AMC10 2006 B

AMC10 2006 B · Q20

AMC10 2006 B · Q20. It mainly tests Coordinate geometry, Distance / midpoint.

In rectangle ABCD, we have A = (6, -22), B = (2006, 178), and D = (8, y), for some integer y. What is the area of rectangle ABCD?

在矩形 ABCD 中，A = (6, -22)，B = (2006, 178)，D = (8, y)，y 为某整数。矩形 ABCD 的面积是多少？

(A) 4000 4000

(B) 4040 4040

(D) 40,000 40,000

(E) 40,400 40,400

Answer

Correct choice: (E)

正确答案：（E）

Solution

(E) The slope of line $AB$ is $(178-(-22))/(2006-6)=1/10$. Since the line $AD$ is perpendicular to the line $AB$, its slope is $-10$. This implies that \[ -10=\frac{y-(-22)}{8-6}, \] so $y=-10(2)-22=-42$, and $D=(8,-42)$. As a consequence, \[ AB=\sqrt{2000^2+200^2}=200\sqrt{101}\ \text{ and }\ AD=\sqrt{2^2+20^2}=2\sqrt{101}. \] Thus \[ \text{Area}(ABCD)=AB\cdot AD=400\cdot 101=40,400. \]

（E）直线 $AB$ 的斜率为 $(178-(-22))/(2006-6)=1/10$。由于直线 $AD$ 与直线 $AB$ 垂直，其斜率为 $-10$。因此 \[ -10=\frac{y-(-22)}{8-6}, \] 所以 $y=-10(2)-22=-42$，并且 $D=(8,-42)$。因此， \[ AB=\sqrt{2000^2+200^2}=200\sqrt{101}\ \text{且}\ AD=\sqrt{2^2+20^2}=2\sqrt{101}. \] 因此 \[ \text{Area}(ABCD)=AB\cdot AD=400\cdot 101=40,400. \]

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