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Answer :

Let \( h\) be the required height of the platform.

The shape of the platform will be like the shape of a cuboid of dimensions \(22 \ m \ × \ 14 \ m \ × \ h \) with a hole in the shape of cylinder of radius \(3.5 m\) and depth \(h\).

The volume of the platform will be equal to the volume of the earth dug out from the well.

Now, the volume of the earth = Volume of the cylindrical well

\(= \ \pi r^2 h \ = \ \frac{22}{7} \ × \ 12.25 \ × \ 20 \ = \ 770 \) m^{3}

Also, the volume of the platform \( = \ 22 \ × \ 14 \ × \ h \) m^{3}

But volume of the platform = Volume of the well

\(\Rightarrow \ 22 \ × \ 14 \ × \ h \ = \ 770 \)

\( \Rightarrow \ h \ = \ \frac{770}{22 \ × \ 14} \ = \ 2.5 \) m

\(\therefore \) Height of the platform = 2.5 m

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