# 2005 AMC 12B Problems/Problem 7

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## 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. $24$

## 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 $3x + 4y = 12.$ Therefore the region is a rhombus, and the area is $$\text{Area} = 4\left(\frac{1}{2}(4\cdot 3)\right) = 24 \rightarrow \boxed{D}$$ $[asy] draw((-5,0)--(5,0),Arrow); draw((0,-4)--(0,4),Arrow); label("x",(5,0),S); label("y",(0,4),E); label("4",(4,0),S); label("-4",(-4,0),S); label("3",(0,3),NW); label("-3",(0,-3),SW); draw((4,0)--(0,3)--(-4,0)--(0,-3)--cycle,linewidth(0.7)); [/asy]$

~Alcumus

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 