# Truncated cone 3

The surface of the truncated rotating cone S = 7697 meters square, the substructure diameter is 56m and 42m, determine the height of the tang.

Correct result:

h =  24.001 m

#### Solution:

$S=7697 \ \text{m}^2 \ \\ D_{1}=56 \ \text{m} \ \\ D_{2}=42 \ \text{m} \ \\ r_{1}=D_{1}/2=56/2=28 \ \text{m} \ \\ r_{2}=D_{2}/2=42/2=21 \ \text{m} \ \\ S_{1}=\pi \cdot \ r_{1}^2=3.1416 \cdot \ 28^2 \doteq 2463.0086 \ \text{m}^2 \ \\ S_{2}=\pi \cdot \ r_{2}^2=3.1416 \cdot \ 21^2 \doteq 1385.4424 \ \text{m}^2 \ \\ S_{3}=S - S_{1}-S_{2}=7697 - 2463.0086-1385.4424 \doteq 3848.549 \ \text{m}^2 \ \\ s=S_{3}/\pi/(r_{1}+r_{2})=3848.549/3.1416/(28+21) \doteq 25.0006 \ \text{m} \ \\ h=\sqrt{ s^2-(r_{1}-r_{2})^2 }=\sqrt{ 25.0006^2-(28-21)^2 }=24.001 \ \text{m}$

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