How much iron sheet (in m$$^2$$) will be needed to construct a rectangular tank measuring 10 m $$\times$$ 8 m $$\times$$ 6 m, if a circular opening of radius one meter is to be left at the top of the tank? (correct to one decimal place)
As per the given question,
The dimension of the tank $$=10 m\times8 m \times 6 m$$
Hence, total surface area of the tank will be $$=2(l\times b +b\times h+h\times l)$$
$$\Rightarrow 2(10\times 8+8\times 6+6\times10)=2(80+48+60)=376$$
The area of the circular part $$=\pi r^2=\pi \times 1^2=\pi$$
Hence the required area $$=376-\pi=372.86 \cong=372.9m^2$$
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