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Q. 3 , 4 and 5 . A canal... Water is desired

A cone of radius..... Two parts

Three cubes .... Three cubes

**Dear Student,**

**Solution for Q3**

**Solution to Q4**

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,

∠AO'C = ∠AOE = 90 deg

∠O'AC = ∠OAE [same angle]

∠ACO' = ∠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

the radius and height of the cone ABC be 2 cm and h/2 cm

therefore the volume of the cone ABC =

volume of the cone ADE =

the volume of the frustum = volume of the cone ADE - volume of the cone ABC

ratio of the volume of the two parts = volume of the cone ABC : volume of the frustum

hope this helps you.

Kindly ask the remaining queries in a separate thread.

**Regards**

**
**