The Solution of Mathematical Pyramid Problem

1, regular triangular prism: a rectangle with a regular triangle at the bottom and congruent sides. And the regular prism must be a regular prism, so the height is the length of the side. The height is 6, the side length of the bottom surface is 4, and the area of one side surface is 4×6=24.

Lateral area = 24× 3 = 72; The volume of all cylinders is the bottom area × height. So the bottom area is the area of an equilateral triangle with a side length of 4,1/2× 4× 2 √ 3 = 4 √ 3v = 4 √ 3× 6 = 24 √ 3.

2. Regular quadrangular pyramid: the bottom is square, the sides are isosceles triangles, and the photography of A and the vertex to the bottom should fall on the diagonal intersection of the bottom square. As shown in the figure: OA = a, BC = 2a, ∴ OH =&; frac 12； BC=a, in RT△AOH, AH=√2a,

Transverse area = 3×&; frac 12； ×2a×√2a=3√2a squared. Total area =3√2a square +4a square =(4+3√2)a square.

Volume = 1/3Sh= 1/3×4a squared× a = 4/3a cubic.