2005 AMC 12B Problems/Problem 7
Problem
What is the area enclosed by the graph of ?
Solution
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
):
$<cmath>\begin{align*} 3x+4y=12 \\ -3x+4y=12 \\ 3x-4y=12 \\ -3x-4y=12 \end{align*}</cmath>$ (Error compiling LaTeX. Unknown error_msg)
We can then put these equations in slope-intercept form in order to graph them.
$<cmath>\begin{align*} 3x+4y=12 \,\implies\, y=-\dfrac{3}{4}x+3\\ -3x+4y=12\,\implies\, y=\dfrac{3}{4}x+3\\ 3x-4y=12\,\implies\, y=\dfrac{3}{4}x-3\\ -3x-4y=12\,\implies\, y=-\dfrac{3}{4}x-3\end{align*}</cmath>$ (Error compiling LaTeX. Unknown error_msg)
Now you can graph the lines to find the shape of the graph:
We can easily see that it is a rhombus with diagonals of and
. The area is
, or
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 |
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