Question

The base and top radius of a truncated cone is 20 mm and 10 mm respectively. The height of the cone is 360 mm. What is the volume of a truncated cone? (Use $\pi$ = 3).

• A. $252,000mm$${}^3$
• B. $242,000mm$${}^3$
• C. $232,000mm$${}^3$
• D. $222,000mm$${}^3$

Answer 1

The correct option is A $252,000mm$${}^3$

Since truncated is in the shape of frustum, so volume of the truncated cone is $V=\frac{\pi h}{3}(R^2+Rr+r^2)$

R = 20 mm, r = 10 mm, h = 360 mm

$V=(3\times 360)/\left(3\right)[{20}^2+20\times 10+{10}^2]$

$=$ 360[400 + 200 + 100]

$=$ 360[700]

$=$ 252,000 mm${}^3$