# The radii of the bases of the truncated cone are 3 dm and 6 dm, and the generatrix is 5 dm.

**The radii of the bases of the truncated cone are 3 dm and 6 dm, and the generatrix is 5 dm. Find: a) the height of the truncated cone; b) the area of its axial section; c) the angle of inclination of the generatrix to the plane of the base.**

Let’s draw the height BH of the cone.

Quadrangle BO1OH is a rectangle, then BO1 = OH = 3 dm.

AH = AO – OH = 6 – 3 = 3 dm.

In a right-angled triangle ABH, according to the Pythagorean theorem, we determine the length of the leg BH.

BH ^ 2 = AB ^ 2 – AH ^ 2 = 5 ^ 2 – 3 ^ 2 = 25 – 9 = 16.

BH = 4 dm.

Determine the angle BAH.

CosBAH = AH / AB = 3/5.

Angle BAH = arcos (3/5) = 53.

The axial section of the truncated cone is an isosceles trapezoid ABCD.

BC = 2 * BO1 = 2 * 3 = 6 dm.

AD = 2 * AO = 2 * 6 = 12 dm.

Determine the cross-sectional area.

Ssech = (ВС + АD) * ВH / 2 = (6 + 12) * 4/2 = 36 dm2.

Answer: The height of the cone is 4 dm, the cross-sectional area is 36 dm2, the angle BAD is 53.