any better way to do this ?

[lament crypt]

my method is split a regilar hexagon into 6 equilateral triangles, and then you can solve for the side lengths (bottom has side length 2/3^(3/4), top has side length 2^(1/2)/3^(3/4)))

i got height as (2^(1/2))(3^(3/4))/(2^(1/2)+1)

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[coarse cargo]

your method seems correct.

is there an answer key, by chance?

[lament crypt]

uhh nop

but i was looking for a better method

[lament crypt]

cuz i feel like this is too "brute forced"

if that even amde sense

[coarse cargo]

i would use some sort of pythagorean theorem here, but i dont really see it.

wait.

since the non-existent part of the pyramid is 1/2 of the total pyramid.

ahhhhhhhhh

[steel bison]

you could maybe use the side lengths to find how much the hexagon shrinks

[coarse cargo]

kind of like similar triangles, which was my alternate approach

[lament crypt]

oh how did u do that

[coarse cargo]

well, the areas are in a ratio of 1:2 for the total hexagon and the truncated hexagon

so the side lengths are naturally in a ratio of 1:sqrt(2)

and thus the volume is naturally in a ratio of 1:2sqrt(2)

let me double check

[amber imp]

I'm actually not sure this follows.

Wait, nevermind.

I get it now.

[coarse cargo]

👍

the sqrt and cubed ratio

i just simplified it a little

[amber imp]

But actually we don't need anything about the volume, we just need the height.

Which... this approach doesn't actually totally help with.

[lament crypt]

thats really smart

i wont need pythagoras for this then

thanks@

+close

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