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unearthly09

What is the surface area of this three-dimensional, symmetrical, capital letter “I” block? Show your work.

  • one year ago
  • one year ago

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  1. unearthly09
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    • one year ago
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  2. dpflan
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    All right, how familiar with surface area are you?

    • one year ago
  3. unearthly09
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    i know how to find the surface area of rectangular prisms just need a little help with the composite figurs

    • one year ago
  4. dpflan
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    We need to determine the number of sides, or specific planes, that the 'I' has. Like a cube, a cube has 6 sides that are in the shape of squares. This 'I' has quite a 'sides'. So, we'll want to find the area of all these different sides and sum them.

    • one year ago
  5. unearthly09
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    why the area?

    • one year ago
  6. dpflan
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    Cool, can you determine the number of sides?|dw:1343144067033:dw|

    • one year ago
  7. dpflan
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    Surface area is just the sum of all the area of the outside of this 3D object. That is the surface, the part you can touch if you were actually holding the 'I', so surface area just means you find the area of the surface, or the outside skin if you will of the object

    • one year ago
  8. unearthly09
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    oh i get it

    • one year ago
  9. dpflan
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    So perhaps the best way to go about this would be to look at the 'I' straight on. If you can imagine that you have this object and can rotate, this will be simple ;)

    • one year ago
  10. dpflan
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    |dw:1343144255358:dw|

    • one year ago
  11. unearthly09
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    ok

    • one year ago
  12. dpflan
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    The best part is that this 'I' is symmetric, so it it has the same properties from different view points. so if we view only the front of the 'I'|dw:1343144361875:dw| this is what the back looks like too

    • one year ago
  13. unearthly09
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    yeah

    • one year ago
  14. dpflan
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    You are given the lengths of all the sides. So, what would be the area of just the front ?

    • one year ago
  15. unearthly09
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    what do u mean the area of the I

    • one year ago
  16. dpflan
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    Find the area of the three rectangles that make up the front. Like how I drew the front, even though the attached picture has the front being one large shape of a capital I, you could break that shape down into smaller shapes: like 2 big rectangles at the top and a square in the middle

    • one year ago
  17. unearthly09
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    would it be 22cm

    • one year ago
  18. dpflan
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    Yeah, nice! So that takes care of the front, well, and the back areas.

    • one year ago
  19. dpflan
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    area of front + area of back + area of left side + area of right side + area of top + area of bottom

    • one year ago
  20. dpflan
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    Due to the symmetry, we can just find the area of the front (=area of back), area of right side (=area of left), and the area of the top (=area of bottom)

    • one year ago
  21. dpflan
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    Now the side is more difficult, it is more complex than the front.

    • one year ago
  22. dpflan
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    How many rectangles comprise a side of the surface?

    • one year ago
  23. unearthly09
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    is it 3 the 3 long ones and the one short one

    • one year ago
  24. unearthly09
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    i meant 2 long ones

    • one year ago
  25. dpflan
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    |dw:1343144879045:dw| The black diagonal lines are for the left side, which we want to work with

    • one year ago
  26. unearthly09
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    i have a question is the answer 104???

    • one year ago
  27. dpflan
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    Heh, cut the chase, eh? Just a second

    • one year ago
  28. unearthly09
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    ha:D

    • one year ago
  29. dpflan
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    No, that's not big enough

    • one year ago
  30. dpflan
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    ...

    • one year ago
  31. unearthly09
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    so where did i go wrong i have no clue what I did

    • one year ago
  32. dpflan
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    Heh, I'm checking my math...

    • one year ago
  33. dpflan
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    because I messed up, 104 is right

    • one year ago
  34. unearthly09
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    really!!!!

    • one year ago
  35. dpflan
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    5 squares for the left side surface, each 2x2 = 20 2 long rectangles, 5x2, and 1 small square, 2x1 for the front = 22 1 long rectangle, 5x2, for the top = 10 That is 52, so due to symmetry, 52x2 is the total surface area

    • one year ago
  36. dpflan
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    Yes, how did you get it? If you wrote down your work,you can take a picture of it with maybe your cell phone and upload your work on here

    • one year ago
  37. unearthly09
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    i really kind of did what u told me than i found the area of the 2 sides

    • one year ago
  38. unearthly09
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    my work is all jumbled up

    • one year ago
  39. dpflan
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    Oh well, very cool though, that's a straightforward way to solve it, so nice work!

    • one year ago
  40. unearthly09
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    does this make sense symmetrical, capital letter i I would find the surface area of one of the "bars" 5 by 2 by 2 (ignoring the fact that the middle connector hides part of the surface). So one bar has 4 sides with 5 by 2 or 4*10= 40 sq cm the 2 ends have area 2*2=4 or 8 total. So the bar has a surface area of 48 Double this for the other bar: 96 Now the middle connector has 4 sides. I used 2 of them to "fill in" the hidden part of the bar We have 2 sides of 2x2 or 2 sides of 4 or 8 sq cm to add in 96 8 104 Therefore the surface area of this figure is 104 centimeters squared.

    • one year ago
  41. unearthly09
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    i meant 96+8= 104

    • one year ago
  42. dpflan
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    The middle part however has 2 2x1 sides, and 2 2x2 sides that are visible, and 2 2x1 sides that not visible, if you break the 'I' in 3 3D objects: 2 prisms on the top, 1 in the middle

    • one year ago
  43. unearthly09
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    yeah

    • one year ago
  44. unearthly09
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    right...???

    • one year ago
  45. dpflan
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    Usually, if you are going to include a part, such as the under side of the top block, which is not part of the 'I''s surface area, that means you need to actually subtract that amount of 'hidden surface area' from the final total.

    • one year ago
  46. dpflan
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    Let's see: 3 blocks: 2 long rectangles and 1 middle connector

    • one year ago
  47. dpflan
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    the long rectangles are similar, so let's just focus on one: We are going to use your concept of building the 'I' from these 3 objects and accounting for hidden area. So, the surface area of one long block is: 4 - 5x2 rectangles 2 - 2x2 rectangles SA = 4(10) + 2(4) = 48 So surface area of both blocks is 48x2 = 96 Now, the middle connector: 2 - 2x2 rectangles 4 - 2x1 rectangles SA = 2(4) + 4(2) = 16 But 2 of those 2x1 rectangles are hidden once all the blocks are connected And, in each long block, an area of 2x1 is hidden too. 96 + 16 - 4(2x1) = 112 - 8 = 104

    • one year ago
  48. dpflan
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    |dw:1343146032157:dw| In the 2 circled parts, we have to account for the hidden parts of the surface when the blocks are all connected. So we you quite close, and I think you're idea of working with the 3 blocks is a little more intuitive

    • one year ago
  49. dpflan
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    Does this make sense?

    • one year ago
  50. unearthly09
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    yes it does....thank u so much :)

    • one year ago
  51. dpflan
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    No problem, I think you can do your other questions if they're similar, you seem to have the right intuition about how to solve the problem

    • one year ago
  52. unearthly09
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    :)thanks again

    • one year ago
  53. unearthly09
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    bye

    • one year ago
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