Truncated cone 3

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

Final Answer:

h =  24.0007 m

Step-by-step explanation:

$$\begin{gathered} S=7697{\text{m}}^2 \\ {D}_1=56\text{m} \\ {D}_2=42\text{m} \\ {r}_1=D_1/2=56/2=28\text{m} \\ {r}_2=D_2/2=42/2=21\text{m} \\ {S}_1=\pi \cdot r_1^2=3.1416\cdot 28^2\doteq 2463.0086{\text{m}}^2 \\ {S}_2=\pi \cdot r_2^2=3.1416\cdot 21^2\doteq 1385.4424{\text{m}}^2 \\ {S}_3=S-S_1-S_2=7697-2463.0086-1385.4424\doteq 3848.549{\text{m}}^2 \\ s=S_3/\pi /(r_1+r_2)=3848.549/3.1416/(28+21)\doteq 25.0006\text{m} \\ h=\sqrt{s^2-(r_1-r_2{)}^2}=\sqrt{25.0006^2-(28-21{)}^2}=24.0007\text{m} \end{gathered}$$

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