Question

The length breadth and height of a cuboidal box are 24 m, 18 m and 10 m. How many cubes
of side 2 m can be placed in the given cuboidal box.​

Answer 1

$\huge \frak {Given}$

Length of cuboid = 24 m

Breadth of cuboid = 18 m.

Height of cuboid = 10 m

Edge of cube = 2 m

$\huge \frak {To \: Find}$

How many cube can be placed

$\huge \frak {Solution}$

Firstly we will find volume of cube and cuboidal box

TSA = 2(lb + bh + lh)

TSA = 2(24×18 +18×10+ 24×10)

TSA = 2(432+180+240)

TSA = 1704 m³

Now,

TSA of cube = 6a²

TSA = 6(2)²

TSA = 6 × 4

TSA = 24 M³

Now,

Total cuboidal box = $\tt\frac{TSA \: of \: cuboid}{TSA\: of\: cube }$

Total box = $\frac{1704}{24}$

Total box = 71 box

Hence :-

$\huge \frak {71\: box}$

Answer 2

Answer :-

• No. of Cubes can be placed in cuboidal box = 540 Cubes

Given :-

• Length of Cuboid = 24 m
• Width of Cuboid = 18 m
• Height of Cuboid = 10 m
• Side of Cube = 2 m

To Find :-

• No. of Cubes can be placed in cuboidal box = ?

Explanation :-

As above given, we have to find the no. of cubes that can be placed in the cuboid Box. It means we need to find the Volume of the Both Solid Shape.

$\purple { \boxed { \bf \bigstar \: Volume \: of \: Cuboid = L \times B \times H \: \bigstar }}$

$\rm : \: \implies (24 \times 18 \times 10) \: m^3$

$\rm : \: \implies (432 \times 10) \: m^3$

$\bf : \: \implies 4320 \: m^3 \quad \star$

$\purple { \boxed { \bf \bigstar \: Volume \: of \: Cube = (Side)^3 \: \bigstar }}$

$\rm : \: \implies (2)^3 \: m^3$

$\rm : \: \implies (2 \times 2 \times 2) \: m^3$

$\rm : \: \implies (4 \times 2) \: m^3$

$\bf : \: \implies 8 \: m^3 \quad \star$

Now, Finding the No. of Cubes

$\pink { \boxed { \bf \bigstar \: No. \: of \: Cubes = \dfrac{Volume \: of \: Cuboid}{Volume \: of \: Cube} \: \bigstar }}$

$\rm : \: \implies \dfrac{4320}{8} \: \: Cubes$

$\red { \bf : \: \implies \star \quad 540 \: \: Cubes \quad \star }$

Hence, the No. of cubes that can be placed in the Cuboidal box  is 540 Cubes.

$\underbrace { \underline { \bf \huge \text {More To Know}}}$

$\rm Cuboid \implies \begin{cases} \rm Volume = L \times B \times H \\ \rm TSA = 2(LB + BH + HL) \\ \rm LSA = 2H(L+B) \\ \rm Diagonal = \sqrt{(l)^2 + (b)^2 + (h)^2} \end{cases}$

$\rm Cube \implies \begin{cases} \rm Volume = (Side)^3 \\ \rm TSA = 6(Side)^2 \\ \rm LSA = 4(Side)^2 \\ \rm Diagonal = \sqrt{3(Side)} \end{cases}$