Question

a solid cone of base radius10cm is cut into two parts through the mid point of its height, by the plane parallel to its base. find the ratio in the volumes of two parts of the cone

Answer 1

Do you mean to say that Cone is divided into 2 parts by a plane , parallel to the base through the mid point of the axis AH ?

In such case the ratio of the volumes of each part can be calculated as follows:

Here, AG = GH = h

And since triangle AGD ~ triangle AHC ( by AAA similarity criterion)

So, (AG/AH) = (GD / HC) ( corresponding sides of similar triangles)

=> h/ 2h = GD /10

=> 2GD = 10

=> GD = 5 cm

So, now, Volume of Cone AED = 1/3 pi r² h

=> 1/3 pi * 25 * h …………(1)

And Volume of the frustum BCDE Of the Cone = Volm (Cone ABC) — Volm ( ConeAED)

= 1/3 * pi* 100* 2h — 1/3 * pi* 25*h

= 1/3* pi* h ( 200 - 25)

= 1/3 * pi * h* 175 ………….(2)

So ratio of 2 parts = (1) ÷ (2)

=> (1/3*pi*25*h) ÷( 1/3*pi*h*175)

= 25/175

= 1/7

So, the volume of the frustum is 7 times the volume of the smaller upper cone.

Answer 2

this the ans. .........