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2008 AMC 12A Problem 14

Problem 14 of 25IntermediateAlgebraGeometry

What is the area of the region defined by the inequality ∣3x−18∣+∣2y+7∣≤3?|3x - 18| + |2y + 7| \le 3?

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Solution

The region is a rhombus centered at (6,−72).\left(6, -\tfrac{7}{2}\right). Setting 2y+7=02y + 7 = 0 gives ∣3x−18∣≤3,|3x - 18| \le 3, so x∈[5,7],x \in [5, 7], a horizontal diagonal of length 2.2. Setting 3x−18=03x - 18 = 0 gives ∣2y+7∣≤3,|2y + 7| \le 3, so y∈[−5,−2],y \in [-5, -2], a vertical diagonal of length 3.3. The area of the rhombus is half the product of its diagonals, 12⋅2⋅3=3. \dfrac{1}{2} \cdot 2 \cdot 3 = 3. Thus, A is the correct answer.
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Tagged: absolute value · rhombus · area

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