Find the cost of digging a cuboidal pit $8m$ long, $6m$ broad and $3m$ deep at the rate of $30$ per ${m^2}$.

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Hint: A cuboid is a three-dimensional structure having six rectangular faces. These six faces of the cuboid exist as a pair of three parallel faces. When the area of the faces of a cuboid is the same, we call this cuboid a cube. The area of all the faces of a cube is the same as they are all squares.
Volume of cuboid: The total surface area of a cuboid is equal to the sum of the areas of the six rectangular faces whereas the Lateral surface area of a cuboid equal to the sum of the four rectangular faces, in which two rectangular faces of the top and bottom faces are excluded.
Since cost is in \[30m\]. Therefore, Volume of cuboidal pit dug is directly proportional to cost.
Volume of Cuboid \[ = length \times breadth \times height\]

Complete step-by-step answer:
 Length of Cuboid \[ = 8m\]
Breadth of Cuboid \[ = 6m\]
Height of Cuboid \[ = 3m\]
According to the Volume of Cuboid Pi \[ + \]
                     \[ = length \times breadth \times height\]
                      \[ = 8 \times 6 \times 3\]
                       \[ = 144\]${m^3}$
Cost of digging cuboidal pit for $1{m^3}$$ = \,Rs.30$
Cost of digging cuboidal pit $144{m^3}$$ = Rs.(144 \times 30)$
                                                                $ = Rs.4320$
$\therefore $ Total Cost of digging the pit $ = Rs.4320$

Note: Since the cost is in ${m^3}$. So, there will be no alternative. But if there was a cost in ${m^2}$ then the formulae of Total Surface Area will be Applied as.
$T.S.A = 2(length \times breadth + breadth \times height + height \times length)$
And then T.S.A is multiplied with the cost to yet the total cost of digging the cuboidal pit.