Question

The base and top radius of a truncated cone is 5 cm and 1.5 cm respectively. The height of the cone is 630 cm. What is the volume of a truncated cone? (Use $\pi$ = 22/7).

• A. 22,935m${}^3$
• B. 22,935mm${}^3$
• C. 22,935cm${}^3$
• D. 22,935m

Answer 1

Answer: (C)

22,935cm${}^3$

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

R = 5 cm, r = 1.5 cm, h = 630 cm

$V=(22\times 630)/(7\times 3)[5^2+5\times 1.5+{1.5}^2]$

$=660[25+7.5+2.25]$

$=660\left[34.75\right]$

$=$ 22,935 cm${}^3$