2008 AMC 12A Problem 14

Attempt Problem 14 of the 2008 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 12A solutions, or check the answer key.

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14.

What is the area of the region defined by the inequality

|3x - 18| + |2y + 7| \le 3?

3

\dfrac{7}{2}

4

\dfrac{9}{2}

5

The region is a rhombus centered where

3x - 18 = 0

and

2y + 7 = 0

Its horizontal diagonal spans

x \in [5, 7]

and its vertical diagonal spans

y \in [-5, -2]

The region is a rhombus centered at

\left(6, -\tfrac{7}{2}\right).

Setting

2y + 7 = 0

gives

|3x - 18| \le 3,

x \in [5, 7],

a horizontal diagonal of length

2.

Setting

3x - 18 = 0

gives

|2y + 7| \le 3,

y \in [-5, -2],

a vertical diagonal of length

3.

The area of the rhombus is half the product of its diagonals,

\dfrac{1}{2} \cdot 2 \cdot 3 = 3.

Thus, A is the correct answer.

Problem 14 in other years