Question

如图，五个单位正方形放在坐标平面上，左下角在原点。斜线从 $(a, 0)$ 延伸到 $(3, 3)$ ，把整个区域分成面积相等的两部分。 $a$ 是多少？

Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extending from $(a, 0)$ to $(3, 3),$ divides the entire region into two regions of equal area. What is $a?$

$\dfrac{1}{2}$

$\dfrac{3}{5}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

$\dfrac{4}{5}$

答案：C

知识点：三角形面积面积分割坐标几何

难度评级：1610

解答：

五个正方形总面积为 $5$ ，所以每部分面积必须为 $\dfrac{5}{2}$ 。

从 $(a, 0)$ 到 $(3, 3)$ 的直线与坐标轴围成一个底为 $3 - a$ 、高为 $3$ 的三角形。直线右下方的区域是这个三角形去掉一个单位正方形，所以得 $3(3 - a) = 7$ ，从而 $a = \dfrac{2}{3}$ 。 $\dfrac{3(3 - a)}{2} - 1 = \dfrac{5}{2}$

所以正确答案是 C。

The five squares have total area $5,$ so each region must have area $\dfrac{5}{2}.$

The line from $(a, 0)$ to $(3, 3)$ together with the axes bounds a triangle of base $3 - a$ and height $3$ ; the region on the lower-right side of the line is this triangle with one unit square removed. Setting $\dfrac{3(3 - a)}{2} - 1 = \dfrac{5}{2}$ gives $3(3 - a) = 7,$ so $a = \dfrac{2}{3}.$

Thus, the correct answer is C.

2009 AMC 12B 第 14 题

答案：C

难度评级：1610

解答：

其他年份的第 14 题