Question

The radius 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.

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Answer 1

Answer:

Here the height of cone = h cm, thus on dividing we get the frustum of the cone with radius, R = 10 cm and radius, r = 5 cm. The height of frustum will be half of the total height =h2. Similarly radius of the smaller cone, r = 5 cm and height of the smaller cone =h2 cm. Thus volume of smaller cone, V=13πr2h=13π(5)2×h2.

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Answer 2

Answer:

Height of cone = H = 10 cm

Let R be the radius of the given cone.

After dividing cone into two parts using a plane parallel to its base at the middle of its height, one part is cone and the other part formed is frustum.

Let r be the top radius of the frustum.

In right angled triangle OCD and OAB,

∠OCD = ∠OAB = 90°

∠COD = ∠AOB (common)

∴ ΔOCD ~ ΔOAB (AA similarity)

⇒ OA/OC = AB/CD = OB/OD

⇒ h/(h/2) = R/r

⇒ R = 2r

Height of the new cone formed = h = 10/2 = 5 cm

Volume of the new cone = (1/3)πr2h = (1/3) × π × (r)2 × (5)

= (5/3)πr2

Volume of the given cone = (1/3)πR2H

= (1/3)π(2r)2H

= (4/3)πr2(10)

= (40/3)πr2

∴ Height of frustum = H’ = 10/2 = 5 cm

Base Radius of frustum = R cm

Top radius of frustum = r = R/2 cm

Volume of frustum = Volume of given cone – Volume of new cone formed

= (40/3)πr2 - (5/3)πr2

= (35/3)πr2

Ratio of volume of two parts = Volume of new cone formed: Volume of frustum

= (5/3)πr2:(35/3)πr2 = 1:7

Therefore, the required ratio is 1:7.