Skip to main content

2005 AMC 12B Problem 7

Problem 7 of 25EasierAlgebraGeometry

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

Answer choices

Show solution

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 12⋅8⋅6=24.\dfrac12 \cdot 8 \cdot 6 = 24. Thus, the correct answer is D.
AoPS wiki

Tagged: absolute value · rhombus

More practice