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如果我们把长方体交于一个顶点的三条棱的长叫做三度,那么就有性质:长方体对角线的长的平方等于三度的平方和。设长方体对角钱的长为l,三度分别为a、b、c,就有l_2=a~2+b~2+c~2。对于正方体来说,如果棱长为a,则有l~2=3a~2。长方体对角线的这个性质,实质上就是两异面直线上两点间的距离公式:l=(m~2+n~2+d~2-2mncosθ)~(1/2)当θ=90°时的特例。看起来如此简单的有关长方体的一个性质,但在1988年高考的四道立体几何题中,却有两题可以用这一性质来解决。可见,长方体对角线性质在应用方面具有一定的广泛性。
If we say that the length of the three edges of a cuboid at a vertex is called third degree, then there is a property: the long square of the cuboid diagonal is equal to the sum of three squares. Let cuboids have a length of l, the third degree is a, b, and c, respectively, and l2=a~2+b~2+c~2. For the cube, if the length is a, then l~2=3a~2. This property of the cuboid diagonal is essentially the distance between two points on two different straight lines: l = (m~2+n~2+d~2-2mncosθ)~(1/2) when θ=90 ° special case. It seems as simple as a cuboid about the nature of the cube, but of the four three-dimensional geometry questions in the 1988 college entrance exam, there are two questions that can be solved with this property. Obviously, the diagonal properties of cuboids have a certain extent in application.