Question

A cubical box has each edge$10cm$ and another cuboidal box is $12.5cm$ long, $10cm$wide and $8cm$ 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?

Answer 1

Given,

Edge of a cubical box = $10cm$

Length of the cuboidal box, l = $12.5cm$

Breadth of the cuboidal box, b = $10cm$

Height of cuboidal box, h = $8cm$

(i) We know that -

The lateral surface area of a cube =$4\times \left({edge}^2\right)$

So, the Lateral surface area of a cubical box =$4\times \left({edge}^2\right)$

=$4\times \left({10}^2\right)$

$=400{cm}^2$

And,

The lateral surface area of a cuboid

So, the Lateral surface area of the cuboidal box

$$\begin{gathered} =2(l+b)\times h \\ =2(12.5+10)\times 8{cm}^2 \\ =2\times 22.5\times 8{cm}^2 \\ =360{cm}^2 \end{gathered}$$

So, the lateral surface area of the cubical box is greater than the cuboidal box by $(400c{m}^2–360c{m}^2)=40c{m}^2$.

(ii) Now,

The total surface area of a cubical box

$$\begin{gathered} =6\times {\left(edge\right)}^2 \\ =6\times {\left(10\right)}^2=6\times 100 \\ =600c{m}^2 \end{gathered}$$

The total surface area of the cuboidal box is

$$\begin{gathered} =2(lb+bh+lh) \\ =2\times (12.5\times 10+10\times 8+8\times 12.5) \\ =2\times (125+80+100) \\ =2\times 305 \\ =610{cm}^2 \end{gathered}$$

Therefore, the total surface area of the cuboidal box is greater than the cubical box by $(610c{m}^2–600c{m}^2)=10c{m}^2$.