Question

please reply fast. fast​

Answer 1

Dimensions of pit :-

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{length \: of \: pit = 10 \: m} \\ &\sf{breadth \: of \: pit \: = 10 \: m}\\ &\sf{height \: of \: pit \: = 5 \: m} \end{cases}\end{gathered}\end{gathered}$

Dimensions of field :-

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{Length \: of \: field \: = 30 \: m} \\ &\sf{Breadth \: of \: field \: = 20 \: m} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find :-  \begin{cases} &\sf{rise \: in \: the \: level \: of \: field}  \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used \::}}}} \end{gathered}$

${\red\bigstar\: { \boxed{{\bold\green{Volume_{(Cuboid)}\:\ = \: Length \times Breadth \times height }}}}}$

${\red\bigstar\: { \boxed{{\bold\purple{Area_{(rectangle)}\:\ = Length \times Breadth }}}}} \:$

${\red\bigstar\: { \boxed{{\bold\pink{\:height \: = \dfrac{Volume_{(Cuboid)}}{\:Base \: Area} }}}}} \:$

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

Dimensions of pit :-

:⟹ ★Length = 10 m

:⟹ ★Breadth = 10 m

:⟹ ★Depth = 5 m

$\tt \:Volume \: of \: the \: earth \: dug \: out =  \: Volume \: of \: pit$

$\tt \: Volume \: of \: the \: earth \: dug \: out =  \: 10 \times 10 \times 5$

$\boxed{ \red{\tt \: Volume \: of \: the \: earth \: dug \: out = \: 500 \: {m}^{3} }}$

Now,

$\tt \: Area_{(field)} \: = \: 30 \times 20 = 600 \: {m}^{2}$

$\tt \: Area_{(pit)} \: = \: 10 \times 10 = 100 \: {m}^{2}$

So,

$\tt \: Area_{(remaining \: field)} \: = Area_{(field)} \: - \: Area_{(pit)}$

$\tt \: Area_{(remaining \: field)} \: =600 - 100$

$\implies \boxed{ \red{\tt \: Area_{(remaining \: field)} \: = \: 500 \: {m}^{2} }}$

:⟹ ★Let rise in level of the field be 'h' m.

:⟹ ★So, rise in level is given by

$\bf \: height \: = \dfrac{\red{\tt \: Volume \: of \: the \: earth \: dug \: out}}{ \red{\tt \: Area_{(remaining \: field)} \: }}$

$:  \implies \bf \: h \: = \dfrac{500}{500}$

$:  \implies \boxed{ \purple{ \bf \: h \: = \: 1 \: m}}$

─━─━─━─━─━─━─━─━─━─━─━─━─

Additional Information

Cuboid :-

• l = Length of Cuboid
• b = Breadth of Cuboid
• h = Height of Cuboid
• Volume = l × b × h
• Curved Surface Area = 2 × (l + b) × h
• Total Surface Area = 2 × (l×b + b×h + h×l)
• Perimeter of Cuboid = 4 × (l + b + h)

Cube :-

• Volume = (side)³
• Curved Surface Area = 4 × (side)²
• Total Surface Area = 6 × (side)²
• Perimeter of cube = 12 × side