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Thread: Helping With Homework

  1. #1
    Old but Crafty RayMac's Avatar
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    Helping With Homework

    When I was down in Kingston yesterday my 12 year old niece was all set with some math questions for me. She's in an enrichment class.
    Here they are:
    • Given a Trapezoid with certain specified dimensions for major side, minor side and height, what are the dimensions of a similar Trapezoid of 1/10th the area?
    • Given a hexagon with specified length of segment, what is the dimension of the same segment on a hexagon of 1/10 the area?

    Well I did them both initially by using a spreadsheet where she could pick a number and refine her guess as she got close to the required area.
    Later on I worked out the algebra which is easy enough for the hexagon but requires some nasty 10th grade math for the trapezoid. Pretty challenging stuff for Grade 7 if you ask me.


    Few things are more delightful than grandchildren fighting over your lap. ~Doug Larson

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    Moderator gnuyork's Avatar
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    The answer is D all of the above

  3. #3
    Quote Originally Posted by gnuyork View Post
    The answer is D all of the above
    Incorrect, the answer is Pi, if it happens not to be the answer the teacher is seeking at least you can eat it later...

    Reminds me, pecan pi(e) from Thanksgiving is calling my name.

  4. #4
    Dinger of Hum Chronopolitano's Avatar
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    Do you have Common Core up in Canada?

    I'm not sure what the controversy is, or if I have grasped the problem (if it is one), but I have seen some examples, and I must admit to being somewhat perplexed as to the logic in this form of pedagogy.

    I kinda get the intention, but I kinda don't.

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    Timewaster jsw41's Avatar
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    Very interesting, Ray. My geometry and algebra skills are rusty; maybe I need to re-do the maths that 7th graders are expected to know:

    http://ca.ixl.com/math/grade-7

    John
    If you come to a fork in the road; take it, and then put it down so someone else can use it.

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    MultiModerator Martin's Avatar
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    As for the trapzoid, as the ratio's stay the same, isn't it just making both sides and height a factor sqrt(10) smaller?
    Old area: 1/2(a+b)xh=A
    New area: 1/2((a/sqrt(10)+b(/sqrt(10))x(h/sqrt(10))=1/sqrt(10) x 1/2(a+b) x 1/sqrt(10) x h = 1/10 x 1/2(a+b)xh

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    Dinger of Hum Chronopolitano's Avatar
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    Quote Originally Posted by Martin View Post
    As for the trapzoid, as the ratio's stay the same, isn't it just making both sides and height a factor sqrt(10) smaller?
    Old area: 1/2(a+b)xh=A
    New area: 1/2((a/sqrt(10)+b(/sqrt(10))x(h/sqrt(10))=1/sqrt(10) x 1/2(a+b) x 1/sqrt(10) x h = 1/10 x 1/2(a+b)xh
    In other words, kinda smallish, yeah?

  8. #8
    MultiModerator Martin's Avatar
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    Quote Originally Posted by Chronopolitano View Post
    In other words, kinda smallish, yeah?
    Well, that depends on the size of your trapezium

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  10. #9
    Old but Crafty RayMac's Avatar
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    Quote Originally Posted by Martin View Post
    As for the trapzoid, as the ratio's stay the same, isn't it just making both sides and height a factor sqrt(10) smaller?
    Old area: 1/2(a+b)xh=A
    New area: 1/2((a/sqrt(10)+b(/sqrt(10))x(h/sqrt(10))=1/sqrt(10) x 1/2(a+b) x 1/sqrt(10) x h = 1/10 x 1/2(a+b)xh
    Basically yes that is how the algebra worked out. You have a better spatial concept than I do.


    Few things are more delightful than grandchildren fighting over your lap. ~Doug Larson

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