An open metal buc

Given: radius of upper circular end of frustum = R = 45 cm

Radius of lower circular end of frustum = radius of cylindrical base = r = 25 cm

Height of bucket = h_b = 40 cm

Height of cylindrical base = h_c = 6 cm

From the figure height of frustum = h_f = h_b - h_c

= 40 - 6

∴ h_f = 34 cm

Volume of cylinder = πr²h_c

Curved surface area of cylinder = 2πrh_c

curved surface area of frustum = π(R + r)l cm²

Where l = slant height

= √(400 + 1156)

= 39.44 cm

Area of metallic sheet used = curved surface area of frustum + curved surface area of base

cylinder + area of base circle of cylinder

now, curved surface area of frustum = π × (R + r) × l cm²

= (22/7) × (45 + 25) × 39.44 cm²

= 22 × 10 × 39.44 cm²

= 8676.8 cm²

Curved surface area of base cylinder = 2πrh_c

= 2 × (22/7) × 25 × 6

= 942.85 cm²

Surface area of base circle of cylinder = πr²

= (22/7) × 25²

= 1964.28 cm²

∴ Area of metallic sheet used = 8676.8 + 942.85 + 1964.28

= 11583.93 cm²

Therefore, area of metallic sheet used to make the bucket is 11583.93 cm² i.e. 1.158393 m²

Volume of water bucket can hold = volume of bas cylinder + volume of frustum

Volume of base cylinder = πr²h_c

= (22/7) × 25² × 6

= 11785.71 cm³

= 35.62 × 3775

= 134465.5 cm³

∴ volume of water bucket can hold = 11785.71 + 134465.5

= 146251.21 cm³

Now 1 litre is 1000 cm³

∴ 146251.21 cm³ = 146251.21/1000 = 146.25121 litres

Volume of water bucket can hold = 146.25121 litres