Question

A container in form of a frustum of a cone is 16 cm in diameter at the open end and 24 cm diameter at the bottom. If the vertical depth of the container is 8 cm calculate the capacity of the container.

Answer 1

Answer:

The capacity of the container is 2546.78 cm³.

Step-by-step explanation:

The volume of the frustum of a cone is:

$\text{Volume}=\frac{\pi h}{3}\cdot[R^{2}+Rr+r^{2}]$

The information provided is:

r = 16/2 = 8 cm

R = 24/2 = 12 cm

h = 8 cm

Compute the capacity of the container as follows:

$\text{Volume}=\frac{\pi h}{3}\cdot[R^{2}+Rr+r^{2}]$

            $=\frac{\pi\cdot8}{3}\cdot[(12)^{2}+(12\cdot 8)+(8)^{2}]\\\\=\frac{8\pi}{3}\times [144+96+64]\\\\=\frac{8\pi}{3}\times304\\\\=2546.784445\\\\\approx 2546.78$

Thus, the capacity of the container is 2546.78 cm³.