A cone of radius 4cm is divided into two parts by drawing a plane through the midpoint of its axis - Maths - Surface Areas and Volumes

Question

A cone of radius 4cm is divided into two parts by drawing a
plane through the midpoint of its axis & parallel to its base,
compare the volume of the two parts

Ajanta Trivedi · Accepted Answer

let the height of the cone be h cm radius of the cone be OE= r = 4cm

the cone is divided into two parts cone ABC and frustum BCED OE = 4 cm, AO' = h/2 cm in the triangles AO'C and AOE, âˆ AO'C = âˆ AOE = 90 deg âˆ O'AC = âˆ OAE [same angle] âˆ ACO' = âˆ AEO [since O'C is parallel to OE] therefore the triangles AO'C and AOE are similar triangles therefore the ratio of the corresponding sides are equal AO'/AO = O'C/OE $\frac{h / 2}{h} = \frac{O' C}{O E} O' C = \frac{1}{2} * O E = \frac{1}{2} * 4 O' C = 2 c m$ the radius and height of the cone ABC be 2 cm and h/2 cm therefore the volume of the cone ABC = $\frac{1}{3} π r^{2} * \frac{h}{2} = \frac{1}{3} π * 2^{2} * \frac{h}{2} = \frac{2}{3} π h c m^{3}$ volume of the cone ADE = $\frac{1}{3} π r^{2} h = \frac{1}{3} π * 4^{2} h = \frac{16}{3} π h c m^{3}$ the volume of the frustum = volume of the cone ADE - volume of the cone ABC $= \frac{16}{3} π h - \frac{2}{3} π h = \frac{14}{3} π h c m^{3}$ ratio of the volume of the two parts = volume of the cone ABC : volume of the frustum $= \frac{2}{3} π h : \frac{14}{3} π h = 1 : 7$ hope this helps you

A cone of radius 4cm is divided into two parts by drawing a plane through the midpoint of its axis & parallel to its base, compare the volume of the two parts.