AMC12 2008 A

AMC12 2008 A · Q14

AMC12 2008 A · Q14. It mainly tests Absolute value, Area & perimeter.

What is the area of the region defined by the inequality $|3x-18|+|2y+7|\le3$?

不等式 $|3x-18|+|2y+7|\le3$ 定义的区域的面积是多少？

(A) 3 3

(B) \frac{7}{2} \frac{7}{2}

(D) \frac{9}{2} \frac{9}{2}

(E) 5 5

Answer

Correct choice: (A)

正确答案：（A）

Solution

Area is invariant under translation, so after translating left $6$ and up $7/2$ units, we have the inequality \[|3x| + |2y|\leq 3\] which forms a diamond centered at the origin and vertices at $(\pm 1, 0), (0, \pm 1.5)$. Thus the diagonals are of length $2$ and $3$. Using the formula $A = \frac 12 d_1 d_2$, the answer is $\frac{1}{2}(2)(3) = 3 \Rightarrow \mathrm{(A)}$.

面积在平移下保持不变，因此向左平移 $6$ 个单位、向上平移 $7/2$ 个单位后，不等式变为 \[|3x| + |2y|\leq 3\] 它形成一个以原点为中心、顶点在 $(\pm 1, 0), (0, \pm 1.5)$ 的菱形。因此两条对角线长度分别为 $2$ 和 $3$。用公式 $A = \frac 12 d_1 d_2$，答案为 $\frac{1}{2}(2)(3) = 3 \Rightarrow \mathrm{(A)}$。

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