Class 10 Frustum of a Cone
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class-10 chapter-13

Frustum of a Cone

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♦ Frustum of a Cone

Frustum of a Cone

We will take a right circular cone and remove a portion of it. There are so many ways in which we can do this.
But one particular case that we are interested in is the removal of a smaller right circular cone by cutting the given cone by a plane parallel to its base. You must have observed that the glasses (tumblers), in general, used for drinking water, are of this shape. (See Fig. 13.19)



`text ( Activity 1 : )` Take some clay, or any other such material (like plasticine, etc.) and form a cone. Cut it with a knife parallel to its base.

Remove the smaller cone. What are you left with?You are left with a solid called a frustum of the cone. You can see that this has two circular ends with different radii.

So, given a cone, when we slice (or cut) through it with a plane parallel to its base (see Fig. 13.20) and remove the cone that is formed on one side of that plane, the part that is now left over on the other side of the plane is called a frustum* of the cone.

Q 3220001811

The radii of the ends of a frustum of a cone 45 cm high are 28 cm and 7 cm
(see Fig. 13.21). Find its volume, the curved surface area and the total suface area

(Take `π = 22/7` ) .
Class 10 Chapter 13 Example 12
Solution:

The frustum can be viewed as a difference
of two right circular cones `OAB` and
`OCD` (see Fig. 13.21). Let the height (in cm)
of the cone `OAB` be `h_1` and its slant height `l_1`,
i.e.,` OP = h1` and `OA = OB = l1`. Let `h_2` be the
height of cone` OCD` and `l_2` its slant height.

We have :` r_1 = 28 cm, r_2 = 7 cm`

and the height of frustum` (h) = 45 cm`. Also,

`h_1 = 45 + h_2`........... (1)
We first need to determine the respective heights `h_1` and `h_2` of the cone `OAB`
and `OCD`.

Since the triangles `OPB` and `OQD` are similar (Why?), we have

` (h_1)/(h_2) = 28/7 = 4/1` ...........(2)

From (1) and (2), we get `h_2 = 15` and `h_1 = 60`.

Now, the volume of the frustum

= volume of the cone `OAB` – volume of the cone `OCD`

` = [1/3 * 22/7 * (28)^2* (60) -1/3 * 22/7 * (7)^2 * (15 ) ] cm^3 = 48510 cm^3`

The respective slant height `l_2` and `l_1` of the cones` OCD` and `OAB` are given
by

`l_2 = sqrt ( (7)^2 + (15)^2 ) = 16.55 cm` (approx)

`l_1 = sqrt ( (28)^2 + (60)^2 ) = 4 sqrt ( (7)^2 + (15)^2 ) = 4 xx 16.55 = 66.20 cm`

Thus, the curved surface area of the frustum `= π r_1 l_1 – π r_2 l_2`

`= 22/7 (28)(66.20) – 22/7 (7) (16.55) = 5461.5 cm^2`

Now, the total surface area of the frustum

= the curved surface area `+ π r_(1)^2 + π r_(2)^2`

` = 5461 .5 cm^2 +22/7 (28)^2 cm^2 + 22/7 (7)^2 cm^2`

`= 5461.5 cm^2 + 2464 cm^2 + 154 cm^2 = 8079.5 cm^2`.

Let `h` be the height, `l` the slant height and `r_1` and `r_2` the radii of the ends
`(r_1 > r_2)` of the frustum of a cone. Then we can directly find the volume, the
curved surace area and the total surface area of frustum by using the formulae
given below :

(i) Volume of the frustum of the cone `= 1/3 pi h ( r_(1)^2 + r_(2)^2 + r_1 r_2)`

(ii) the curved surface area of the frustum of the cone `= π(r_1 + r_2)l`

where ` l = sqrt ( h^2 + ( r_1 - r_2 )^2 )`

(iii) Total surface area of the frustum of the cone `= πl ( r_1 + r_2) + π r_(1)^2 + π r_(2)^2`,

where` l = sqrt ( h^2 + ( r_1 - r_2)^2 )`.

These formulae can be derived using the idea of similarity of triangles but we
shall not be doing derivations here.

Let us solve Example 12, using these formulae :

(i) Volume of the frustum `= 1/3 pi h ( r_(1)^2 + r_(2)^2 + r_1 r_2)`

` = 1/3 * 22/7 * 45 * [ (28)^2 + (7)^2 + (28) (7) ] cm^3`

` =48510 cm^3`

(ii) We have ` l = sqrt ( h^2 + ( r_1 -r_2)^2 ) = sqrt ( (45)^2 + (28 - 7 )^2 ) cm`

` = 3 sqrt ( (15)^2 + (7)^2 ) = 49.65 cm`

So, the curved surface area of the frustum

` = pi ( r_1 + r_2) l = 22/7 (28 + 7) (49 .65 ) = 5461.5 cm^2`


(iii) Total curved surface area of the frustum

` = pi ( r_1 + r_2 ) l + pi r_(1)^2 + pi r_(2)^2`

` = [ 5461 .5 +22/7 (28)^2 + 22/7 (7)^2] cm^2 = 8079.5 cm^2`

Let us apply these formulae in some examples.
Q 3240001813

Hanumappa and his wife Gangamma are busy making jaggery out of sugarcane juice. They have
processed the sugarcane juice to make the molasses, which is poured into moulds in the shape of a frustum of
a cone having the diameters of its two circular faces as `30 cm` and `35 cm` and the vertical height of the mould is
`14 cm` (see Fig. 13.22). If each `cm^3` of molasses has mass about 1.2 g, find the mass of the molasses that can
be poured into each mould. (Take ` pi = 22/7 ` )
Class 10 Chapter 13 Example 13
Solution:

Since the mould is in the shape of a frustum of a cone, the quantity (volume)

of molasses that can be poured into it `= pi/3 h ( r_(1)^2 + r_(2)^2 + r_1 r_2)` ,

where `r_1` is the radius of the larger base and `r_2` is the radius of the smaller base.

` = 1/3 xx 22/7 xx 14 [ ( 35/2)^2 + ( 30/2)^2 + (35/2 xx 30/2) ] cm^3 = 11641.7 cm^3` .

It is given that `1 cm^3` of molasses has mass `1.2g`. So, the mass of the molasses that can
be poured into each mould `= (11641.7 × 1.2) g`

`= 13970.04 g = 13.97 kg = 14 kg` (approx.)
Q 3260001815

An open metal bucket is in the shape of a frustum of a cone, mounted on a
hollow cylindrical base made of the same metallic sheet (see Fig. 13.23). The diameters of
the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket
is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used
to make the bucket, where we do not take into account the handle of the bucket. Also, find
the volume of water the bucket can hold.

(Take ` pi = 22/7` ) .
Class 10 Chapter 13 Example 14
Solution:

The total height of the bucket `= 40 cm`, which includes the height of the
base. So, the height of the frustum of the cone `= (40 – 6) cm = 34 cm`.

Therefore, the slant height of the frustum,` l = sqrt ( h^2 + (r_1 - r_2 )^2 )`

where `r_1 = 22.5 cm, r_2 = 12.5 cm` and `h = 34 cm`.

So, ` l = sqrt ( 34^2 + ( 22.5 -12.5)^2 ) cm`

` = sqrt ( 34^2+ 10^2) = 35.44 cm`

The area of metallic sheet used = curved surface area of frustum of cone
+ area of circular base
+ curved surface area of cylinder

` = [ pi xx 35.44 (22.5 +12.5) + pi xx (12.5)^2 +2 pi xx 12.5 xx 6 ] cm^2`

`= 22/7 (1240.4 +156.25 +150 ) cm^2`

` = 4860.9 cm^2`

Now, the volume of water that the bucket can hold (also, known as the capacity
of the bucket)

` =( pi xx h)/3 xx ( r_(1)^2 + r_(2)^2 + r_1 r_2)`

`= 22/7 xx 34/3 xx [ (22.5)^2 + (12.5)^2 + 22.5 xx 12.5 ] cm^3`

`= 22/7 xx 34/3 xx 943.75 =33615.48 cm^3`

`= 33.62 ` liters (approx)
 
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