SSC (English Medium) Class 10th Board ExamMaharashtra State Board
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Frustum of a Cone

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definition

A frustum may be formed from a right circular cone by cutting off the tip of the cone with a cut perpendicular to the height, forming a lower base and an upper base that are circular and parallel.

 

notes

 

In ΔAFE and ΔAGC

ΔAFE ΔAGC (Angle-Angle rule)

 

`"AF"/"AG"= "FE"/"GC"= "AE"/"AC"`

 

`(h_1-h)/(h_1)= r_2/r_1= (l_1-l)/l_1 `

 

`(r_2)/(r_1)= (l_1-l)/(l_1)`

 

`r_2l_1= r_1(l_1-l)`

 

`r_2l_1= r_1l_1- r_1l`

 

`r_1l= r_1l_1- r_2l_1`

 

`r_1l= l_1 (r_1-r_2)`

 

`l_1= (r_1l)/ (r_1-r_2)` ...........eq1

 

`(h_1-h)/h_1= r_2/r_1= (l_1-l)/l_1`

 

`(h_1-h)/h_1= r_2/r_1`

 

`r_1(h_1-h)= h_1r_2`

 

`r_1h_1- r_1h= h_1r_2`

 

`r_1h_1- r_2h_1= r_1h`

 

`h_1(r_1-r_2)= r_1h`

 

`h_1= (r_1h)/(r_1-r_2)` ............eq2

 

1) Curved surface area of frustum= Curved surface area of ABC- Curved surface area of ADE

 

= `pi r_1l_1- pi r_2 (l_1-l)`

 

= `pi r_1l_1- pi r_2l_1 + pi r_2l`

 

= `pi l_1 (r_1- r_2) + pi r_2l`

 

= `pi r_1/(r_1-r_2) (r_1-r_2) +pi r_2l`       (from eq1)

 

= `pi r_1l + pi r_2l`

 

Curved surface area of frustum = `pi l(r_1+r_2)`

 

2) Total surface area of frustum= Curved surface area of frustum + area of the circle above+ area of the circle below

 

Total surface area of frustum = `pi l (r_1+r_2) + pi r_1^2+ pi r_2^2`

 

3) Volume of frustum= Volume of cone ABC- Volume of cone ADE

 

= `1/3  pi r_1^2h_1- 1/3  pi r_2^2 (h_1-h)`

 

=`1/3   pi r_1^2h_1- 1/3   pi r_2^2h_1+ 1/3 pi r_2^2h`

 

= `1/3   pi h_1 (r_1^2- r_2^2) +1/3   pi r_2^2h`

 

= `1/3  pi (r_1h)/(r_1-r_2) (r_1^2-r_2^2)+ 1/3  pi r_2^2h`     (from eq2)

 

= `1/3   pi (r_1h)/(r_1-r_2) (r_1-r_2) (r_1-r_2)+ 1/3 pi r_2^2h`

 

= `1/3   pi r_1h (r_1+r_2)+ 1/3   pi r_2^2h`

 

= `1/3 pi h [r_1 (r_1+r_2)+ r_2^2]`

 

Volume of frustum= `1/3   pi h (r_1^2+ r_2^2+ r_1r_2)`

Example- 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. 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)`

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_1r_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`

 

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 xx 1.2) g`

 

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



 

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Shaalaa.com | Surface Area and Volume part 15 (Frustum Concepts)

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Surface Area and Volume part 15 (Frustum Concepts) [00:03:15]
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