A hollow cone is cut by a plane parallel to the base and the upper portion is removed If - Maths - Surface Areas and Volumes

Question

A hollow cone is cut by a plane parallel to the base and the
upper portion is removed If the curved surface area of the
remainder is 8/9th of the curved surface of the whole cone, find
the ratio of the line segments into which the cone's altitude is
divided by the plane

Aakash Sharma · Accepted Answer

Let R, H and L be the radius, height and slant height of the original cone respectively and r, h and l be the radius, height and slant height of the smaller cone respectively.

In ∆OAB and ∆OCD,

∠OAB = ∠OCD (90°)

∠AOB = ∠COD (Common)

∴ ∆ OAB ≅ ∆OCD (AA Similarity)

$\Rightarrow \frac{OB}{OD} = \frac{AB}{CD} = \frac{OA}{OC} (Corresponding sides are proportional) \Rightarrow \frac{l}{L} = \frac{r}{R} = \frac{h}{H} ... (1)$

Curved surface area of the smaller cone

= Curved surface area of cone – Curved surface area of the frustum

$= (1 - \frac{8}{9}) \times Curved surface area of cone = \frac{1}{9} \times Curved surface area of cone ∴ \frac{Curved surface area of smaller cone}{Curved surface area of cone} = \frac{1}{9} \Rightarrow \frac{π r l}{π R L} = \frac{1}{9} \Rightarrow \frac{r}{R} \times \frac{l}{L} = \frac{1}{9} \Rightarrow \frac{h}{H} \times \frac{h}{H} = \frac{1}{9} [U \sin g (1)] \Rightarrow (\frac{h}{H})^{2} = (\frac{1}{3})^{2} \Rightarrow \frac{h}{H} = \frac{1}{3} \Rightarrow h = \frac{H}{3} \frac{OA}{AC} = \frac{h}{H - h} = \frac{\frac{H}{3}}{H - \frac{H}{3}} = \frac{\frac{H}{3}}{\frac{2 H}{3}} = \frac{1}{2}$

Thus, the cones altitude is divided in the ratio 1 : 2.

375

A hollow cone is cut by a plane parallel to the base and the upper portion is removed. If the curved surface area of the remainder is 8/9th of the curved surface of the whole cone, find the ratio of the line segments into which the cone's altitude is divided by the plane.