Question

Imagine a very huge industrial metallic bucket is in the shape of the frustum of a right circular cone whose top and bottom diameters are 10 m and 4 m respectively and a height of 4 m. a) what is the curved surface area of the bucket? and b)what is the volume of water if it’s to be filled to the brim if the thickness of the metal is 0.60 meters including the base?

Answer 1

refer to attachment...

hope it helps

Answer 2

Answer:

The volume of water in the bucket is  133.52 $m^{3}$

Step-by-step explanation:

From the above question,

They have given :

Imagine a very huge industrial metallic bucket is in the shape of the frustum of a right circular cone whose top and bottom diameters are 10 m and 4 m respectively and a height of 4 m.

a) what is the curved surface area of the bucket

The curved surface area of the bucket can be calculated using the formula for the surface area of a frustum of a right circular cone.

The formula is A = π((r1 + r2) * √(r1 - r2)$.^{2}$ + h)$.^{2}$ ).

In this case, r1 = 5 m (the average of the two radii) and

                     r2 = 2 m, and

                      h = 4 m.

Therefore,

         The curved surface area of the bucket is,

          A = π((5 + 2) * √(5 - 2)$.^{2}$  + 4)$.^{2}$ ) = 276.48 m$.^{2}$.

b) what is the volume of water if it’s to be filled to the brim if the thickness of the metal is 0.60 meters including the base

The volume of water in the bucket can be calculated using the formula for the volume of a frustum of a right circular cone.

The formula is,

        V = ($\frac{1}{3}$) π (h(r1 + r1r2 + r2$.^{2}$)). In this case,

        r1 = 5 m,

        r2 = 2 m, and

         h = 4 m.

Therefore, the volume of water in the bucket is,

                  V = ( $\frac{1}{3}$ )π(4(5$.^{2}$ + 5 * 2 + 2$.^{2}$))

                     = 133.52 $m^{3}$

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