An open metal bucket is in the shape of a frustrating of a cone,mounted on a hollow cylinder Base made of the same metallic sheet. 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.Also,find the volume of water of the bucket can hold,in litres.
shashwat9792:
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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
please mark as BRAINliest answer
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
please mark as BRAINliest answer
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