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2005 AMC 12B Problem 7

Problem 7 of 25EasierAlgebraGeometry

What is the area enclosed by the graph of 3x+4y=12?|3x| + |4y| = 12?

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Solution

Setting y=0y = 0 gives 3x=12,|3x| = 12, so x=±4.x = \pm 4. Setting x=0x = 0 gives 4y=12,|4y| = 12, so y=±3.y = \pm 3. The graph is a rhombus with vertices (±4,0)(\pm 4, 0) and (0,±3),(0, \pm 3), so its diagonals have lengths 88 and 6.6. Its area is 1286=24.\dfrac12 \cdot 8 \cdot 6 = 24. Thus, the correct answer is D.

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Concepts: absolute value · rhombus

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