2005 AMC 12B Problems/Problem 7

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 1

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 $\boxed{\mathrm{(D)}\ 24}$

Solution 2

You can also assign $x$ and $y$ to be $0$ . Then you can easily see that the diagonals are $6$ and $8$ . Multiply and divide by $2$ to get D. $24$ --zoevv

2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AMC 12 Problems and Solutions

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2005 AMC 12B Problems/Problem 7

Contents

Problem

Solution 1

Solution 2

See also