Question

A cubical box has each edge 8 cm and another cuboidal box is 8cm long, 6 cm wide and 4 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?

Answer 1

Answer:

Given :-

• A cubical box has each edge of 8 cm and another cuboidal box is 8 cm long, 6 cm wide and 4 cm high.

To Find :-

• (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 ?

Solution :-

(i) Which box has the greater lateral surface area and by how much :-

☆ First, we have to find the lateral surface area of a cubical box :

Given :

• Edge of cube = 8 cm

According to the question by using the formula we get,

$\footnotesize \implies \sf\boxed{\bold{\pink{Lateral\: Surface\: Area_{(Cube)} =\: 4(a)^2}}}$

where,

• a = Edge

So, by putting the values we get,

$\implies \sf L.S.A_{(Cubical\: Box)} =\: 4 \times (8)^2$

$\implies \sf L.S.A_{(Cubical\: Box)} =\: 4 \times 64$

$\implies \sf\bold{\purple{L.S.A_{(Cubical\: Box)} =\: 256\: cm^2}}$

☆ Again, we have to find the lateral surface area of cuboid :

Given :

• Length = 8 cm
• Breadth = 6 cm
• Height = 4 cm

According to the question by using the formula we get,

$\footnotesize \implies \sf\boxed{\bold{\pink{Lateral\: Surface\: Area_{(Cuboid)} =\: 2(lh + bh)}}}$

where,

• l = Length
• h = Height
• b = Breadth

By putting those values we get,

$\implies \sf L.S.A_{(Cuboidal\: Box)} =\: 2\{(8 \times 4) + (6 \times 4)\}$

$\implies \sf L.S.A_{(Cuboidal\: Box)} =\: 2(32 + 24)$

$\implies \sf L.S.A_{(Cuboidal\: Box)} =\: 2(56)$

$\implies \sf\bold{\purple{L.S.A_{(Cuboidal\: Box)} =\: 112\: cm^2}}$

Now, we have to find which box has the greater lateral surface area and by how much :

$\footnotesize \dashrightarrow \bf L.S.A_{(Cubical\: Box)} > L.S.A_{(Cuboidal\: Box)}$

$\footnotesize \dashrightarrow \sf L.S.A_{(Cubical\: Box)} - L.S.A_{(Cuboidal\: Box)}$

$\dashrightarrow \sf 256\: cm^2 - 112\: cm^2$

$\dashrightarrow \sf\bold{\red{144\: cm^2}}$

$\therefore$ The lateral surface area of a cubical box is greater than the lateral surface area of a cuboidal box by 144 cm² .

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(ii) Which box has the smaller total surface area and by how much :

☆ First, we have to find the total surface area of a cubical box :

$\footnotesize \implies \sf\boxed{\bold{\pink{Total\: Surface\: Area_{(Cube)} =\: 6(a)^2}}}$

$\implies \sf T.S.A_{(Cubical\: Box)} =\: 6 \times (8)^2$

$\implies \sf T.S.A_{(Cubical: Box)} =\: 6 \times 64$

$\implies \sf\bold{\purple{T.S.A_{(Cubical\: Box)} =\: 384\: cm^2}}$

☆ Now, we have to find total surface area of the cuboidal box :

$\footnotesize \implies \sf\boxed{\bold{\pink{Total\: Surface\: Area_{(Cuboid)} =\: 2(lb + bh + hl)}}}$

$\implies \sf T.S.A_{(Cuboidal\: Box)} =\: 2\{(8 \times 6) + (6 \times 4) + (4 \times 8)\}$

$\implies \sf T.S.A_{(Cuboidal\: Box)} =\: 2(48 + 24 + 32)$

$\implies \sf T.S.A_{(Cuboidal\: Box)} =\: 2(104)$

$\implies \sf\bold{\purple{T.S.A_{(Cuboidal\: Box)} =\: 208\: cm^2}}$

Now, we have to find which box has the smaller total surface area and by how much :

$\footnotesize \dashrightarrow \bf T.S.A_{(Cubical\: Box)} > T.S.A_{(Cuboidal\: Box)}$

$\footnotesize \dashrightarrow \sf T.S.A_{(Cubical\: Box)} - T.S.A_{(Cuboidal\: Box)}$

$\dashrightarrow \sf 384\: cm^2 - 208\: cm^2$

$\dashrightarrow \sf\bold{\red{176\: cm^2}}$

$\therefore$ The total surface area of a cuboidal box is smaller than the total surface area of a cubical box by 176 cm² .

Answer 2

(i) Lateral surface area of cube = 4edge²

= 4(10)²

= 400

Lateral surface area of cuboid = 2h ( l + b )

2 × 8 ( 12.5 + 10 )

= 16 × 22.5

= 360

So, the lateral surface area of the cube is larger by (400 − 360 = 40)cm²

So, the lateral surface area of the cube is larger by (400 − 360 = 40)cm²

(ii) Total surface area of cube = 6edge²

= 6(10)²

= 600

lateral surface area of cuboid = 2(lb + bh + hl)

= 2 (12.5×10 + 10×8 + 8×12.5)

= 2 (125 + 80 + 100)

= 610

So, the total surface area of cuboid is larger by (610 − 600 = 10) cm²