A small cone of height 8cm is cut off from a bigger cone to leave a frustrum of height 16cm. If the volume of smaller cone is 160cm³, Find the volume of the frustrum?

Let the radius of the smaller cone be r and the radius of the frustrum be R.

The volume of a cone is given by the formula V = (1/3)πr²h, where r is the radius and h is the height.

Given that the height of the smaller cone is 8cm and the volume is 160cm³, we have:

160 = (1/3)πr²(8)
r² = 60/π
r = √(60/π)

The height of the frustrum is 16cm, and the volume of the frustrum is given by the formula V = (1/3)π(R² + Rr + r²)h.

Substitute the values of r, R, and h into the formula:

V = (1/3)π(R² + R√(60/π) + 60/π)(16)

Given that the volume of the frustrum is the volume of the bigger cone minus the volume of the smaller cone:

V = (1/3)πR²(16) - 160

Equating the two expressions for V, we get:

(1/3)π(R² + R√(60/π) + 60/π)(16) = (1/3)πR²(16) - 160

Solving for R, we get:

R = 4√(60/π)

Now, substitute the value of R back into the formula for the volume of the frustrum:

V = (1/3)π(16)(16 + 4√(60/π) + 60/π)

V = (1/3)π(256 + 64√(60/π) + 60)

V = 256π + 64π√(60/π) + 60π

Therefore, the volume of the frustrum is 256π + 64π√(60/π) + 60π cm³.

Let the radius of the base of the bigger cone be R cm and that of the smaller be r cm. Then,

Volume of the smaller cone = (1/3)πr² x 8 = 160 cm³
⇒ πr² = 60 cm²
⇒ r² = 60/π cm²

Volume of the frustrum = Volume of the bigger cone - Volume of the smaller cone
= (1/3)πR² x 24 - (1/3)πr² x 8
= (1/3)π(R² x 24 - r² x 8)
= (1/3)π(R² - r²) x 24
= (1/3)π x 60 x 24
= 480π cm³
= 480 x 22/7 cm³
= 1507.96 cm³

Therefore, the volume of the frustrum is 1507.96 cm³