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题目一个四面体的所有棱长都为2~(1/2),四个顶点在同一个球面上,则此球的表面积为( ) (A)3π. (B)4π. (C)33~(1/2)π. (D)6π. 分析这个正四面体可以想象是由棱长为1的正方体砍去四个角所得(实验修订本第二册下53页第8题),而这个正方体8个顶点所在的球面,也正是这个正四面体四个顶点所在的球面,而这个正方体对角线的长就是球的直径,显然,应选(A). 由题目条件想象到构造相应的正方体,这种转化使思路变清晰,各种线面位置关系也容易观察,
The head of a tetrahedron has a length of 2~(1/2), and the four vertices are on the same sphere. The surface area of the ball is ()(A)3π. (B)4π. (C)33~ (1/2)π. (D)6π. The analysis of this regular tetrahedron can be imagined by cutting out the four corners of a cube with an edge length of 1 (Experimental revised volume 2, page 53, item 8). The sphere where the eight vertices of the cube are located is also the sphere where the four vertices of the regular tetrahedron are located. The length of the diagonal of this cube is the diameter of the ball. Obviously, it should be selected (A). The square shape, this transformation makes the idea clearer, and the positional relationship between various lines is also easy to observe.