Question

A cone has its
radius(R) equal to its
height. What is the
volume of its frustum
if the smaller cone's
radius is
0
R
A​

Answer 1

Given:

A cone has its  radius(R) equal to its  height.

The smaller cone's  radius is r

To find:

The  volume of its frustum

Solution:

Referring to the figure attached below, we have

"H" → OB → represents the height of the bigger cone

"R" → BC → represents the radius of the bigger cone

"r" → AD → represents the radius of the smaller cone

"h" → OA → represents the height of the smaller cone

We are given that the radius of the bigger cone is equal to its height

∴ R = H   ..... (i)

If we cut a smaller cone of radius "r" and height "h" from the bigger, we will get a frustum as shown in the figure attached below.

Considering, ∆OBC & ∆OAD

∠AOD = ∠BOC ...... [common angles]

∠OAD = ∠OBC ....... [corresponding angles]

∴ ∆OBC ~ ∆OAD ..... [by AA Similarity]

⇒ $\frac{OB}{OA} = \frac{BC}{AD}$  ..... [∵ the corresponding sides of similar triangles are proportional to each other]

substituting the values from the attached figure

⇒ $\frac{H}{h} = \frac{R}{r}$

∵ R = H (given)

⇒ $\bold{h = r}$ ...... (ii)

Now,

The volume of the frustum is given by,

= [The volume of the bigger cone] - [The volume of the smaller cone ]

= $[\frac{1}{3} \pi R^2 H] - [\frac{1}{3} \pi r^2 h]$

from (i) & (ii), we get

= $[\frac{1}{3}\times \pi\times R^2\times R] - [\frac{1}{3}\times \pi\times r^2 \times r]$

=  $[\frac{1}{3}\times \pi\times R^3] - [\frac{1}{3}\times \pi\times r^3 ]$

= $\frac{1}{3}\times \pi\times [R^3 - r^3 ]$

we know from identity → a³ - b³ = (a - b)(a² + ab + b²)

= $\bold{\frac{1}{3} \pi (R - r) (R^2 + Rr +r^2)}$

Thus, the volume of its frustum is →  $\boxed{\underline{\bold{\frac{1}{3} \pi (R - r) (R^2 + Rr +r^2)}}}$.

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