The dimensions of a field are 20 text{ m} by | vedclass

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(D) Volume of earth removed $= 10 	ext{ m} 	imes 4.5 	ext{ m} 	imes 3 	ext{ m} = 135 	ext{ m}^3$.Total area of the field $= 20 	ext{ m} 	imes

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$1$

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(D) Volume of earth removed $= 10 	ext{ m} 	imes 4.5 	ext{ m} 	imes 3 	ext{ m} = 135 	ext{ m}^3$.Total area of the field $= 20 	ext{ m} 	imes 9 	ext{ m} = 180 	ext{ m}^2$.Area of the pit $= 10 	ext{ m} 	imes 4.5 	ext{ m} = 45 	ext{ m}^2$.Remaining area of the field $= 180 	ext{ m}^2 - 45 	ext{ m}^2 = 135 	ext{ m}^2$.Let the rise in the height of the field be $h 	ext{ m}$.Since the earth removed is spread over the remaining area,the volume of the earth removed equals the volume of the raised layer:$135 	ext{ m}^2 	imes h 	ext{ m} = 135 	ext{ m}^3$.Solving for $h$:$h = \frac{135}{135} = 1 	ext{ m}$.Therefore,the rise in the height of the field is $1 	ext{ m}$.

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