Question

A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long , 10. cm wide and 8cm high. (1) which box has the greater lateral surface area and by how much ? ( 2) which box has the smaller total surface area and by how much ?

Answer 1

It is given that

Edge of cube=10cm

Lateral surface area of cube=4l^2

So, area of cube =4×10×10=400cm^2

And

In cuboidal ,length=12.5cm

Breadth=10cm,height=8cm

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

So ,here area= 2×8(12.5+10)

=36cm^2

Hence lateral surface area of cube is more than cuboid

By364cm^2

Now ,total surface area of cube=6l^2

6×10×10=600^2

And t.s.a of cuboid = 2(lb+bh+hl)

2(12.5×10+10×8+8×12.5)

=610cm^2

Hence total surface are of cube is less than cuboid by 6 cm^2

Answer 2

AnswEr:

(i) Let the total surface areas of the cuboid and cuboidal box be $\sf{S}_{1}$ and $\sf{S}_{2}$ respectively. Then,

$\sf{S}_{1}=$$\sf{6(Edge)^2=6\times\:10^2\:cm^2=600\:cm^2}$

And, $\sf{S}_{2}=$$\sf{2(12.5\times\:10+10\times\:8+8\times\:12.5)^2}$

$\implies$$\sf{S}_{2}=$$\sf{2(125+80+100)cm^2=610\:cm^2}$

$\therefore$$\sf{S}_{2}-$$\sf{S}_{1}=$$\sf{(610-600)cm^2=10\:cm^2}$

$\huge\star$ Thus, the cuboidal box has greater total surface and is greater by 10 cm².

___________________________

(ii) Let the lateral surface area of the cuboidal and cubical boxes be $\sf{l}_{1}$ and $\sf{l}_{2}$ respectively. Then,

$\sf{l}_{1}=$ $\sf{4(Edge)^2=4\times\:10^2\:cm^2=400\:cm^2}$

and, $\sf{l}_{2}=$$\sf{2(length+breadth)\times\:height}$

$\implies$$\sf{l}_{2}=$$\sf{2(12.5+10)\times\:8\:cm^2}$

$= \sf \: 2 \times 22.5 \times 8 \: {cm}^{2} \\ \\ = \sf \: 360 \: {cm}^{2} \\ \\ \sf{l}_{1}={l}_{2}=(400 - 360) {cm}^{2} = 40 \: {cm}^{2}$

$\huge\star$ Thus, the cubical box has larger Lateral surface area and is greater by 40 cm².

#BAL

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