Question

The earth taken out from a pit is evenly spread over a rectangular field of length 90 m, width 60 m. If the, volume of the earth dug is 3078 m³ . Find the height of the field raised.​

Answer 1

Step-by-step explanation:

Volume of earth dug out =3696m

3

Length of rectangular field=80m

width of rectangular field=60m

∴ Height raised

=

Areaofrectangularpit

Volumeofearthdugout

=

80m×60m

3696m

3

=

4800

3696

m

=0.77m=77cm

Answer 2

$\large\underline{ \underline{ \sf \maltese{ \: Question:- }}}$

• The earth taken out from a pit is evenly spread over a rectangular field of length 90 m, width 60 m. If the, volume of the earth dug is 3078 m³ . Find the height of the field raised.

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$\large\underline{ \underline{ \sf \maltese{ \: Given:- }}}$

• Length = 90m
• Width = 60m
• Volume = 3078m^3

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$\large\underline{ \underline{ \sf \maltese{ \:To \: Find:- }}}$

• The height of the field .

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$\large\underline{ \underline{ \sf \maltese{ \: Solution:- }}}$

Volume of the Earth Dug = ( L × W × H ) m^3

$\qquad \quad {:} \longrightarrow \sf{\sf{3078 \: = \: (90 \: \times \: 60 \: \times \: h \: ) {m} }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ \frac{3078}{90 \: \times \: 60}m \: = \: \: height \: \: }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ \frac{3078}{5400} m \: = \: \: height \: \: }}$

1 metre = 100 cm

$\qquad \quad {:} \longrightarrow \sf{\sf{ \frac{3078}{5400} \: \times 100 \: = \: \: height \: \: }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ \frac{3078}{54 \cancel{00}} \: \times \: 1 \cancel{00}\: cm \: = \: \: height \: \: }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ \frac{3078}{54 }cm \: = \: \: height \: \: }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ 57cm \: = \: \: height \: \: }}$

$\qquad\quad {:} \longrightarrow \underline \red{\boxed{\sf{Height \: = \: 57cm }}}$

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★ Additional Information :

• 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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