Question

12. A rectangular field is 70 m x 60 m. A well of dimensions 14 mx 5 mx 6 m is dug outside the field
and the earth dug out from this well is evenly spread over the rectangular field. How much will
the earth level of the field rise?​

Answer 1

Given :

• Rectangular field's dimensions = 70 m * 60 m
• Dimensions of wall = 14m * 5m * 6m
• Earth dugout from this wall is evenly spread over the rectangular field.

To find :

• How much earth level of field will rise

Solution :

Let the rise in earth level be x m

Now volume of wall = Volume of earth dugout

⇒ Volume of earth dugout = 14 * 5 * 6

⇒ Volume of earth dugout = 420 m³

Now volume of earth dugout is equal to volume of cuboid formed using length and breadth of rectangular field and height as earth level of field rise.

So according to question,

⇒ 420 = 70 * 60 * x

⇒ 420 = 4200 * x

⇒ x = 420/4200

⇒ x = 1/10 m

⇒ x = 0.1 m

Therefore,

Earth level of field will rise by 0.1 m

Answer 2

Answer:

Given

Length of rectangular field = 70m

Breadth of rectangular field= 60m

Dimensions of well = 14 m x 5 m x 6 m

Find

How much will the earth level of the field rise?

Solution

${ \to{ \sf{volume \: \: of \: \: well = 14 \times 5 \times 6}}}$

${ \to{ \sf{volume \: = 420}}}$

Volume of rectangular field = volume of well

${ \to{ \sf{lbh = 420}}}$

${ \to{ \sf{70 \times 60 \times x = 420}}}$

${ \to{ \sf{4200 \times x = 420}}}$

${ \to{ \sf{x = \frac{420}{4200} }}}$

${ \to{ \sf{x = \frac{1}{10}m }}}$

${ \to{ \sf{x = 0.1m}}}$

Therefore, 0.1m the earth level of field rise.