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

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 as shown in the figure. 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 π = 22/7)

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2 thoughts on “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”

  1. Answer:

    Answer:-

    The Area of the metallic sheet is 4860.9 CM²

    and Volume of Bucket is 33615.78 cm³

    Step-by-step explaination:-

    Area of the metallic sheet = CSA of frustum + CSA of cylinder + area of circle

    CSA of frustum :- π (r 1 + r 2 ) l

    h = 34 cm because height of cylinder is 6 cm.

    => l = √h²+(r1²-r2² )

    => l = √34²+ ( \frac{45²}{2}

    2

    45²

    – \frac{25²}{2}

    2

    25²

    )

    => 35.44 cm

    r1 = \frac{45}{2}

    2

    45

    and r2= \frac{25}{2}

    2

    25

    )

    => CSA = π ( 45/2 + 25/2 ) 35.44

    => 1240.4 π CM²

    CSA of cylinder :- 2πrh

    => 2 π 25/2 x 6

    => 150π CM²

    A of circle :- πr² = π x (25/2)²

    = 156.25π CM²

    Area of the metallic sheet = 1240.4 π + 150π + 156.25π

    = > π ( 1240.4+ 150 +156.25 )

    => 4860.9 CM²

    Volume of Bucket = Volume of frustum = 1/3 π h( r1 ²+r2²+r1xr2)

    => 1/3 x 22/7 x 34 ( 45/2 x 45/2 + 25/2 x 25/2 + 45 x 25/4 )

    => 33615.78 cm³

    Hence , the Area of the metallic sheet is 4860.9 CM² and Volume of Bucket is 33615.78 cm³

    Step-by-step explanation:

    Hope this answer will help you.✌️

    Reply
  2. Answer:

    • The Area of the metallic sheet is 4860.9 CM²
    • and Volume of Bucket is 33615.78 cm³

    [tex] \\ [/tex]

    Step-by-step explaination:

    Area of the metallic sheet = CSA of frustum + CSA of cylinder + area of circle

    CSA of frustum :- π  (r 1 + r 2 ) l

    h = 34 cm because height of cylinder is 6 cm.

    • => l = √h²+(r1²-r2² )
    • => l = √34²+ ( [tex] \frac{45²}{2}[/tex] – [tex] \frac{25²}{2}[/tex] )
    • => 35.44 cm

    r1 = [tex] \frac{45}{2}[/tex] and r2= [tex] \frac{25}{2}[/tex] )

    • => CSA =  π ( 45/2 + 25/2 ) 35.44
    • => 1240.4  π  CM²

    CSA of cylinder :- 2πrh

    • => 2 π 25/2 x 6
    • => 150π CM²

    A of circle :- πr² = π x (25/2)²

    = 156.25π CM²

    • Area of the metallic sheet = 1240.4  π + 150π + 156.25π
    • = > π ( 1240.4+ 150 +156.25 )
    • => 4860.9 CM²

    • Volume of Bucket = Volume of frustum = 1/3  π h( r1 ²+r2²+r1xr2)
    • => 1/3 x 22/7 x 34 ( 45/2 x 45/2 + 25/2 x 25/2 + 45 x 25/4 )
    • => 33615.78 cm³

    Hence , the Area of the metallic sheet is 4860.9 CM² and Volume of Bucket is 33615.78 cm³

    Reply

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