Answer
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Hint: We will first derive the formula for the surface area of a cube, knowing it has 6 faces and all squares. Using this formula, we will equate it with the given surface area and find the value of the edge of the cube. As the volume of the cube can be calculated by the formula, $V = {a^3}$, We will find the volume by substituting the value of edge length that we calculated in the previous step.
Complete step by step solution: A cube is a 3D object having six faces with all edge lengths equal and all faces as squares.
Now, for a square, we know that its area is equal to the square of its side.
Consider a cube of edge length ‘a’ units, as each face of the cube is a square of same side length, so the surface area of the cube will be 6 times the area of a single face.
So, the area of a single face with side ‘a’ units is ${a^2}$sq. units
Thus, the surface area of the cube will be 6 times ${a^2}$sq. units.
Now, the given surface area is 1734$c{m^2}$
Therefore, 1734 $c{m^2} = 6{a^2}$
\[\Rightarrow{a^2} = \dfrac{{1734}}{6}\]$c{m^2}$
\[\Rightarrow{a^2} = 289\]$c{m^2}$
$\Rightarrow{a} = 17$ cm
Thus, the side of the cube is 17 cm.
Now, for a cube of edge length ‘a’ units, the volume of the cube is equal to the cube (power 3) of its edge length.
Thus, for the cube with edge length $a = 17$ units, let V be the required volume
$V = {a^3}$
$\Rightarrow{V} = {\left( {17} \right)^3}c{m^3}$
$\Rightarrow{V} = 4913c{m^3}$
Hence, the volume of the cube is $4913c{m^3}$, so option B is correct.
Note: Students must not get confused with the units associated with different physical quantities. If an edge length has a unit of cm then the area associated with it will have the unit of $c{m^2}$and a volume associated with it will have the unit of $c{m^3}$.
Students must remember the formula of volume of different 3 D objects carefully, and do not get confused with the radius and diameter of an object. The volumes of various 3D objects are:
Sphere with radius r: $\dfrac{4}{3}\pi {r^3}$
Cuboid with dimensions l, b, h: $l \times b \times h$
Cylinder with height h and radius r: $\pi {r^2}h$
Cone with height h and radius r: $\dfrac{1}{3}\pi {r^2}h$
These formulas must be remembered to solve these type of problems.
Complete step by step solution: A cube is a 3D object having six faces with all edge lengths equal and all faces as squares.
Now, for a square, we know that its area is equal to the square of its side.
Consider a cube of edge length ‘a’ units, as each face of the cube is a square of same side length, so the surface area of the cube will be 6 times the area of a single face.
So, the area of a single face with side ‘a’ units is ${a^2}$sq. units
Thus, the surface area of the cube will be 6 times ${a^2}$sq. units.
Now, the given surface area is 1734$c{m^2}$
Therefore, 1734 $c{m^2} = 6{a^2}$
\[\Rightarrow{a^2} = \dfrac{{1734}}{6}\]$c{m^2}$
\[\Rightarrow{a^2} = 289\]$c{m^2}$
$\Rightarrow{a} = 17$ cm
Thus, the side of the cube is 17 cm.
Now, for a cube of edge length ‘a’ units, the volume of the cube is equal to the cube (power 3) of its edge length.
Thus, for the cube with edge length $a = 17$ units, let V be the required volume
$V = {a^3}$
$\Rightarrow{V} = {\left( {17} \right)^3}c{m^3}$
$\Rightarrow{V} = 4913c{m^3}$
Hence, the volume of the cube is $4913c{m^3}$, so option B is correct.
Note: Students must not get confused with the units associated with different physical quantities. If an edge length has a unit of cm then the area associated with it will have the unit of $c{m^2}$and a volume associated with it will have the unit of $c{m^3}$.
Students must remember the formula of volume of different 3 D objects carefully, and do not get confused with the radius and diameter of an object. The volumes of various 3D objects are:
Sphere with radius r: $\dfrac{4}{3}\pi {r^3}$
Cuboid with dimensions l, b, h: $l \times b \times h$
Cylinder with height h and radius r: $\pi {r^2}h$
Cone with height h and radius r: $\dfrac{1}{3}\pi {r^2}h$
These formulas must be remembered to solve these type of problems.
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