Your 10m explanation has not a scientific explanation but just your thoughts.

No, it isn't just my thoughts.

It is based upon the accuracy of the data provided.

You can do a more accurate error analysis if you want to determine a better value, but it will be no where near what you need.

This is how error works in science.

They don't just pull a number out of thin air and declare that to be the error limit.

Instead they use the uncertainty of the data they have. That can be limitations in the instruments or a precision limit in values reported in the literature.

The only part I am leaving out is combining the multiple error ranges to produce the final uncertainty.

But if you want, I can do that as well.

Lets use the small area shall we?

I'll even be nice and overestimate the error by adding them rather than adding in quadrature.

I will even be generous and take the error as 10 m, rather than the far more common approach of the error being half the least significant digit reported without uncertainty.

However, you just kept adding on to your distances.

So that means each reported value (except the first, but for simplicity I will include that) has an uncertainty of 20 m.

Now we could just go and try to add the errors up for each step, but that will likely over estimate it. (admittedly, that is what I first went to do, but I stopped myself when I got to y and started to reuse values.) Let's do it the more scientific way.

First, what is our final formula?

Well, if the top length is a, and the bottom length is b, and the side length is c, what do we do?

First we find x, the extra bit on the side of the bottom. This is (b-a)/2.

Then we find the height h, by taking the square root of the different between c^2 and x^2.

i.e. h^2=c^2-x^2.

Then we find y, the length of the top, plus the extra bit on the bottom. This is a+x=a+(b-a)/2=(a+b)/2.

Now we find d, the diagonal, by finding the square root of h^2+y^2

d^2=h^2+y^2

=c^2-x^2+y^2

=c^2-(b-a)^2/4+(a+b)^2/4

=c^2-(b^2-2ab+a^2)/4+(a^2+2ab+b^2)/4

=c^2+(-b^2+2ab-a^2)/4+(a^2+2ab+b^2)/4

=c^2+(-b^2+2ab-a^2+a^2+2ab+b^2)/4

=c^2+(4ab)/4

=c^2+ab

So d=sqrt(c^2+ab)

Note: While I am saying percentage error, I will express the percentage error in fractional form. i.e. 100%=1, 5%=0.05

Also note: While I am providing the values to some number of s.f. Excel will use more than reported.

Now this is represented nice and simply as a function of the original values.

So the first step to figuring out the errors, we note that we are multiplying values (even if it is just by itself). When multiplying, the percentage errors add.

This means for c^2, we first find the percentage error of 3.00E-05, and multiply it by 2 to get the percentage error in c^2 of 6.00E-05.

Then we multiply this by c^2 to get the absolute error in c^2, so we end up with 445129 km^2 with an uncertainty of 27 km^2.

Now we add the percentage errors in a and b of 1.91E-05 and 1.70E-05 to get 3.61E-05, the percentage error in ab.

This gives us ab as 1234051 km^2 with an uncertainty of 45 km^2.

Then as we add the 2 values together, we add the absolute errors, to get d^2 as 1679180.799 km^2 ith an uncertainty of 71 km^2.

Now for the last step we find the square root.

For this we need to use percentage errors again and similar to how squaring doubled the percentage error, square rooting will cut it in half.

That means we take our percentage error of 4.24E-05 and cut in half to get 2.12E-05.

This means we end up with d as 1295.832 km with an uncertainty of 0.027 km, or 0.03 km.

Quite close to the 0.01 km I started with, much smaller than the actual difference between the 2 values, confirming that these values are in fact different and much smaller than your magic 0.5%, which is pulled from nowhere.

If you take the more common approach of adding errors in quadrature and using 0.005 km as the starting point as half the least significant digit, you end up with a final error of 0.01 km, just like I started with.

So no, I am following the scientific approach to errors.

Using this approach, the values for the diagonals assuming Earth is flat and off the map are different.

However the values for the diagonals assuming Earth is round and off the map are equal within uncertainty.

You are using a completely unscientific method of just asserting the error is 0.5% without paying any attention to the accuracy of the initial data.

In some cases that will be a massive overestimate. In others it will be a massive underestimate.

There is nothing scientific about just asserting it is 0.5%.

You repeatedly insulting me wont change how uncertainties work.

Now how about you justify your 0.5%?

Are you sure you aren't confusing it with a p value of 0.05?

If so, that is to expand the uncertainty based upon more complex statistics to produce a 95% confidence interval.

Some fields will use other confidence intervals, such as 99% or 99.5%.

But there is no widely held 0.5%

So again, can you justify your 0.5%?

Me not answering about my education doesn't mean it is absent, it means it is irrelevant.

I don't need to appeal to credentials to back up my claims.

I don't need to appeal to a PhD or multiple scientific publications in scientific journals.

If someone needs to appeal to academic credentials it shows they can't defend their position.