how 😫

[amber knot]

Recall the formula for the area of a regular n-gon with side length a.

[stiff fiber]

No idea sorry

It’s not taught for gcse maths 😭

[amber knot]

Don't know what gcse is.
The area of a regular n-gon with side length a is S(n) = (1/4)na^2 cot(π/n).

[cosmic elm]

are you retared you autistic monkey this shi ez

[amber knot]

Woah, no need to be so rude!

[cosmic elm]

ok

sorry

[stiff fiber]

Bro I am 15 💀

How tf this guy a helper

[hidden spoke]

I know Hemu already warned you, do this again and I will mute you

[stiff fiber]

Plus I live in the uk so we learn different formulas 🤷‍♂️

[amber knot]

Well, still. You need to apply that formula here.
Of course, the ratio of half the areas of these polygons is the same as the ratio of full areas, so you need to find S(6)/S(8).

Note, however, that the side lengths are different.

You can circumvent that by recalling the variant of this formula in terms of the circumradius R instead of side length, though.

[verbal river]

aye

im pretty sure

ur supposed to use the formula

1/2 ab sinC

since this is "GCSE"

[verbal river]

find the angle of each triangle

the area should be just sine of the angle

add the areas of the triangles up to find the ratio

3sin(60) : 4sin(45)

so

3root3/2 : 4/root2

3root3/2 : 2root2

3*3/2:2root6

9:4root6

therefore

1:4/9root6

@stiff fiber

this would be the gcse method

[amber knot]

As I said, I don't know what GCSE is.

[verbal river]

it's below a levels

therefore

[amber knot]

Below what?

[verbal river]

bro

a levels

[amber knot]

I don't understand what you mean. Did you mean "a level" or "levels"?

[verbal river]

did you forget that the uk exists 💀

[amber knot]

Well, I know that the UK exists, but what does that have to do with this?

[verbal river]

cause that's how it works in the uk

[amber knot]

Hm. Well, ok.

Still not sure why you write "a levels".

[verbal river]

what are you on about

[amber knot]

Oh. OHH. You meant A-levels!

Now I get it.

Well, I still don't know how that works, but whatever. This would be my approach. I think it's appropriate for grade 8 and above.
The area of a regular n-gon with circumradius R is S(n, R) = (1/2)nR^2 sin(2π/n). In our case the circumradius is the same, so we get:
S(8)/S(6) = 8sin(π/4)/(6sin(π/3)) = 8(√(2)/2)/(6(√(3)/2)) = 4√(6)/9

[verbal river]

it is definitely not in the UK 🤣

education is such a joke here

do nothing in 7,8,9

[amber knot]

Oh, that's a shame...