an open metallic bucket is in the shape of frustum of a cone mounted on hollow cylindrical base made of metallic sheet if the diameter of two circular and of bucket are 45 cm and 25 cm and the total vertical height of the bucket is 30 cm and that of cylindrical portion is 6 centimetre find the volume of water it can hold...???
Answers
Answer:
Step-by-step explanation:
Area of metallic sheet used = CSA of Frustum + CSA of Cylinder + CSA of Base
CSA of Frustum -
Diameter of the bigger circular end = 45 cm
Radius = 45/2 = 22.5 cm
Diameter of the smaller circular end = 25 cm
Radius = 25/2 = 12.5 cm
Height of the frustum = Total height of the bucket - Height of the circular base
⇒ 40 - 6 = 34 cm
Slant Height = l √h² + (r1² - r2²)²
⇒ √34² + (22.5 - 12.5)²
⇒ √1156 + (10)²
⇒ √1156 + 100
⇒ √1256
⇒ Slant Height = 35.44 cm
CSA of Frustum = π(r1 + r2)l
⇒ 22/7*(22.5 + 12.5)*35.44
⇒ 22/7*35*35.44
= 3898.4 cm²
Area of Circular Base -
Base is a circular part with radius 25/2 = 12.5 cm
Area of circular base = πr²
⇒ 22/7*12.5*12.5
491.07 cm²
CSA of Cylinder = 2πrh
⇒ 2*22/7*12.5*6
⇒ 471.428
Area of metallic sheet used = 3898.4 cm² + 491.07 cm² + 471.428
= 4860.898 cm²