Question

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.

A cone of radius 10 cm is divided into two parts by a plane parallel to its base through the mid-point of its height. Compare the volumes of the two parts

Answer 1

We have,

Radius of the solid cone, R = CP

Height of the solid cone, AP = H

Radius of the smaller cone, QD = r

Height of the smaller cone, AQ = h

Also, `"AQ" = "AP"/2` i.e, `"h" = "H"/2` or H = 2h   ...(1) 

Now, in ∆AQD and ∆APC,

∠QAD = ∠PAC   ...(Common angle)

∠AQD = ∠APC = 90°

So, by AA criteria

∆AQD ~ APC

⇒ `"AQ"/"AP" = "QD"/"PC"`

⇒ `h/H = r/R`

⇒ `h/(2h) = r/R      ["Using (i)"]`

⇒ `1/2 = r/R`

⇒ R = 2r   ...(ii)

As Volume of smaller cone = `1/3 πr^2h`

And Volume of solid cone = `1/3 πR^2H`

= `1/3 π(2r)^2 xx (2h)`   ...[Using i and ii]

= `8/3 πr^2h`

So, Volume of frustum = Volume of solid cone – Volume of smaller cone

= `8/3 πr^2h - 1/3 πr^2h`

= `7/3 πr^2h`

Now, the ratio of the volumes of the two parts

= `"Volume of the smaller cone"/"Volume of the frustum"`

= `(1/3 πr^2h)/(7/3 πr^2h)`

= `1/7`

= 1 : 7

So, the ratio of the volume of the two parts of the cone is 1 : 7.

Answer 2

Let the radius of cone be r² and cut of cone r¹

Height of the cone = 10 cm

And the height the cone cut off = 5 cm

ΔAOC ∼ ΔAO’D

∴ `"AO"/"AO"^' = r_2/r_1 = 10/5`

⇒ r₂ = 2r₁

Volume of cut off cone = `1/3 πr_1^2 xx 5`

= `5/3 πr_1^2` sq. units

Volume of original cone = `1/3 π(2r_1)^2 xx 10`

= `40/3 πr_1^2` sq. units

Volume of frustum = Volume of original cone – Volume of cut of cone

= `40/3 πr_1^2 - 5/3 πr_1^2`

= `35/3 πr_1^2` sq. units

Ratio of two parts = `(35 πr_1^2)/(5 πr_1^2) = 7/1`

Hence, the ratio of two parts = 7 : 1.