Question

在图示长方体中， $AB = 3$ 、 $BC = 1$ ，且 $CG = 2$ 。点 $M$ 是 $FG$ 的中点。以 $BCHE$ 为底、 $M$ 为顶点的棱锥体积是多少？

In the rectangular parallelepiped shown, $AB = 3,$ $BC = 1,$ and $CG = 2.$ Point $M$ is the midpoint of $FG.$ What is the volume of the rectangular pyramid with base $BCHE$ and apex $M?$

$1$

$\dfrac{4}{3}$

$\dfrac{3}{2}$

$\dfrac{5}{3}$

$2$

答案：E

知识点：体积棱锥坐标几何

难度评级：1570

解答：

将 $A$ 放在原点，并令三条棱沿坐标轴： $A = (0,0,0)$ 、 $B = (3,0,0)$ 、 $C = (3,1,0)$ 、 $E = (0,0,2)$ 、 $H = (0,1,2)$ 、 $F = (3,0,2)$ 、 $G = (3,1,2)$ ，所以 $M = (3, \tfrac12, 2)$ 。底面 $BCHE$ 是矩形，且 $BC = 1$ ， $BE = \sqrt{3^2 + 2^2} = \sqrt{13}$ ，所以底面积为 $\sqrt{13}$ 。底面所在平面为 $2x + 3z = 6$ ，点 $M$ 到该平面的距离为 $\frac{|2\cdot3 + 3\cdot2 - 6|}{\sqrt{13}} = \frac{6}{\sqrt{13}}$ 。体积为 $\tfrac13 \cdot \sqrt{13} \cdot \frac{6}{\sqrt{13}} = 2$ 。正确答案是 E。

Put $A$ at the origin with edges along the axes: $A = (0,0,0),$ $B = (3,0,0),$ $C = (3,1,0),$ $E = (0,0,2),$ $H = (0,1,2),$ $F = (3,0,2),$ $G = (3,1,2),$ so $M = (3, \tfrac12, 2).$ The base $BCHE$ is a rectangle with $BC = 1$ and $BE = \sqrt{3^2 + 2^2} = \sqrt{13},$ hence area $\sqrt{13}.$ Its plane is $2x + 3z = 6,$ and $M$ sits at distance $\frac{|2\cdot3 + 3\cdot2 - 6|}{\sqrt{13}} = \frac{6}{\sqrt{13}}$ from it. The volume is $\tfrac13 \cdot \sqrt{13} \cdot \frac{6}{\sqrt{13}} = 2.$ Therefore, the answer is E.

2018 AMC 10B 第 10 题

答案：E

难度评级：1570

解答：

其他年份的第 10 题