AMC10 2009 B

AMC10 2009 B · Q17

AMC10 2009 B · Q17. It mainly tests Area & perimeter, Coordinate geometry.

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 $a$?

五个单位正方形如图在坐标平面中排列，左下角位于原点。从 $(a, 0)$ 到 $(3, 3)$ 的斜线将整个区域分为两个面积相等的区域。求 $a$？

(A) $\frac{1}{2}$ $\frac{1}{2}$

(B) $\frac{3}{5}$ $\frac{3}{5}$

(D) $\frac{3}{4}$ $\frac{3}{4}$

(E) $\frac{4}{5}$ $\frac{4}{5}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): The area of the entire region is 5. The shaded region consists of a triangle with base $3-a$ and altitude $3$, with one unit square removed. Therefore $$ \frac{3(3-a)}{2}-1=\frac{5}{2}. $$ Solving this equation yields $a=\frac{2}{3}$.

答案（C）：整个区域的面积为 $5$。阴影区域由一个底为 $3-a$、高为 $3$ 的三角形组成，并去掉了一个单位正方形。因此 $$ \frac{3(3-a)}{2}-1=\frac{5}{2}. $$ 解此方程得到 $a=\frac{2}{3}$。

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