Find the total surface area of a truncated cone if it is formed by rotating a rectangular trapezoid with bases of 13 and 18 cm

Find the total surface area of a truncated cone if it is formed by rotating a rectangular trapezoid with bases of 13 and 18 cm around the smaller side equal to 12 cm.

The total surface area of ​​a truncated cone consists of the area of ​​its lateral surface and the areas of its two bases:
Sp.p = Sb.p. + Sbasn.1 + Sbasn.2;
Sb.p. = π (r1 + r2) l;
Sbn.1 = πr1 ^ 2;
Sbasn 2 = πr2 ^ 2;
You need to find the generator of the cone. To do this, consider this trapezoid, designating it as AVSD.
VK – height. The BC and CD segments are equal. Therefore:
AK = HELL – KD;
AK = 18 – 13 = 5 cm.
AB ^ 2 = VK ^ 2 + AK ^ 2;
AB ^ 2 = 12 ^ 2 + 5 ^ 2 = 144 + 25 = 169;
AB = √169 = 13 cm.
Sb.p. = 3.14 * (13 + 18) * 13 = 3.14 * 31 * 13 = 1265.42 cm2;
Sb. 1 = 3.14 * 132 = 3.14 * 169 = 530.66 cm2;
Sb. 2 = 3.14 * 182 = 3.14 * 324 = 1017.36 cm2;
Sp.p = 1265.42 + 530.66 + 1017.36 = 2813.44 cm2.
Answer: the total surface area of ​​the truncated cone is 2813.44 cm2.