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

**The radii of the bases of the truncated cone are 3 m and 6 m, and the generatrix is 5 m. Find the area of the truncated cone.**

The lateral surface area of the cone is calculated by the formula:

Sb.p. = π (r1 + r2) h.

Let’s find the height of the HB. Take the triangle ΔАВН.

Since HM is equal to BK, then:

AH = AM – HM;

AH = 6 – 3 = 3 m.

According to the Pythagorean theorem:

AB ^ 2 = BH ^ 2 + AH ^ 2;

BH ^ 2 = AB ^ 2 – AH ^ 2;

BH ^ 2 = 5 ^ 2 – 3 ^ 2 = 25 – 9 = 16;

BH = √16 = 4 m.

Sb.p. = 3.14 * (3 + 6) * 4 = 3.14 * 9 * 4 = 113.04 cm2.

The total surface area of the truncated cone is equal to the sum of the areas of both bases and the area of the lateral surface:

Sp.p. = Sbn.1 + Sbn.2 + Sb.p ..

Since the bases are a circle, their area is equal to:

Sbn.1 = πr1^2;

Sbn. 1 = 3.14 * 3^2 = 3.14 * 9 = 28.26 cm2;

Sbn. 2 = πr2^2;

Sb. 2 = 3.14 · 6^2 = 3.14 · 36 = 113.04 cm2;

Sp.p. = 28.26 + 113.04 + 113.04 = 254.34 cm2.

Answer: the lateral surface area is 113.04 cm2, the total surface area is 254.34 cm2.