Question

A cone of radius 10cm is cut into two parts by a plane through the mid-point of its vertical axis parallel to the base. Find the ratio of the volumes of the smaller cone and frustum of the cone

Answer 1

The ratio of the volume of the smaller cone and the volume of the frustum of the cone is 1:7.

Step-by-step explanation:

Let’s assume from the attached figure below,

“2h” = OB = height of the large cone  

“R” = BC = radius of the base of the large cone = 10 cm (given)

When the cone is cut into 2 by the plane through the midpoint of its vertical axis parallel to its base, then we will get a smaller cone and a frustum.

Then,

“r” = AD = radius of the base of the smaller cone  

“h” = OA = height of the smaller = AB = height of the frustum = ½ * OB

Now, first, let’s consider ∆ OAD and ∆ OBC, we have

∠A = ∠A …… [common angle]

∠OAD = ∠OBC ……. [corresponding angles, since AD//BC]

∴ By AA similarity, ∆ OAD ~ ∆ OBC

Since the corresponding sides of two similar triangles are proportional to each other.

∴ OA/OB = AD/BC

⇒ h/2h = r/R

⇒ r/R = ½

⇒ r = ½ * 10 ….. [since R = 10 cm]

⇒ r = 5 cm

Therefore,

The volume of the smaller cone is given by,

= 1/3 * π * (r)²(h)

= 1/3 * π * (5)²(h) …… (i)

And,

The volume of the large cone is given by,  

= 1/3 * π * (R)²(2h)

= 1/3 * π * (10)²(2h) …… (ii)

So, by subtracting (i) from (ii), we get

The volume of the frustum as,

=  [1/3 * π * (10)² * (2h)] - [1/3 * π * (5)² * (h)]

= 1/3 * π * h * (200 – 25)

= 1/3 * π * h * 175 ……. (iii)

Thus, from (i) & (iii), we get  

The ratio of the volume of the smaller cone to the volume of the frustum as,

= [volume of the smaller cone] / [volume of the frustum]

= [1/3 * π * (5)² * (h)] / [1/3 * π * h * 175]

= [1/3 * π * (5)²(h)] / [1/3 * π * h * 175]

Cancelling all the similar terms

= (25) / (175)

= 1/7

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