# Truncated cone

Calculate the height of the rotating truncated cone with volume V = 1115 cm3 and a base radii r1 = 7.9 cm and r2 = 9.7 cm.

Result

h =  4.57 cm

#### Solution:

$r_{1}=7.9 \ \text{cm} \ \\ r_{2}=9.7 \ \text{cm} \ \\ V=1115 \ \text{cm}^3 \ \\ \ \\ V=\dfrac{ 1 }{ 3 } \pi h (r_{1}^2+r_{1} \ r_{2}+r_{2}^2) \ \\ \ \\ h=\dfrac{ 3 \cdot \ V }{ \pi \cdot \ (r_{1}^2+r_{1} \cdot \ r_{2}+r_{2}^2) }=\dfrac{ 3 \cdot \ 1115 }{ 3.1416 \cdot \ (7.9^2+7.9 \cdot \ 9.7+9.7^2) } \doteq 4.5672 \doteq 4.57 \ \text{cm}$

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Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):

Math student
can u explain why you do (r2 + rxR + R2) in the first step

1 year ago  3 Likes
Dr Math

Kukoslav
but to prove formula, you need to know how to solve integral

Kukoslav
need to solve a cubic equation, as obtained above, to find rises in heights... integral

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