Solution: (1) Because the sides of a regular triangular pyramid are known to be perpendicular to each other, one side including two sides can be regarded as the bottom, and the other side is equivalent to the height. So the volume v =1/2a * 2× a×1/3 =1/6a * 3.

You can also find out that the side length at the bottom of the regular triangular pyramid is √2a, then the height of the regular triangle at the bottom (which is also the midline) is √6/2a, then the height of the regular triangular pyramid = √ [a * 2-(√ 6/2a× 2/3) * 2] = √ 3/3a Bottom area =12. √2a？ √2a？ sin60=√3/2？ A * 2 so V = 1/3? √3/3a？ √3/2? a*2 = 1/6a*3

(2) When the side length of an equilateral triangle is a, the height is √3/2a, which is the radius of the bottom circle.

The figure obtained after rotating around the straight line where one side is located is the synthesis of two cones. The radius of the bottom surface of each cone is √3/2a, and the height is half of the side length a of an equilateral triangle, that is, 1/2a.

So v = 1/3 ∏ (√ 3/2a) * 2? 1/2a？ 2= 1/4∏a*3

(3) Given that the generatrix of the cone is 5cm long and 4cm high, the radius of the bottom circle is 3cm according to Pythagorean theorem.

V= 1/3∏？ 3*2? 4= 12∏