Question

There are two boxes. One is a
cube of side 50 cm. and the
other is a cuboid of length 60
cm, breadth 50 cm and height
40cm. Find the total surface area
of both Which box has less
surface area.​

Answer 1

Answer:

For cuboid, let it's length l=60 cm, breadth b=40 cm and height h=50 cm. It's TSA =2(lb+bh+lh)=2(60*40+40*50+60*50) =2*(7400) =14800 sq. cm. For cuboid, it's side length, a=50 cm.

Step-by-step explanation:

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Answer 2

This question is quite simple.

Given,

For 1st cuboidal box,

Length(l) = 60cm

Breadth(b) = 40cm

Height(h) = 50cm

Now material used for making of this box is equal to TOTAL SURFACE AREA OF CUBOIDAL.

Therefore,

Material required = 2*(l*b+b*h+h*l)

= 2(60*40+40*50+50*60)

=2(2400+2000+3000)

=2(7400)

= 14800 sq cm

Similarly use this method for other cuboidal,

Given,

Length(l)= 50 cm

Breadth(b)= 50 cm

Height(h)= 50 cm

Material required = 2(l*b+b*h+h*l)

=2(50*50+50*50+50*50)

=2(7500)

= 15000 sq cm

Comparing both values we came to know that First cuboidal box requires less material .

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