Question

The base and top radius of a truncated cone is 3.5 m and 3 m respectively. The height of the cone is 630 m. What is the volume of a truncated cone? (Use $\pi$ = 22/7).

• A. 21,955 m${}^3$
• B. 20,955 m${}^3$
• C. 22,955 m${}^3$
• D. 23,955 m${}^3$

Answer 1

Answer: (B)

20,955 m${}^3$

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

R = 3.5 m, r = 3 m, h = 630 m

$V=(22\times 630)/(7\times 3)[{3.5}^2+3.5\times 3+3^2]$

$=660[12.25+10.5+9]$

$=660\left[31.75\right]$

$=20,955m^3$