Difference between revisions of "2008 AMC 12A Problems/Problem 14"

Revision as of 01:37, 26 April 2008

Problem

What is the area of the region defined by the inequality $|3x-18|+|2y+7|\le3$ ?

$\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ \frac {7}{2}\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ \frac{9}{2}\qquad\mathrm{(E)}\ 5$

Solution

Area is invariant under translation, so after translating left $6$ and up $3/2$ units, we have the inequality

$|3x| + |2y|\leq 3$

which forms a diamond centered at the origin and vertices at $(\pm 1, 0), (0, \pm 1.5)$ . Thus the diagonals are of length $2$ and $3$ . Using the formula $A = \frac 12 d_1 d_2$ , the answer is $\frac{1}{2}(2)(3) = 3 \Rightarrow \mathrm{(A)}$ .

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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

@@ Line 1: / Line 1: @@
 == Problem ==
-What is the area of the region defined by the [[inequality]] <math>|3x - 18| + |2y + 7|\le 3</math>?
+What is the area of the region defined by the [[inequality]] <math>|3x-18|+|2y+7|\le3</math>?
-<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac {7}{2} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac {9}{2} \qquad \textbf{(E)}\ 5</math>
+<math>\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ \frac {7}{2}\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ \frac{9}{2}\qquad\mathrm{(E)}\ 5</math>
 == Solution ==

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Difference between revisions of "2008 AMC 12A Problems/Problem 14"

Revision as of 01:37, 26 April 2008

Problem

Solution

See also