Question

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/)​

Answer 1

Answer:

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

Step-by-step explanation:

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+ ( 45/2 ² - 25/2 ² )

=35.44 cm

r1 = 45/2 and r2= 25/2

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

______☺️Brainliest please ☺️______

Answer 2

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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=√ h^2+(r1−r2)^2

where r1 = 22.5 cm, r2 = 12.5 cm and h = 34 cm.

_______________

l =√34^2+(22.5–12.5)^2

________

  = √ 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

= [π × 35.44 (22.5 + 12.5) + π × (12.5)^2 + 2π × 12.5 × 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

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

= (πh)/3 × (r1^2 + r2^2 + r1r2)

=(22/7) × (34/3) × [(22.5)2 + (12.5)2 + (22.5)(12.5)]

=(22/7) × (34/3) × 943.75

= 33615.48 cm^3

= 33.62 litres (approx.)