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Math Help - total surface area of prism

  1. #1
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    total surface area of prism

    hi,

    i need to work out the total surface area of this prism:

    total surface area of prism-prism.jpg

    i know its not draw well but it gets the point across. i was going to divide it up into 3 parts and times the final answer by 3 because there are 3 parts.

    however i don't know how to work out the surface area of a cuboid but i know it would contain the length of 8cm, width 3cm and height 3cm.

    so if somone could help i would be very very grateful!!

    thanks
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  2. #2
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    Quote Originally Posted by andyboy179 View Post
    hi,

    i need to work out the total surface area of this prism:

    Click image for larger version. 

Name:	prism.jpg 
Views:	85 
Size:	19.0 KB 
ID:	21372


    i know its not draw well but it gets the point across. i was going to divide it up into 3 parts and times the final answer by 3 because there are 3 parts.

    however i don't know how to work out the surface area of a cuboid but i know it would contain the length of 8cm, width 3cm and height 3cm.

    so if somone could help i would be very very grateful!!

    thanks
    Hi andyboy179,

    if the 3 blocks have identical dimensions,
    then their volumes are equal,
    so multiplying by 3 would give total volume
    (assuming the side with no dimension given is 3 units).

    However, one block is on top of another, removing 2 sides from the surface area
    and the 2 base blocks are side-by-side removing another 2 sides from the total surface
    of the shape.
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  3. #3
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    Hi,
    I understand but how would I actually work out the total surface area of the prism?
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  4. #4
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    Quote Originally Posted by andyboy179 View Post
    Hi,
    I understand but how would I actually work out the total surface area of the prism?
    Have a look at your drawing again.

    It seems that all "long" faces are 8X3, according to your dimensions.
    We can consider your figure as 3 identical blocks joined together.

    3 long faces are exposed on the bottom right block.
    2 long faces are exposed on the bottom left block.
    3 long faces are exposed on the top block.

    All six 3X3 edges are exposed.

    If you add all those independent parts of the surface area,
    you will have the total surface area.

    You should get the exact same answer if you calculate the surface area of all six flat sides of a block,
    then multiply the answer by 3,
    then subtract the areas of the 4 hidden flat sides.

    Try experimenting with that.
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  5. #5
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    8 long faces are exposed and 6 3x3 ones. so would i do 8x8=64 and 3x6=18 64+18=82 82x3=246 ??
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  6. #6
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    Quote Originally Posted by andyboy179 View Post
    8 long faces are exposed and 6 3x3 ones. so would i do 8x8=64 and 3x6=18 64+18=82 82x3=246 ??
    It's a roundabout way of getting there.
    I wouldn't recommend those calculations by any means though.

    To get the surface area of an 8X3 rectangle, how many 1X1 "unit" squares would fit inside ?
    Also, for a 3X3 square, how many 1X1 "unit" squares fit inside ?
    Those calculations tell you how many "unit" squares are in those sides.
    The total surface area is the number of "unit" squares that cover the entire surface
    (imagine the surface being tiled with 1X1 squares).
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  7. #7
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    would there be 11 and 6 1x1 squares?
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  8. #8
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    Quote Originally Posted by andyboy179 View Post
    would there be 11 and 6 1x1 squares?
    If you had a 3X1 rectangle, then you could say it's made of 1 row of 3 "unit" 1X1 squares.
    Hence it's area is 3 "square units".
    If you had a 2X3 rectangle, then you could say it's made of 2 rows of 3 "unit" 1X1 squares.
    Hence it's area is 2X3=6 "square units".
    We are counting the "unit squares" within the rectangles.

    If you continue in this way, how many "unit" squares fit inside a 3X3 square and an 8X3 rectangle ?

    It looks like you were guessing above.
    Guessing gets you nowhere in maths.
    You should try to get the ideas instead.
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  9. #9
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    oh i understand so it would be 24 and 9 square units
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  10. #10
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    Quote Originally Posted by andyboy179 View Post
    oh i understand so it would be 24 and 9 square units
    Perfect!

    So you only need discover how many of the 24 "square unit" sides are exposed
    and how many 9 "square unit" squares are also exposed.

    You will get the result you had earlier, which was correct,
    but you needed a much clearer way to calculate it.
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  11. #11
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    how do i actually find out if they are exposed or not?
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  12. #12
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    Quote Originally Posted by andyboy179 View Post
    how do i actually find out if they are exposed or not?
    What I mean by "exposed" is....

    Look at the block on the bottom right.
    It's upper face is exposed.
    So is it's rightmost face.
    It's underneath face is exposed also
    (think of the entire object suspended on a piece of string
    attached to the celing or something like that).
    It's leftmost face is not exposed since it is against another block.

    Therefore, those 3 faces contribute to the overall surface area.
    That block also has two 9 "square unit" faces exposed.

    Notice that the bottom-left block has only 2 of the long faces exposed
    and two 9 "square unit" square sides exposed.

    What about the block at the top ?
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  13. #13
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    the block at the top has two 9 square unit sides exposed and 3 long faces exposed.
    so in total the amount of 9 square unit sides exposed are 6 and 24 square unit has 8 so the total surface area is 14cm^2?
    Last edited by andyboy179; April 6th 2011 at 09:03 AM.
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  14. #14
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    There are 8 of the 8X3=24 sq unit sides,
    hence the number of "square units" altogether in those 8 sides is 8X24.
    Then the 6 square sides each have 9 "square units",
    so that's another 6X9 "square units" to be counted.

    Grand Total is ?
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  15. #15
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    8x24=192
    6x9= 54
    192+54= 246

    grand total of 246?
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