Difference between revisions of "2005 AMC 12B Problems/Problem 7"
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− | 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 | + | 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 |
− | \text{Area} = 4\left(\frac{1}{2}(4\cdot 3)\right) = 24 \rightarrow \boxed{D} | + | |
− | + | <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((-5,0)--(5,0),Arrow); | ||
draw((0,-4)--(0,4),Arrow); | draw((0,-4)--(0,4),Arrow); |
Latest revision as of 19:15, 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.