Question

Find the cost of digging a cuboidal pit 8m long, 6m broad and 3 m deep at the rate of ₹ 30 per m³

Answer 1

Answer:

Hope it helps ☺️☺️☺️☺️ ............

Answer 2

Answer:

${\large{\underline{\underline{\pmb{\frak{Given:-}}}}}}$

A Cuboidal Pit

• ★ 8 m long
• ★ 6 m broad
• ★ 3 m deep

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\pmb{\frak{To \: Find:-}}}}}}$

• ★ The cost of digging cuboidal pit at the rate of ₹ 30 per m³.

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\pmb{\frak{Using \: Formula :-}}}}}}$

$\quad\bigstar{\underline{\boxed{\sf{Volume \: of \: Cuboid = lbh }}}}$

$\quad\bigstar{\underline{\boxed{\sf{Total \: Cost = Volume \: of \: cuboidal \: pit \times Cost \: of \: digging \: per \: {m}^{3}}}}}$

Where

• ★ L = Lenght
• ★ B = Breadth
• ★ H = Height

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\pmb{\frak{Diagram:-}}}}}}$

$\setlength{\unitlength}{0.74 cm}\begin{picture}\thicklines\put(5.6,5.4){\bf}\put(11.1,5.4){\bf}\put(11.2,9){\bf}\put(5.3,8.6){\bf}\put(3.3,10.2){\bf}\put(3.3,7){\bf}\put(9.25,10.35){\bf}\put(9.35,7.35){\bf}\put(3.5,6.1){\sf 6\:m}\put(7.7,6.3){\sf 8\:m}\put(11.3,7.45){\sf 3\:m}\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\end{picture}$

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\pmb{\frak{Solution:-}}}}}}$

$\red\bigstar$ Here :-

• ★ Lenght of cuboid = 8 m
• ★ Breadth of cuboid = 6 m
• ★ Height of cuboid = 3 m
• ★ Cost of digging per m³ = ₹30

$\begin{gathered}\end{gathered}$

$\red\bigstar$ Finding the Volume of Cuboidal Pit :-

$\quad{\dashrightarrow{\sf{Volume \: of \: cuboid = l \times b \times h }}}$

• Substuting the values

$\quad{\dashrightarrow{\sf{Volume \: of \: cuboid = 8 \times 6 \times 3 }}}$

$\quad{\dashrightarrow{\sf{Volume \: of \: cuboid = 144 \: {m}^{3}}}}$

$\quad{\bigstar{\underline{\boxed{\sf{\purple{Volume \: of \: cuboid = 144 \: {m}^{3}}}}}}}$

∴ The Volume of cuboidal pit is 144 m³.

$\begin{gathered}\end{gathered}$

$\red\bigstar$ Now, Calculating the cost of digging cuboidal pit at the rate of ₹ 30 per m³ :-

$\quad{\dashrightarrow{\sf{Total \: Cost = Volume \: of \: cuboidal \: pit \times Cost \: of \: digging \: per \: {m}^{3}}}}$

• Substuting the values

$\quad{\dashrightarrow{\sf{Total \: Cost = 144\times 30}}}$

$\quad{\dashrightarrow{\sf{Total \: Cost = 4320}}}$

$\quad{\bigstar{\underline{\boxed{\sf{\purple{Total \: Cost = 4320}}}}}}$

∴ The cost of digging cuboidal pit is ₹4320.

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\pmb{\frak{Learn \: More :-}}}}}}$

$\red\bigstar$ Formulas related to SA & Volume :-

$\begin{array}{|c|c|c|}\cline{1-3}\bf Shape&\bf Volume\ formula&\bf Surface\ area formula\\\cline{1-3}\sf Cube&\tt l^3}&\tt 6l^2\\\cline{1-3}\sf Cuboid&\tt lbh&\tt 2(lb+bh+lh)\\\cline{1-3}\sf Cylinder&\tt {\pi}r^2h&\tt 2\pi{r}(r+h)\\\cline{1-3}\sf Hollow\ cylinder&\tt \pi{h}(R^2-r^2)&\tt 2\pi{rh}+2\pi{Rh}+2\pi(R^2-r^2)\\\cline{1-3}\sf Cone&\tt 1/3\ \pi{r^2}h&\tt \pi{r}(r+s)\\\cline{1-3}\sf Sphere&\tt 4/3\ \pi{r}^3&\tt 4\pi{r}^2\\\cline{1-3}\sf Hemisphere&\tt 2/3\ \pi{r^3}&\tt 3\pi{r}^2\\\cline{1-3}\end{array}$

$\begin{gathered}\end{gathered}$

${\large{\underline{\underline{\pmb{\frak{Request:-}}}}}}$

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