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