Question

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 $\pi$ = 3).

• A. 555 m${}^3$
• B. 545 m${}^3$
• C. 535 m${}^3$
• D. 525 m${}^3$

Answer 1

Answer: (A)

555 m${}^3$

Volume of the truncated cone is $V=\frac{\pi h}{3}(R^2+Rr+r^2)$

R = 1.5 m, r = 2 m, h = 60 m

$V=(3\times 60)/\left(3\right)[{1.5}^2+1.5\times 2+2^2]$

$=60[2.25+3+4]$

$=60\left[9.25\right]$

$=555m^3$