An open metal bucket is in the shape of a frustrum 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 45cm and 25cm,the total vertical height of the bucket is 40cm and that of the cylindrical base is 6cm Find the area of metallic sheet used to make a bucket.
Answers
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
Height of frustum, h = 21 cm
Radius of lower end, r = 10 cm
Radius of upper end, R = 20 cm
Volume of frustum = 1/3 * pi * h * (R2 + r2 + Rr)
Put in the values to get the volume as 15.4 litre.
Thus, the cost of milk that can completely fill the bucket is Rs (30 x 15.4) = Rs 462