The base and top radius of a truncated cone is 1.5 m and 2 m respectively. The height of the cone is 60 m. What is the volume of a truncated cone? (Use π = 3).
545 m3
555 m3
535 m3
525 m3
A
535 m3
B
555 m3
C
525 m3
D
545 m3
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Solution
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Volume of the truncated cone is V=πh3(R2+Rr+r2)
R = 1.5 m, r = 2 m, h = 60 m
V=(3×60)/(3)[1.52+1.5×2+22]
=60[2.25+3+4]
=60[9.25]
=555m3
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