Relationship between perfect splitting and surface contact area (Cub(e/oids)

Surface area is the surface which is in contact with air/the surrounding (if it is supposedly a vacuum)

For every cub(e/oid), the formula to calculate the surface area is:

\(\boxed{2lb + 2hl + 2bh}\)

This is also used:

2(lb+hl+bh)\boxed{2(lb + hl + bh)}

When we split a cub(e/oid) in half, any one of those three variables will double [when considering both cub(es/iods)]

When I say variable, I mean any of the three values (lb,hl,bhlb, hl, bh

For an instance, I say that I want to slice a cub(e/iod) downwards

So, one cub(e/iod) will have the surface area of:

2hl2+2lb2+2bh\boxed{\frac{2hl}{2} + \frac{2lb}{2} + 2bh}

Both cub(es/oids), when considered a one, will have the surface area of:

4hl2+4lb2+4bh\frac{4hl}{2} + \frac{4lb}{2} + 4bh

Which will become:

2hl+2lb+4bh2hl + 2lb + \boxed{4bh}

As you can see, the bhbh variable has doubled...

Which means surface area increases as more splitting is done

Note by A Former Brilliant Member
1 year ago

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@Vinayak Srivastava, the quest of birds and human surface area has come to a halt

Check this and see whether you have understood the post...

Or say if you learnt something new

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I know these formulas, I just feel that the question was not clear: It asked which will eat more calories, not which will tend to lose more heat, 75 kg birds which don't do anything for a day will need far less calories than 75 kg birds who work all through the day. I hope you understand my point.

Vin_Math 91 - 1 year ago

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And humans work the same way too. When both were to do the same work at the same environment, the birds will need more calories, @Vinayak Srivastava

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@A Former Brilliant Member That should have been stated though...

Vin_Math 91 - 1 year ago

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@Vin_Math 91 Yes...

I thought about that too

But I can't help, Brilliant sometimes makes unclear questions

Mistakes do happen

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@A Former Brilliant Member Yes, and we have to move on, that challenge is obviously complete now.

Vin_Math 91 - 1 year ago

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