how do i find the dimensions of a rectangle inside of a semi-cirble?

- anonymous

how do i find the dimensions of a rectangle inside of a semi-cirble?

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- anonymous

Hello, how are you doing?

- anonymous

not so good

- anonymous

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

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## More answers

- anonymous

yes please

- anonymous

Is there any hint you can give me about your rectangle? So many dimensions can fit into any semi circle. Are the corners touching anything on the circle that you can tell me about?

- anonymous

the flat edge of the semi circle's dimension is 72 and the height or radius of the semi circle is 36. the rectangle fits into the semi circle so that the length of the rectangle is touching the flat edge of the semi circle and two top corners of the rectangle are touching the curve of the semi circle...

- anonymous

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

- anonymous

yes, the rectangle does lie on the full 72 length of the semi circle. i forgot how to find the dimensions of inscribed retangles. so long ago and its late

- anonymous

I am sorry, but from the specs you gave me, it is impossible for the rectangle to have anything but 72x0. The top two corners cannot be going up anywhere near 36.
And you are certain this is a rectangle in a semi circle and not the other way around?

- anonymous

If you draw the picture, the circle is possible, but if the rectangle's base is 72, it cannot be inscribed within the semi circle.

- anonymous

yes, i'm certain that this is a rectangle in a semi circle and not the other way around.
problem: you are to design an office block inside a structure witht the following specifications: the building has a rectangular base 150m long and 72m wide. the maximum height of the structure should not exceed 75% of it width for stability or be less than half the width for aesthetic purposes. the maximum height of a room in a public building is 2.5m.
create a model for the curved roof structure when the height is 36m. (i did that)
now, it's telling me to find the dimensions of the cuboid with maximum volume which would fit inside this roof structure

- anonymous

- anonymous

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

- anonymous

i know that one of the dimensions of the cuboid will be 150, i'm just having a hard time finding the other two dimensions that happen to be inside the semi circle

- anonymous

Hmm.. This still troubles me because the problem is written so strange. I can't find the connection between the curved roof and the office block.
And if the radius of the semi circle is 36, the biggest length you would be able to build is 72, not 150. I guess I am not understanding the problem clearly here.

- anonymous

this is a three dimensional problem. the office block is shaped like a cube inside the arc-shaped building. i know the cube will be 150m long because that's how long the arc-shaped building is. i'm just having trouble finding the height and width of the cube which can be found on the face of the cube and the face of the cube is touching the face of the arc. when you just look at the front of the building, you will see the semi circle and the rectangle inside the semi circle. i need to know the height and the width of the rectangle if the rectangle's base is fully touching the base of the semi circle and the length of the semi circle is 72m and the height of the semi circle is 36m.

- anonymous

I am trying to dissect this as slowly as I can, sorry for such clueless responses. If it's a cuboid that's 150m, all lengths, widths, and heights are going to be 150m..

- anonymous

well, ok. if its a cuboid all the lengths must be the same. so what would that be because the cube cant fit all the way to the top of the arc-shaped building nor can it stretch all the way equal to the 72 because of the arc...

- anonymous

The max we can do is a cube, and if we are talking angles, that's 45 degrees from the center of the circle, or at 36m.

- anonymous

but 36 doesn't make since because whenever you put a rectangle or square inside a circle or semi circle, there's always gonna be room between the arc and the line of the rectangle/square. therefore, it cant be 36 because that's how high the arc is

- anonymous

Oh no, I meant that the origin is at 36m. If the base is 72m.
Sorry for the absence. Ok. So we know that our radius is 36. So if we go 45 degrees from the center to the circle, that's 36m.

- anonymous

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

- anonymous

ok

- anonymous

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

- anonymous

cosine

- anonymous

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

- anonymous

yeah, i got it

- anonymous

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

- anonymous

yeah, i got it

- anonymous

i got 18.911

- anonymous

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

- anonymous

lol ok

- anonymous

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

- anonymous

yup!

- anonymous

If you square that number, you get 648. But that's only for one quarter of our circle. We have 2 quarters in our semi circle so we'll need to double that area for a total max area of 1296.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

And we were cubing because if the width value is equal to 25.456, then both altitude and length are also 25.456.

- anonymous

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

- anonymous

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

- anonymous

lol ok.
so we cube 18.9

- anonymous

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

- anonymous

and then double it.

- anonymous

and why do we double it?

- anonymous

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

- anonymous

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

- anonymous

We have two quarters here.

- anonymous

hmmm... ok

- anonymous

hmmm... ok

- anonymous

Does that not make sense?

- anonymous

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

- anonymous

If you draw a picture of the semi circle and draw 45 degrees out both ways from the point of origin and then draw the triangles we used to find the length, you will understand why we had to double.

- anonymous

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

- anonymous

So double that double.

- anonymous

so co(45)*36 = 18.9. then we cube that because of the three dimensions, then we quadruple that for the volume???

- anonymous

yes, for the total volume.

- anonymous

okay, thank you for all your help!

- anonymous

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

- anonymous

yeah, enough i guess

- anonymous

But draw a 2d side cutout of it. If you draw lines from the origin, at 36 m going 45 degrees out, you are only accounting for half the semi circle.

- anonymous

ok, i get it

- anonymous

That's why you have to double the volume of one quarter, because a semicircle is really two quarters. And then when you realize that this is a dome, you have to have that depth, so you multiply it by two again.

- anonymous

ok

- anonymous

Great. :D I have to do my math homework now (can't believe this is what I choose to do when I want to procrastinate) and it's 1:30AM, so I will see you later. Have a goodnight!

- anonymous

lol okay! night!

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