2005 AMC 12B Problem 7

Attempt Problem 7 of the 2005 AMC 12B 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 2005 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

7.

What is the area enclosed by the graph of ∣3x∣+∣4y∣=12?|3x| + |4y| = 12?
66
1212
1616
2424
2525
Answer: D
Concepts:absolute valuerhombus
Small hint:
Find the intercepts on the two axes
Big hint:
The graph is a rhombus with diagonals along the axes
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.
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