Question

21. 160 m3 of water is to be used to irrigate a rectangular field whose area is 800 m2. What will be the height of the water level in the field?

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Answer 1

Answer:

20 cm

Step-by-step explanation:

given area of rectangular field ,A= 800 m2

volume of water to irrigate the field ,V=160m3

volume = area of rectangular field (A)* height of water level (h)

V=A*H

on solving the above equation for h, we get h =V/A

=1/2m=20cm.

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Answer 2

Given :

• Volume of water used = 160m³
• Area of rectangle field = 800m²

To Find :

• The hieght of water level in the field.

Formula To be Applied:

• Area = l×b
• Volume = l×b×h
• Hieght = volume/l×b
• hieght = volume/area

Process :

• First we will calculate the area of field (Already given)
• Then, volume of water level (Already given)
• As the water level remains same in all areas, this will form a water cuboid.
• Then using the formula of volume of cuboid, we will find the hieght of water level.

Solution :

Area = l×b

So, Area of rectangle field = 800 m².

Volume of water used = 160m³ .

This water will spread equally in all areas of field evenly.

Hence, Volume of water = l×b×h

$\implies \tt{160 {m}^{3} = (l \times b) \times h}$

$\implies \tt{160 {m}^{3} = 800 \times h}$

$\implies \tt{h = \frac{160}{800}}$

$\red{ \underline{ \boxed{ \tt{ \bigstar \: h = 0.2m}}}}$

Hence, Hieght of Water level = 0.2 m

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Additional Information:

Some formula related to mensuration:

• Area of square = side²
• perimeter of Square = 4side
• Volume of cube = Side³
• TSA of cube = 6side²

• Area of rectangle = l×b
• perimeter of rectangle = 2(l+b)
• Volume of cuboid = l×b×h
• TSA of cuboid= 2(lb+lh+bh)

• Area of circle = (pi)r²
• Perimeter of circle = 2(pi)r
• Sphere volume = 4/3 (pi)r³
• Cylinder volume =(pi)r²h