Question

please solve this question friends
Q1 please fast

Answer 1

Let the height and radius of the given cone be H and R respectively.
The cone is divided into two parts by drawing a plane through the mid point of its
axis and parallel to the base.
Upper part is a smaller cone and the bottom part is the frustum of the cone.
⇒ OC = CA = h/2
Let the radius of smaller cone be r cm.
In ΔOCD and ΔOAB,
∠OCD = ∠OAB = 90°
∠COD = ∠AOB (common)
∴ ΔOCD ∼ ΔOAB (AA similarity)

⇒ OA/OC = AB/CD = OB/OD⇒ h / h/2 = R/r⇒ R = 2rthe radius and height of the cone OCD are r cm and h/2 cm
therefore the volume of the cone OCD = 1/3 x π x r2 x h/2 = 1/6 πr2h
Volume of the cone OAD = 1/3 x π x R2 x h = 1/3 x π x 4r2 x h
The volume of the frustum = Volume of the cone OAD - Volume of the cone OCD
                                                 = (1/3 x π x 4r2 x h) – (1/3 x π x r2 x h/2)
                                                 = 7/6 πr2h
Ratio of the volume of the two parts = Volume of the cone OCD : volume of the frustum
                                                                  = 1/6 πr2h : 7/6 πr2h
                                                                  = 1 : 7
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Answer 2

the volume of frustum is 7 times of volume of small cone....