Now you can graph the lines to find the shape of the graph:
 
Now you can graph the lines to find the shape of the graph:
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We can easily see that it is a rhombus with diagonals of <math>6</math> and <math>8</math>. The area is <math>\dfrac{1}{2}\times 6\times8</math>, or <math>\boxed{\mathrm{(D)}\ 24}</math>
 
We can easily see that it is a rhombus with diagonals of <math>6</math> and <math>8</math>. The area is <math>\dfrac{1}{2}\times 6\times8</math>, or <math>\boxed{\mathrm{(D)}\ 24}</math>
 +
 
== 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 13:57, 1 June 2015

Problem

What is the area enclosed by the graph of $|3x|+|4y|=12$?

$\mathrm{(A)}\ 6      \qquad \mathrm{(B)}\ 12      \qquad \mathrm{(C)}\ 16      \qquad \mathrm{(D)}\ 24      \qquad \mathrm{(E)}\ 25$

Solution

If we get rid of the absolute values, we are left with the following 4 equations (using the logic that if $|a|=b$, then $a$ is either $b$ or $-b$):

\begin{align*} 3x+4y=12 \\ -3x+4y=12 \\ 3x-4y=12 \\ -3x-4y=12 \end{align*}

We can then put these equations in slope-intercept form in order to graph them.

\begin{align*} 3x+4y=12 \,\implies\, y=-\dfrac{3}{4}x+3\\ -3x+4y=12\,\implies\, y=\dfrac{3}{4}x+3\\ 3x-4y=12\,\implies\, y=\dfrac{3}{4}x-3\\ -3x-4y=12\,\implies\, y=-\dfrac{3}{4}x-3\end{align*}

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]

We can easily see that it is a rhombus with diagonals of $6$ and $8$. The area is $\dfrac{1}{2}\times 6\times8$, or $\boxed{\mathrm{(D)}\ 24}$

See also

2005 AMC 12B (ProblemsAnswer KeyResources)
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

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