Difference between revisions of "2005 AMC 12B Problems/Problem 7"
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− | == Solution == | + | == Solution 1== |
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: | ||
Line 33: | Line 33: | ||
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> | ||
+ | |||
+ | == Solution 2== | ||
+ | You can also assign <math>x</math> and <math>y</math> to be <math>0</math>. Then you can easily see that the diagonals are <math>6</math> and <math>8</math>. Multiply and divide by <math>2</math> to get D. <math>24</math> | ||
+ | |||
+ | ==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 <math>3x + 4y = 12.</math> Therefore the region is a rhombus, and the area is | ||
+ | |||
+ | <cmath>\text{Area} = 4\left(\frac{1}{2}(4\cdot 3)\right) = 24 \rightarrow \boxed{D}</cmath> | ||
+ | |||
+ | <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 | ||
+ | |||
== 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}} |
Latest revision as of 20:15, 27 December 2020
Problem
What is the area enclosed by the graph of ?
Solution 1
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
Solution 2
You can also assign and
to be
. Then you can easily see that the diagonals are
and
. Multiply and divide by
to get D.
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 Therefore the region is a rhombus, and the area is
~Alcumus
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.