Question

正方形 $ABCD$ 的边长为 $3$ 。线段 $CM$ 和 $CN$ 将正方形面积分成三个相等部分。线段 $CM$ 有多长？

Square $ABCD$ has sides of length $3.$ Segments $CM$ and $CN$ divide the square’s area into three equal parts. How long is segment $CM?$

$\sqrt{10}$

$\sqrt{12}$

$\sqrt{13}$

$\sqrt{14}$

$\sqrt{15}$

三个区域各占正方形面积的三分之一

Each of the three regions has area one-third of the square

用 $\triangle BMC$ 的面积求 $BM$

Use the area of $\triangle BMC$ to find $BM$

正方形面积为 $3^2 = 9$ ，所以每个区域的面积是 $9 \div 3 = 3$ 。

因此 $\triangle BMC$ 的面积为 $3$ ，所以 $\dfrac{1}{2} \cdot 3 \cdot BM = 3$ $BM = 2\text{。}$

因为 $\triangle BMC$ 是直角三角形， $CM = \sqrt{2^2 + 3^2} = \sqrt{13}\text{。}$

所以正确答案是 C。

The area of the square is $3^2 = 9,$ which means that the area of one region is $9 \div 3 = 3.$

This means the area of $\triangle BMC$ is $3,$ which means that $\dfrac{1}{2} \cdot 3 \cdot BM = 3$ $BM = 2.$

Since $\triangle BMC$ is right, we have $CM = \sqrt{2^2 + 3^2} = \sqrt{13}.$

Thus, C is the correct answer.