Question

A cubical box with dimensions 8cm by 6cm by 12cm is melted into another cube whose width is 16 cm . Find the length and height of the new cube formed if l=h

Answer 1

The length and height of the new cube is 6 cm.

Step-by-step explanation:

Given : A cubical box with dimensions 8 cm by 6 cm by 12 cm is melted into another cube whose width is 16 cm.

To Find : The length and height of the new cube formed if l=h ?

Solution :

The volume of a rectangular cuboid is

$V_1=l\times b\times h$

$V_{1}=8\times 6\times 12\ cm^{3}$

$V_{1}=576\ cm^{3}$

After the cuboid is melted into another cube whose width is 16 cm then the volume will be,

$V_{2}=16\times l\times h$

If l=h,

$V_{2}=16\times l\times l$

$V_{2}=16l^2$

Their volumes are equal i.e. $V_1=V_2$

$576=16l^2$

$36=l^2$

$l=6\ cm$

The length of the new cube is 6 cm.

The height of the new cube is 6 cm.

#Learn more

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Answer 2

Answer:

The length and height of the new cube are $6 \mathrm{~cm}$.

Step-by-step explanation:

Given:

A cubical box with dimensions $8 \mathrm{~cm}$ by $6 \mathrm{~cm}$ by $12 \mathrm{~cm}$ exists melted into another cube whose width exists $16 \mathrm{~cm}$.

To Find:

Length and height of the new cube made if $I=h$ ?

Step 1

The volume of a rectangular cuboid exists

$&V_{1}=l \times b \times h$

$&V_{1}=8 \times 6 \times 12 \mathrm{~cm}^{3}$

$&V_{1}=576 \mathrm{~cm}^{3}$

Step 2

After the cuboid exists melted into another cube whose width exists $16 \mathrm{~cm}$ then the volume will be,

$V_{2}=16 \times l \times h$

If $\mathrm{l}=\mathrm{h}$,

$&V_{2}=16 \times l \times l$

$&V_{2}=16 l^{2}$

Hence, volumes are equal.

Therefore, $V_{1}=V_{2}$

Step 3

Let, $&576=16 l^{2}$

$&36=l^{2}$

$&l=6 \mathrm{~cm}$

Length of the new cube = $6 \mathrm{~cm}$.

Height of the new cube = $6 \mathrm{~cm}$.

#SPJ2