AMC10 2007 A

AMC10 2007 A · Q21

AMC10 2007 A · Q21. It mainly tests Pythagorean theorem, 3D geometry (volume).

A sphere is inscribed in a cube that has a surface area of $24$ square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?

一个球体内切于一个表面积为$24$平方米的立方体。然后，一个第二立方体内切于该球体。内立方体的表面积是多少平方米？

(A) 3 3

(B) 6 6

(D) 9 9

(E) 12 12

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Since the surface area of the original cube is 24 square meters, each face of the cube has a surface area of 24/6 = 4 square meters, and the side length of this cube is 2 meters. The sphere inscribed within the cube has diameter 2 meters, which is also the length of the diagonal of the cube inscribed in the sphere. Let $l$ represent the side length of the inscribed cube. Applying the Pythagorean Theorem twice gives $l^2 + l^2 + l^2 = 2^2 = 4.$ Hence each face has surface area $l^2 = \frac{4}{3}$ square meters. So the surface area of the inscribed cube is $6 \cdot (4/3) = 8$ square meters.

答案（C）：由于原立方体的表面积为 24 平方米，所以立方体每个面的面积为 $24/6 = 4$ 平方米，因此该立方体的棱长为 2 米。内接于该立方体的球的直径为 2 米，而这也等于内接于该球的立方体的体对角线长度。设 $l$ 表示该内接立方体的棱长。两次应用勾股定理可得 $l^2 + l^2 + l^2 = 2^2 = 4.$ 因此每个面的面积为 $l^2 = \frac{4}{3}$ 平方米。所以该内接立方体的表面积为 $6 \cdot (4/3) = 8$ 平方米。

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