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In the middle of the $13.5 m × 2.5 m$ area, a rectangular tank measuring the dimensions $5 m × 4.5 m × 2.1 m$ is dug. Over the remaining area of the field, the dug-out earth is scattered. Then the field's level is raised by ____ m.

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Detailed Solution

In the middle of the

13.5 m × 2.5 m

area, a rectangular tank measuring the dimensions

5 m × 4.5 m × 2.1 m

is dug. Over the remaining area of the field, the dug-out earth is scattered. Then the field's level is raised by 4.2 m.
Given that the tank's dimension is

5 m × 4.5 m × 2.1 m,

and the dimension of the middle of the area is

13.5 m × 2.5 m .

We know that the volume of the cuboid is

l × b × h,

where

l

is the length,

b

is the breadth and

h

is its height.
Determine the earth's volume after the hole was dug as depicted.

V = 5 \times 4.5 \times 2.1

\Rightarrow V = 5 \times \frac{45}{10} \times \frac{21}{10}

\Rightarrow V = 5 \times \frac{9}{2} \times \frac{21}{10}

\Rightarrow V = 5 \times \frac{45}{10} \times \frac{21}{10}

\Rightarrow V = \frac{9}{2} \times \frac{21}{2}

\Rightarrow V = \frac{189}{4}

Therefore the area that the earth occupies can be calculated as,

A = (13.5 \times 2.5) - (5 \times 4.5)

\Rightarrow A = (\frac{135}{10} \times \frac{25}{10}) - (5 \times \frac{45}{10})

\Rightarrow A = (\frac{27}{2} \times \frac{5}{2}) - (5 \times \frac{9}{2})

\Rightarrow A = (\frac{135}{4} - \frac{45}{2})

\Rightarrow A = \frac{45}{4}

The raised field's level can be calculated as,

h = volume Area = 1894 × 445 = 4.2 m

Therefore the field's level is raised by 4.2 m.

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