The truncated right circular cone has a large base radius 8 cm and a small base radius of 4 cm. The height of the truncated cone is 6 cm. How many cm^3 are in the volume of this solid?
The truncated right circular cone has a large base radius 8 cm and a small base radius of 4 cm. The height of the truncated cone is 6 cm. How many cm^3 are in the volume of this solid?
This is known as a "frustum" of a cone
The volume is given by
pi * ( height / 3 ) ( R^2 + Rr + r^2)
Where R is the larger base radius and r is the smaller base radius
So we have
pi * ( 6/3) ( 8^2 + 8*4 + 4^2) =
pi * (2) ( 64 + 32 + 16) =
2pi * ( 112) =
224 pi cm^3 ≈ 703.7 cm^3