# The radius of the smaller base of the truncated cone is 6 cm, and the generatrix is 5 cm

**The radius of the smaller base of the truncated cone is 6 cm, and the generatrix is 5 cm, inclined to the base plane at an angle of 60 degrees. Find S side, S full.**

To determine the radius of the circle of the larger base, draw the height of BB1. Radius AO = OB + AB. In a right-angled triangle, the angle AB1B = 180 – 90 – 60 = 30. Then the leg AB lies opposite the angle 30 and is equal to half the length of the hypotenuse AB1. AB = 5/2 = 2.5 cm.Then AO = 6 + 2.5 = 8.5 cm.

Let’s define the lateral surface area.

Sside = n * (AO + B1O1) * AB1 = n * (8.5 + 6) * 5 = n * 72.5 cm2.

Determine the areas of the bases.

S1 = n * OA ^ 2 = n * 8.5 ^ 2 = n * 72.25 cm2.

S2 = n * O1B1 ^ 2 = n * 6 ^ 2 = n * 36 cm2.

Determine the total area of the cone.

S floor = S side + S1 + S2 = n * 72.5 + n * 72.25 + n * 36 = n * 180.75 cm2.

Answer: Sside = n * 72.5 cm2, Spol = n * 180.75 cm2.