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# A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high. i) Which box has the greater lateral surface area and by how much? ii) Which box has the smaller total surface area and by how much?

Given:
Edge of cube = 10 cm
Length (l) of box = 12.5 cm
Breadth (b) of box = 10 cm
Height (h) of box = 8 cm

i) We know that, Lateral surface area of cubical box
= $$4 × (Edge)^2$$
= $$4 × (10)^2 {cm}^2$$
= 400 $${cm}^2$$

Also, We have,

Lateral surface area of cuboidal box
= 2 [lh + bh]
= [2 (12.5 x 10 x 8)] $${cm}^2$$
= [2 x 180] $${cm}^2$$
= 360 $${cm}^2$$

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

Now, Lateral surface area of cubical box - Lateral surface area of cuboidal box
= 400 $${cm}^2$$ - 360 $${cm}^2$$ = 40 $${cm}^2$$

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

ii)Similarly, Total surface area of cubical box
=$$6 × (Edge)^2$$
= $$6 × (10)^2 {cm}^2$$
= 600 $${cm}^2$$

Also, We have,

Total surface area of cuboidal box
= 2(lh + bh + lb)
=[2 (12.5 x 8 + 10 x 8 + 12.5 x 100)] $${cm}^2$$
= 610 $${cm}^2$$

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

Now, Total surface area of cuboidal box - Total surface area of cubical box
= 610 $${cm}^2$$ - 600$${cm}^2$$
= 10 $${cm}^2$$

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