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

Revision as of 20:36, 3 July 2013

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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

Revision as of 20:36, 3 July 2013

Problem

Solution

See also