Answer:
Given, the radii of top and bottom circular ends are 20 cm and 12 cm respectively. And, volume of frustum (bucket) \[=12308.8\text{ }c{{m}^{3}}\] \[\Rightarrow \frac{\pi h}{3}[{{R}^{2}}+{{r}^{2}}+Rr]=12308.8\] \[\frac{3.14\times h}{3}[400+144+240]=12308.8\] \[\therefore \] Height \[(h)=\frac{12308.8\times 3}{3.14\times 784}\] \[=\frac{36926.4}{2461.76}=15\,cm\] Slant height of the bucket \[(l)=\sqrt{{{h}^{2}}+{{(R-r)}^{2}}}\] \[=\sqrt{{{(15)}^{2}}+{{(20-12)}^{2}}}\] \[=\sqrt{225+64}=\sqrt{289}\] \[=17\,cm\] \[\therefore \] Area of metal sheet used in making the bucket = Curved surface area of frustum + Base area \[=\pi l(R+r)+\pi {{r}^{2}}\] \[=3.14\times 17\times (20+12)+3.14\times 12\times 12\] \[=3.14\times 17\times 32+3.14\times 144\] \[=3.14\text{ (}544+144)\] \[=3.14\times 688\] \[=2160.32\text{ }c{{m}^{2}}\]
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