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High school mathematics compulsory two space geometry problems.
Let ABC-a1b1c1,then according to the meaning of the question, AB=2, A 1B 1=5, AA 1=5.

Let O and O 1 be the centers of gravity of the regular triangle ABC and A1b1respectively, then the height of the frustum OO 1 is perpendicular to ABC and in the same plane as AB, and AO=2/3*√3, a/kloc.

Therefore, oo12 = aa12-(a1o1-ao) 2 = 25-3 = 22.

Prism height OO 1=√22

As shown in the figure, the projection triangle A''B''C'' of triangle ABC on the bottom surface A'B'C' it

The plan is as shown on the right, because it is a regular triangular prism, so the centers of the two coincide.

It is easy to find the root number 3 of A'A''=2.

Therefore, the height AA''= root number (aa' 2+aa'' 2) = root number 13cm.

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