Question

an an open metallic bucket is in the shape of a frustum of a cone if the diameter of the two circular ends of the bucket are 45 cm and 25 cm and the vertical height of the bucket is 24 cm find the area of the metallic sheet used to make the bucket also find the volume of water it can hold ​

Answer 1

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Answer 2

Answer:

Total area of Metalic sheet used = Ab = 3348.025 cm²

Total volume of water bucket can hold = V = 23707 cm³

Step-by-step explanation:

To find:

1) Total Surface area of the bucket = Ab

2) Volume of the bucket = Vb

Given: Bottom Diameter of the Bucket = d = 25 cm

Top Diameter of the ucket = D = 45 cm

Height of the bucket = H = 24 cm

1) Total area of metallic sheet used = Lateral area of frustum of the cone + Area of the bottom of the bucket (circular plate)

∴ Area of the frustum of the bucket is given by

Af = $\pi(R+r) \sqrt{(R-r)^{2} + H^{2}  }$

∴Af = $\pi(45/2+25/2) \sqrt{(45/2-25/2)^{2} + 24^{2}  }$

∴Af = 3.14×35×[√(10²+24²)]

∴Af = 109.9 × √(100+ 576)

∴ Af = 109.9 × √676 = 109.9 × 26

∴ Af = 2857.4  cm²     ..... (1)

Area of the bottom of the circular plate = Ac

Ac = ($\pi$×d²) / 4

∴ Ac = 3.14 × 25² / 4 = 3.14 × 625 / 4

∴ Ac = 490.625 cm²    ...... (2)

From eq. (1) and (2) we get the total area of the bucket, i.e. Area of metalic sheet required

Ab = Af + Ac

Ab = 2857.4 +  490.625

Ab = 3348.025 cm²

2) Volume of the Bucket: V

Volume of the frustum of the cone is given by

V = (1/3) * H * (R² + r² + R*r)

∴ V = (π/3) * 24 * {(45/2)² + (25/2)² + (45/2*25/2)}

∴V = π8 * (2025 + 625 + 1125)/4

∴ V = 3.14 × 8 × (3775 / 4)

∴ V = 23707 cm³  

Volume of the water bucket can hold = 23707 cm³