In a truncated cone, the radius of the smaller base = 2cm. The height of the cone

In a truncated cone, the radius of the smaller base = 2cm. The height of the cone = 3cm, and its generatrix makes an angle of 45 with the plane of the larger base. Find the volume of the cone.

From point B we draw the height BH to the segment AD. In a right-angled triangle ABН, the angle ВAН, by condition, is equal to 45, then the angle ABН = 45, and the triangle ABН is rectangular and isosceles, AH = BH = 3 cm.

Then the length of the segment AO = AH + OH = 3 + 2 = 5 cm.

The volume of the truncated cone is determined by the formula: V = n * h * (r ^ 2 + R ^ 2 + r * R) / 3, where r, R are the radii of the smaller and larger bases.

V = n * 3 * (4 + 25 + 10) / 3 = n * 39 cm3.

Answer: The volume of the truncated cone is n * 39 cm3.