Difference between revisions of "2005 AMC 12B Problems/Problem 7"
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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>): | ||
− | + | <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. | ||
− | + | <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> | ||
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== 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 ?
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 ):
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:
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.