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 =  24.0007 m

Solution:

S=7697 m2 D1=56 m D2=42 m r1=D1/2=56/2=28 m r2=D2/2=42/2=21 m S1=π⋅ r12=3.1416⋅ 282≐2463.0086 m2 S2=π⋅ r22=3.1416⋅ 212≐1385.4424 m2 S3=S−S1−S2=7697−2463.0086−1385.4424≐3848.549 m2 s=S3/π/(r1+r2)=3848.549/3.1416/(28+21)≐25.0006 m h=s2−(r1−r2)2=25.00062−(28−21)2=24.0007 m

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