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Hello, how are you doing?

not so good

:/ I am sorry. Shall we try and get rid of this problem then?

yes please

Does the rectangle lie on the full 72 length of the semi circle? I'll need more clues I am afraid.

Right, that's much clearer now. I'll just deal with the second part of the problem.

Now we have a triangle with hyp = 36 and angle 45. We can use trig to solve for the other sides.

ok

Let's solve for the adjacent. Know what trig function to use if we have the hyp and need the adj?

cosine

yes, exactly. Can you solve for the adj now?

yeah, i got it

So you got cos(45)*36, correct? =19.911

yeah, i got it

i got 18.911

Oops, Yes, my fault. I was in radians.

lol ok

So we have our x value, and it's equal to our y value. We have a square here.

yup!

But that's in 2d, we get to cube 25.456 and then double it for a total of 32,990.773 units cubed.

wait i'm lost. why are we squaring thr 18.9?

uh oh.. You aren't getting 25.456 for cos(45)*36?

nope, i'm in radians and i'm getting 18.9

right, that's what i thought, but you said squaring not cubing!

Haha, I am sorry. It is late. I'm also not a pro.

lol ok.
so we cube 18.9

But yes, I'm getting 18.91 too. Sorry.

and then double it.

and why do we double it?

And if the theory works out, that should be the maximum volume for the rectangle inside the arch.

Oh, because we only found one quarter of the full circle, half of the semi circle.

We have two quarters here.

hmmm... ok

hmmm... ok

Does that not make sense?

I wish I could show you all the drawings I did trying to figure this one out lol

Shoot.. I take it back. You have to quadrupedal. I keep forgetting that we are in 3d.

So double that double.

yes, for the total volume.

okay, thank you for all your help!

No problem, so sorry I kept making mistakes. I hope I was somewhat clear in the end.

yeah, enough i guess

ok, i get it

ok

lol okay! night!