1996 AMC 12 第 23 题

Question

一个长方体的十二条棱长之和为

140

，从一个顶点到最远顶点的距离为

21

。该长方体的总表面积是

The sum of the lengths of the twelve edges of a rectangular box is

140,

and the distance from one corner of the box to the farthest corner is

21.

The total surface area of the box is

776

784

798

800

812

若三条边长为

a,b,c

，把棱长之和与空间对角线分别写成方程

If the side lengths are

a,b,c

, translate the edge sum and space diagonal into equations

展开

(a+b+c)^2

，直接求得

2(ab+bc+ca)

Expand

(a+b+c)^2

to obtain

2(ab+bc+ca)

directly

棱长之和给出

4(a+b+c)=140

，所以

a+b+c=35

。空间对角线给出

a^2+b^2+c^2=21^2=441

。因此总表面积为

\begin{gathered} 2(ab+bc+ca)\\ {}=(a+b+c)^2\\ {}-(a^2+b^2+c^2)\\ {}=35^2-441=784 \end{gathered}\text{。}

所以正确答案是 B。

The edge sum gives

4(a+b+c)=140,

so

a+b+c=35.

The space diagonal gives

a^2+b^2+c^2=21^2=441.

Therefore the surface area is

\begin{gathered} 2(ab+bc+ca)\\ {}=(a+b+c)^2\\ {}-(a^2+b^2+c^2)\\ {}=35^2-441=784. \end{gathered}

Thus the correct answer is B.

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