Question

A pit of dimensions 20 m x 5.5 m x 4 m is dug at one corner of a field having dimensions 30 m x 12 m. If the earth that was removed was spread evenly over the remaining area of the field, then determine the rise in the height of the field.

Answer 1

Answer:

11÷9

Step-by-step explanation:

20×5•5×4=440

12×30=360

440÷360=11/9

Answer 2

The rise in the height of field is 1.76 m.

Given:

• A pit of dimensions 20 m x 5.5 m x 4 m is dug at one corner of a field having dimensions 30 m x 12 m.
• If the earth that was removed was spread evenly over the remaining area of the field.

To find:

• Determine the rise in the height of the field.

Solution:

Formula/ Concept to be used:

• Volume of cuboid=l×b×h; where these are Length,breadth and height respectively.
• Area of rectangle= Length×breadth
• Remember that earth taken out will not be spreading on the area of dig.
• The rectangular field will become cuboid after spreading the earth upto a particular height.

Step 1:

Find the volume of earth taken out.

After diging pit of 20 m x 5.5 m x 4 m, the earth taken out is equal to the volume of the dig.

Volume of earth taken out $= 20 \times 5.5 \times 4$

Volume of earth taken out= 440 m³.

Step 2:

Find the rise in height of earth at rectangular field.

Let the rise in the level of field is h meters.

Soil is spread over the field leaving the area of dig.

Area of top surface of dig= 20×5.5 m²

Area of top surface of dig= 110 m²

Area of remaining field= (30×12)-110

Area of remaining field= 250 m²

Volume of the remaining field= $250 \times h$

Total volume of earth taken out= Volume of the field.

$250 \times h = 440$

or

$h = \frac{440}{250}$

or

$h = \frac{44}{25}$

or

$\bf \red{h = 1.76 \: m}$

Thus,

The rise in the height of field is 1.76 m.

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