2005 AMC 12B Problem 7 — Solution | LIVE by Po-Shen Loh

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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 $|3x| + |4y| = 12?$

$6$

$12$

$16$

$24$

$25$

Answer: D

Concepts:absolute value rhombus

Difficulty rating: 1270

Solution:

Setting $y = 0$ gives $|3x| = 12,$ so $x = \pm 4.$ Setting $x = 0$ gives $|4y| = 12,$ so $y = \pm 3.$

The graph is a rhombus with vertices $(\pm 4, 0)$ and $(0, \pm 3),$ so its diagonals have lengths $8$ and $6.$

Its area is $\dfrac12 \cdot 8 \cdot 6 = 24.$

Thus, the correct answer is D.

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