Question

Example 5. A cubical box has each edge 10
cm and another cuboidal box is 12.5 cm long, 10 cm
(i) Which box has the greater lateral surface
wide and 8 cm high.
area and by how much?
(1) Which box has the smaller total surface area
and by how much?
​

Answer 1

❍ $\Large{\underline{\underline{\bf{Answer:}}}}$

$\large{\boxed{\sf{\red{(i)\:40cm^{2}}}}}$

$\large{\boxed{\sf{\red{(ii)\:10cm^{2}}}}}$

❍ $\Large{\underline{\underline{\bf{Explanation:}}}}$

(i) Each edge of the cubical box (α) = 10 cm

${\therefore}$ Lateral surface area of the cubical box

⠀⠀⠀⠀⠀⠀= 4α² = 4(10)² = 4(100) = 400 cm² .

For Cuboidal box

$\displaystyle{\sf{l=12.5cm,\:b=10cm,h=8cm}}$

${\therefore}$ Lateral surface area of the cuboidal box

⠀⠀⠀⠀⠀⠀:$\implies{\sf{2(l+b)h}}$

⠀⠀⠀⠀⠀⠀:$\implies{\sf{2(12.5+10)(8)}}$

⠀⠀⠀⠀⠀⠀:$\implies{\sf{360cm^{2}.}}$

${\therefore}$ Cuboidal box has the greater lateral surface area than the cuboidal box by ( 400 - 360 ) cm², i.e., 40cm².

$\large{\boxed{\sf{\orange{(i)\:40cm^{2}}}}}$

(ii) Total surface area of the cubical box

⠀⠀⠀⠀⠀⠀ = 6α² = 6(10)² = 6(100) = 600cm²

Total surface area of the cuboidal box

⠀⠀⠀⠀⠀⠀ :$\implies{\sf{2(lh+bh+hl)}}$

⠀⠀⠀⠀⠀⠀ :$\implies{\sf{2\big[(12.5)(10)+(10)(8)+(8)(12.5)\big]}}$

⠀⠀⠀⠀⠀⠀ :$\implies{\sf{2\big[125+80+100\big]}}$

⠀⠀⠀⠀⠀⠀ :$\implies{\sf{610cm^{2}}}$

${\therefore}$ Cubical box has the smaller total surface area than the cuboidal box by (610-600) cm² , i.e., 10cm².

$\large{\boxed{\sf{\orange{(ii)\:10cm^{2}}}}}$

__________________________

Answer 2

Given,

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

To find :

(1) Which box has the greater lateral surface  wide and 8 cm high area and by how much?

(2) Which box has the smaller total surface area  and by how much?

Solution :

(1) Edge of cubical box = 10 cm

⇒ LSA of cube = 4 * (Edge)²

⇒ LSA of cubical box = 4 * (10)²

⇒ LSA of cubical box = 4 * 100

⇒ LSA of cubical box = 400 cm²

Now , dimensions of cuboidal box = 12.5 cm long, 10 cm wide and 8 cm high.

⇒ LSA of cuboid = 2(l + b) * h

⇒ LSA of cuboidal box = 2 * (12.5 + 10) * 8

⇒ LSA of cuboidal box = 2 * 22.5 * 8

⇒ LSA of cuboidal box = 360 cm²

Therefore,

⇒ Difference b/w LSA = LSA of cubical box - LSA of cuboidal box

⇒ Difference b/w LSA = 400 - 360

⇒ Difference b/w LSA = 40 cm²

∴ Cubical box has greater lateral surface area and by 40 cm²

___________________________

(2) Edge of cubical box, a = 10 cm

⇒ TSA of cube = 6 * (Edge)²

⇒ TSA of cubical box = 6 * (10)²

⇒ TSA of cubical box = 6 * 100

⇒ TSA of cubical box = 600 cm²

Now dimensions of cuboidal box = 12.5 cm long, 10 cm wide and 8 cm high.

⇒ TSA of cuboid = 2(lb + bh + hl)

⇒ TSA of cuboidal box = 2(12.5 * 10 + 10 * 8 + 8 * 12.5)

⇒ TSA of cuboidal box = 2(125 + 80 + 100)

⇒ TSA of cuboidal box = 2 * 305

⇒ TSA of cuboidal box = 610 cm²

Now,

⇒ Difference b/w TSA = TSA of cuboidal box - TSA of cubical box

⇒ Difference b/w TSA = 610 - 600

⇒ Difference b/w TSA = 10 cm²

∴ Cubical box has smaller total surface area and by 10 cm²