Question

please answer this question (WITH STATEMENTS ) ASAP

Answer 1

Length of field = 30m
Breadth of field = 18m
So area of field = 30×18 = 540 m²

Length of pit = 6m
Breadth = 4m
Depth of pit = 3m
Area of pit = 24 m²
So volume of pit = 6×4×3 = 72 m³
Volume of pit = Volume of mud dug out

Volume of mud dug out will be spread on the field but not on the pit
So it will be spread on the remaining area which will be 540 - 24 = 516 m²

Let rise in the height be h
so h×516 = 72
so h = 0.1395 m
h = 13.95 cm
13.95 cm will be rise in height of field

Answer 2

Given that Length of the Field is 30 m

Breadth of the Field is 18 m

As it is Rectangle : Area of the Rectangle = Length × Breadth

Area of the Field = 30 × 18 = 540 m²

Given the Length of the Pit is 6 m

Breadth of the Pit is 4 m

Depth of the Pit is 3 m

As it is in shape of Cuboid :

Volume of Cuboid = Length × Breadth × Depth

Volume of the Pit = 6 × 4 × 3 = 72 m³

The Quantity of Earth removed from the Pit = 72 m³

Now this Quantity of Earth is spread evenly all over the Field.

But the Pit Area does not include in this Area

Area of the Field over which the Earth Removed from the Pit is :

Area of the Field - Area of the Pit

Area of the Pit is : Length of the Pit × Breadth of the Pit

Area of the Pit = 6 × 4 = 24 m²

Area of Field over which Earth is Spread = 540 - 24 = 516 m²

We should notice One thing that Volume of the Quantity of Earth removed from the Pit should be Equal to the Area of the Field over which Earth is Spread × Rise in the level of Field.

Rise in the level of Field × 516 = 72

Rise in the Level of the Field = $\frac{72}{516} = 0.139\;Meter$

Converting into Centimeters :

Rise in the Level of the Field = 13.9 cm