2018 AMC 10B 第 10 题

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

小提示：

将

A

放在原点，三条棱沿坐标轴，求矩形

BCHE

及其面积。

Place

A

at the origin with the edges along the axes; find the rectangle

BCHE

and its area

大提示：

体积

= \tfrac13

\times (\text{底面积})

\times (\text{从 } M \text{ 到底面的距离})

。

Volume

= \tfrac13

\times (\text{base area})

\times (\text{distance from } M \text{ to base})

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

小提示：

大提示：

解答：

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