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:

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

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 ($lb, hl, bh$

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:

$\boxed{\frac{2hl}{2} + \frac{2lb}{2} + 2bh}$

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

$\frac{4hl}{2} + \frac{4lb}{2} + 4bh$

Which will become:

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

As you can see, the $bh$ 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

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.

- 1 year ago

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

That should have been stated though...

- 1 year ago

Yes...

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

Mistakes do happen

Yes, and we have to move on, that challenge is obviously complete now.

- 1 year ago