The radius of the truncated cone ratio is 4: 7, but the generatrix of the cone, the length of which is 6 cm

The radius of the truncated cone ratio is 4: 7, but the generatrix of the cone, the length of which is 6 cm, forms a 30 degree angle of the truncated base of the cylinder. Determine the surface area and volume of the truncated cone.

Since the radius of the larger base (AK) and the generator (AB) form an angle of 30 °, using the theorem of sines, you can find the height (OK = BN):

Sin α = h ÷ l;

h = Sin α ∙ l;

Sin 30 ° = ½ = 0.5;

BN = OK = ½ ∙ 6 = 3 cm.

Now, applying the Pythagorean theorem, we find the length of the segment AN:

AB2 = BN2 + AN2;

AN2 = AB2 – BN2;

AN2 = 62 – 32 = 36 – 9 = 25;

AN = √25 = 5 cm.

Since the radii of the bases are related as 4: 7, then we express:

4x – radius of the smaller base AO;

7x – radius of the larger base of the AK.

Since the segment AN = 5 cm, then:

7x – 4x = 5;

3x = 5;

x = 5/3;

VO = 5/3 4 = 20/3 = 6.67 cm;

AK = 5/3 7 = 35/3 = 11.7 cm.

Find the surface area:

S = π (r1 + r2) l;

S = 3.14 * (6.67 + 11.7) * 6 = 3.14 * 18.3 * 6 = 345.4 cm2.

Find the volume of the truncated cone:

V = 1/3 π h (r12 + r1 r2 + r22);

V = 1/3 π 3 (6.672 + 6.67 11.7 + 11.72) = π (44.49 + 6.67 11.7 + 136.89) = 3.14 (44.49 + 78.04 +136.89) = 3.14 259.42 = 814.58 cm3.

Answer: the surface area is 345.4 cm2, the volume is 814.58 cm3.