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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    Hello, how are you doing?

  2. anonymous
    • 5 years ago
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    not so good

  3. anonymous
    • 5 years ago
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    :/ I am sorry. Shall we try and get rid of this problem then?

  4. anonymous
    • 5 years ago
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    yes please

  5. anonymous
    • 5 years ago
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    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?

  6. anonymous
    • 5 years ago
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    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...

  7. anonymous
    • 5 years ago
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    Does the rectangle lie on the full 72 length of the semi circle? I'll need more clues I am afraid.

  8. anonymous
    • 5 years ago
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    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

  9. anonymous
    • 5 years ago
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    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?

  10. anonymous
    • 5 years ago
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    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.

  11. anonymous
    • 5 years ago
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    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

  12. anonymous
    • 5 years ago
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    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

  13. anonymous
    • 5 years ago
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    Right, that's much clearer now. I'll just deal with the second part of the problem.

  14. anonymous
    • 5 years ago
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    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

  15. anonymous
    • 5 years ago
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    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.

  16. anonymous
    • 5 years ago
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    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.

  17. anonymous
    • 5 years ago
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    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..

  18. anonymous
    • 5 years ago
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    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...

  19. anonymous
    • 5 years ago
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    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.

  20. anonymous
    • 5 years ago
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    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

  21. anonymous
    • 5 years ago
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    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.

  22. anonymous
    • 5 years ago
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    Now we have a triangle with hyp = 36 and angle 45. We can use trig to solve for the other sides.

  23. anonymous
    • 5 years ago
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    ok

  24. anonymous
    • 5 years ago
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    Let's solve for the adjacent. Know what trig function to use if we have the hyp and need the adj?

  25. anonymous
    • 5 years ago
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    cosine

  26. anonymous
    • 5 years ago
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    yes, exactly. Can you solve for the adj now?

  27. anonymous
    • 5 years ago
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    yeah, i got it

  28. anonymous
    • 5 years ago
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    So you got cos(45)*36, correct? =19.911

  29. anonymous
    • 5 years ago
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    yeah, i got it

  30. anonymous
    • 5 years ago
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    i got 18.911

  31. anonymous
    • 5 years ago
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    Oops, Yes, my fault. I was in radians.

  32. anonymous
    • 5 years ago
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    lol ok

  33. anonymous
    • 5 years ago
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    So we have our x value, and it's equal to our y value. We have a square here.

  34. anonymous
    • 5 years ago
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    yup!

  35. anonymous
    • 5 years ago
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    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.

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

  37. anonymous
    • 5 years ago
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    wait i'm lost. why are we squaring thr 18.9?

  38. anonymous
    • 5 years ago
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    uh oh.. You aren't getting 25.456 for cos(45)*36?

  39. anonymous
    • 5 years ago
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    nope, i'm in radians and i'm getting 18.9

  40. anonymous
    • 5 years ago
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    And we were cubing because if the width value is equal to 25.456, then both altitude and length are also 25.456.

  41. anonymous
    • 5 years ago
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    right, that's what i thought, but you said squaring not cubing!

  42. anonymous
    • 5 years ago
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    Haha, I am sorry. It is late. I'm also not a pro.

  43. anonymous
    • 5 years ago
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    lol ok. so we cube 18.9

  44. anonymous
    • 5 years ago
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    But yes, I'm getting 18.91 too. Sorry.

  45. anonymous
    • 5 years ago
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    and then double it.

  46. anonymous
    • 5 years ago
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    and why do we double it?

  47. anonymous
    • 5 years ago
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    And if the theory works out, that should be the maximum volume for the rectangle inside the arch.

  48. anonymous
    • 5 years ago
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    Oh, because we only found one quarter of the full circle, half of the semi circle.

  49. anonymous
    • 5 years ago
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    We have two quarters here.

  50. anonymous
    • 5 years ago
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    hmmm... ok

  51. anonymous
    • 5 years ago
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    hmmm... ok

  52. anonymous
    • 5 years ago
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    Does that not make sense?

  53. anonymous
    • 5 years ago
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    I wish I could show you all the drawings I did trying to figure this one out lol

  54. anonymous
    • 5 years ago
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    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.

  55. anonymous
    • 5 years ago
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    Shoot.. I take it back. You have to quadrupedal. I keep forgetting that we are in 3d.

  56. anonymous
    • 5 years ago
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    So double that double.

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

  58. anonymous
    • 5 years ago
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    yes, for the total volume.

  59. anonymous
    • 5 years ago
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    okay, thank you for all your help!

  60. anonymous
    • 5 years ago
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    No problem, so sorry I kept making mistakes. I hope I was somewhat clear in the end.

  61. anonymous
    • 5 years ago
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    yeah, enough i guess

  62. anonymous
    • 5 years ago
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    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.

  63. anonymous
    • 5 years ago
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    ok, i get it

  64. anonymous
    • 5 years ago
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    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.

  65. anonymous
    • 5 years ago
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    ok

  66. anonymous
    • 5 years ago
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    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!

  67. anonymous
    • 5 years ago
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    lol okay! night!

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