# Truncated cone

Calculate the volume of a truncated cone with base radiuses r1=13 cm, r2 = 10 cm and height v = 8 cm.

Result

V =  3342.7 cm3

#### Solution:

$v_2 = \dfrac{r_2}{r_1-r_2}v = \dfrac{ 10}{ 13-10} 8 = 26.67 \ \\ V = V_1-V_2 = \dfrac{1}{3} \pi (r_1^2(v+v_2) - r_2^2v_2) = 3342.7 \ \text{cm}^3 \ \\$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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