31.

The height of a cone is 10 cm. The cone is divided into two parts using a plane parallel to its base at the middle of its height. Find the ratio of the volumes of the two parts.

Let the height and radius of the given cone be H and R respectively.

The cone is divided into two parts by drawing a plane through the mid point of its axis and parallel to the base.

Upper part is a smaller cone and the bottom part is the frustum of the cone.
⇒ OC = CA = h/2

Let the radius of smaller cone be r cm.

In Δ OCD and Δ OAB,
∠OCD = ∠OAB = 90°
∠COD = ∠AOB (common)
∴ Δ OCD ∼ Δ OAB (AA similarity)

⇒ OA/OC = AB/CD = OB/OD
⇒ h / h/2 = R/r
⇒ R = 2r

The radius and height of the cone OCD are r cm and h/2 cm

Therefore the volume of the cone OCD = 1/3 x π x r² x h/2 = 1/6 πr²h

Volume of the cone OAD = 1/3 x π x R² x h = 1/3 x π x 4r² x h

The volume of the frustum = Volume of the cone OAD - Volume of the cone OCD
= (1/3 x π x 4r² x h) – (1/3 x π x r² x h/2)
= 7/6 πr²h

Ratio of the volume of the two parts = Volume of the cone OCD : volume of the frustum
= 1/6 πr²h : 7/6 πr²h
= 1 : 7