2014 AMC 12B 第 14 题

Question

一个长方体的总表面积为

94

平方英寸。它所有棱长之和为

48

英寸。它所有体对角线长度之和是多少英寸？

A rectangular box has a total surface area of

94

square inches. The sum of the lengths of all its edges is

48

inches. What is the sum of the lengths in inches of all of its interior diagonals?

8\sqrt{3}

10\sqrt{2}

16\sqrt{3}

20\sqrt{2}

40\sqrt{2}

设三条边长为

x, y, z

，则

2(xy+yz+zx)=94

，且

4(x+y+z)=48

。

Let the edges be

x, y, z,

so

2(xy+yz+zx)=94

and

4(x+y+z)=48

用

(x+y+z)^2 =

x^2+y^2+z^2

+ 2(xy+yz+zx)

求每条体对角线

\sqrt{x^2+y^2+z^2}

。

Use

(x+y+z)^2 =

x^2+y^2+z^2

+ 2(xy+yz+zx)

to find each diagonal

\sqrt{x^2+y^2+z^2}

设三条边长为

x, y, z

，则

xy+yz+zx = 47

，且

x+y+z = 12

。因此

\begin{gathered} x^2+y^2+z^2 = (x+y+z)^2 \\ {}- 2(xy+yz+zx) \\ = 144 - 94 \\ = 50 \end{gathered}\text{。}

4

条体对角线中的每一条长度为

\sqrt{x^2+y^2+z^2} = \sqrt{50} = 5\sqrt2

，所以总长度为

4 \cdot 5\sqrt2 = 20\sqrt2

。

所以正确答案是 D。

Let the edges be

x, y, z.

Then

xy+yz+zx = 47

and

x+y+z = 12.

Therefore

\begin{gathered} x^2+y^2+z^2 = (x+y+z)^2 \\ {}- 2(xy+yz+zx) \\ = 144 - 94 \\ = 50. \end{gathered}

Each of the

4

interior diagonals has length

\sqrt{x^2+y^2+z^2} = \sqrt{50} = 5\sqrt2,

so their total length is

4 \cdot 5\sqrt2 = 20\sqrt2.

Thus, the correct answer is D.

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