A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long,

Solution:

(i) Edge of cube $=10 \mathrm{~cm}$

Length $(l)$ of box $=12.5 \mathrm{~cm}$

Breadth $(b)$ of box $=10 \mathrm{~cm}$

Height $(h)$ of box $=8 \mathrm{~cm}$

Lateral surface area of cubical box $=4(\text { edge })^{2}$

$=4(10 \mathrm{~cm})^{2}$

$=400 \mathrm{~cm}^{2}$

Lateral surface area of cuboidal box $=2[/ h+b h]$

$=[2(12.5 \times 8+10 \times 8)] \mathrm{cm}^{2}$

$=(2 \times 180) \mathrm{cm}^{2}$

$=360 \mathrm{~cm}^{2}$

Clearly, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box.

Lateral surface area of cubical box - Lateral surface area of cuboidal box $=400 \mathrm{~cm}^{2}-360 \mathrm{~cm}^{2}=40 \mathrm{~cm}^{2}$

Therefore, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box by $40 \mathrm{~cm}^{2}$.

(ii) Total surface area of cubical box $=6(\text { edge })^{2}=6(10 \mathrm{~cm})^{2}=600 \mathrm{~cm}^{2}$

Total surface area of cuboidal box

$=2[/ h+b h+l b]$

$=\left[2(12.5 \times 8+10 \times 8+12.5 \times 10] \mathrm{cm}^{2}\right.$

$=610 \mathrm{~cm}^{2}$

Clearly, the total surface area of the cubical box is smaller than that of the cuboidal box.

Total surface area of cuboidal box $-$ Total surface area of cubical box $=610 \mathrm{~cm}^{2}-600 \mathrm{~cm}^{2}=10 \mathrm{~cm}^{2}$

Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by $10 \mathrm{~cm}^{2}$.