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:
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.✌️
Answer:–
[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.
r1 = [tex] \frac{45}{2}[/tex] and r2= [tex] \frac{25}{2}[/tex] )
CSA of cylinder :- 2πrh
A of circle :- πr² = π x (25/2)²
= 156.25π CM²
Hence , the Area of the metallic sheet is 4860.9 CM² and Volume of Bucket is 33615.78 cm³