Question

The perimeters of the ends of a frustum of a right circular cone are 44 cm and 33 cm. If the height of the frustum be 16 cm, find its volume, the slant surface and the total surface.

 

Answer 1

The height of the frustum of the cone is h= 16 cm. The perimeters of the circular ends are 44 cm and 33 cm. Let the radii of the bottom and top circles are r₁ cm and r₂ cm respectively. Then, we have

2πr₁= 44

⇒ πr₁= 22

⇒ `r_1+(22xx7)/22`

⇒ r₁ = 7

2πr₂ = 33

⇒ `pi r_2=33/2`

⇒ `r_2=33/2xx7/22`

⇒ `r_2=21/4`

The slant height of the bucket is

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

`=sqrt((7_21/4)^2+h^2)`

= 16.1 cm

The curved/slant surface area of the frustum cone is

`= pi(r_1+r_2)xxl`

`=(pir_1+pir_2)xxl`

= (22 + 16.5)xx16.1

= 619.85 cm²

Hence Curved surface area = 619.85cm²

The volume of the frustum of the cone is

`V=1/3pi(r_1^2+r_1r_2+r_2^2)xxh`

`=1/3pi(7^2+7xx5.25+5.25^2)xx16`

= 1900 cm³

Hence Volume of frustum = 1900cm³

The total surface area of the frustum cone is

`=pi(r_1+r_2)xxl+pi r_1^2+pi r_2^2`

`=619.85+22/7xx7^2+22/7xx5.25^2`

= 860.25 Square cm

Hence Total surface area = 860.25 cm²