Question

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

Answer 1

Given :-

• Area of the rectangular field = 800m²

• Volume of the water to irrigate 800m² rectangular field = 160m³

To Find :-

• Height of the water level in the rectangular field

Solution :-

➻ let the length of the rectangular field be ' l ' meter

➻ And, The breadth of the field be ' b ' meter

As we know that,

Area of the rectangular field = length × breadth

$\therefore\bf{l × b = 800m²}$

Let the height of the water level in field be ' h 'meter

Now,

Volume = l × b × h

By substituting values,

$\implies\sf{l × b × h = 160m³}$

$\implies\sf{800 × h =160m³(l \times b = 800 )}$

$\implies\sf{h = \dfrac{160}{800}m }$

$\implies\sf{h = \dfrac{1}{5} m}$

$\implies\mathtt{h =0.2m \: or \: 20cm \: }$

Hence, The height of the water level in the field is 0.2m or 20cm

Answer 2

$\begin{gathered}\begin{gathered}\bf Given - \begin{cases} &\sf{Area \: of \: the \: rectangular \: field = \: 800 \: m²} \\ &\sf{Volume \: of \: the \: water \: to \: irrigate = \: 160 \: m³} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf To \: Find :- \begin{cases} &\sf{Height \: of \: the \: water \: level} \end{cases}\end{gathered}\end{gathered}$

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$\large\underline\purple{\bold{Solution :- }}$

$\begin{gathered}\begin{gathered}\bf Let :- \begin{cases} &\sf{height \: of \: the \: water \: level \: be \: ' h ' \: m} \end{cases}\end{gathered}\end{gathered}$

$\bf \:  \bf \:   ✬ \: As \: we \: know \: that,$

$\bf \:  ⟼ ✬ \: Volume \: = area \: \times \: height$

$\bf\implies \: \: 160 \: = \: 800 \: \times \: h$

$\bf\implies \:h \: = \: \dfrac{ \cancel{160}}{ \cancel{800}} = \dfrac{1}{5} \: m$

$\bf\implies \:h \: = \dfrac{1}{5} \times 100 = 20 \: cm$

☆ Hence, The height of the water level in the field is 20cm

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$\large \red{\bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: Explore \: more } ✍$

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✬ More information ✬

Perimeter of rectangle = 2(length× breadth)

Diagonal of rectangle = √(length ²+breadth ²)

Area of square = side²

Perimeter of square = 4× side

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²

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