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

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

Latest revision as of 09:28, 13 March 2016

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 $7/2$ 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 $\frac{1}{2}(2)(3) = 3 \Rightarrow \mathrm{(A)}$.

See also

2008 AMC 12A (ProblemsAnswer KeyResources)
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

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