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 (see figure). 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,where we do not take into account the handle of the bucket. Also,find the volume of water the bucket can hold. (Take $\pi = \frac{22}{7}$)

(N/A) The total height of the bucket is $40 \, cm$,which includes the height of the base. So,the height of the frustum of the cone is $h = 40 - 6 = 34 \, cm$.
The radii are $r_1 = \frac{45}{2} = 22.5 \, cm$ and $r_2 = \frac{25}{2} = 12.5 \, cm$.
The slant height of the frustum is $l = \sqrt{h^2 + (r_1 - r_2)^2} = \sqrt{34^2 + (22.5 - 12.5)^2} = \sqrt{1156 + 100} = \sqrt{1256} \approx 35.44 \, cm$.
The area of the metallic sheet used is the sum of the curved surface area of the frustum,the area of the circular base (bottom),and the curved surface area of the cylindrical base.
Area $= \pi l(r_1 + r_2) + \pi r_2^2 + 2 \pi r_2 h_{base}$
$= \frac{22}{7} \times 35.44 \times (22.5 + 12.5) + \frac{22}{7} \times (12.5)^2 + 2 \times \frac{22}{7} \times 12.5 \times 6$
$= \frac{22}{7} \times (1240.4 + 156.25 + 150) = \frac{22}{7} \times 1546.65 \approx 4860.9 \, cm^2$.
The volume of water the bucket can hold is the volume of the frustum:
$V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$
$= \frac{1}{3} \times \frac{22}{7} \times 34 \times (22.5^2 + 12.5^2 + 22.5 \times 12.5)$
$= \frac{1}{3} \times \frac{22}{7} \times 34 \times (506.25 + 156.25 + 281.25) = \frac{1}{3} \times \frac{22}{7} \times 34 \times 943.75 \approx 33615.48 \, cm^3$.