Truncated cone 3

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

Correct answer:

h =  24.0007 m

Step-by-step explanation:

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 2822463.0086 m2 S2=π r22=3.1416 2121385.4424 m2 S3=SS1S2=76972463.00861385.44243848.549 m2 s=S3/π/(r1+r2)=3848.549/3.1416/(28+21)25.0006 m h=s2(r1r2)2=25.00062(2821)2=24.0007 m



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