Answer:
The capacity of the container is 2546.78 cm³.
Step-by-step explanation:
The volume of the frustum of a cone is:
[tex]\text{Volume}=\frac{\pi h}{3}\cdot[R^{2}+Rr+r^{2}][/tex]
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:
[tex]\text{Volume}=\frac{\pi h}{3}\cdot[R^{2}+Rr+r^{2}][/tex]
[tex]=\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[/tex]
Thus, the capacity of the container is 2546.78 cm³.
