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. 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 the bucket can hold, in litres.​

Answer 1

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

Diameter of the smaller circular end = 25 cm

$Radius = \frac{45}{2} = 22.5 cm$

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

Amannnscharlie