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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.
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Solution
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
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