Truncated cone

A truncated cone has a bases radiuses 40 cm and 10 cm and a height of 25 cm. Calculate its surface area and volume.

Correct result:

S =  11474.8633 cm2
V =  54977.8714 cm3

Solution:

r1=10 cm r2=40 cm v=25 cm  S1=π r12=3.1416 102314.1593 cm2 S2=π r22=3.1416 4025026.5482 cm2  x=r1r2=1040=30 cm  s=v2+x2=252+302=5 61 cm239.0512 cm2  S3=π (r1+r2) s=3.1416 (10+40) 39.05126134.1558 cm2 S=S1+S2+S3=314.1593+5026.5482+6134.1558=11474.8633 cm2=1.147104 cm2
S4=r12+r1 r2+r22=102+10 40+402=2100 cm2  V=13 π v S4=13 3.1416 25 2100=54977.8714 cm3=5.498104 cm3



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