Results 1 to 13 of 13

Thread: area of rectangle

  1. Nov 4th, 2003, 06:44 PM #1
    dolor
    dolor is offline

    Thread Starter
    Addicted Member dolor's Avatar
    Join Date
    Sep 2002
    Location
    in a place seldom found on a map
    Posts
    220

    area of rectangle

    A rectangular field is twice as long as it is wide. Express the area of the rectangle as a function of the perimeter, P.
    - you've been privileged to read a post by Miz
    Reply With Quote Reply With Quote
  2. Nov 4th, 2003, 07:38 PM #2
    jemidiah
    jemidiah is offline
    Only Slightly Obsessive jemidiah's Avatar
    Join Date
    Apr 2002
    Posts
    2,431
    Perimeter = 2*Length + 2*Width

    2W = L

    Perimeter = 6*Width

    Area = Length*Width = Width*2*Width = 2W2

    (6*Width)2/18 = 2W2 so (Perimeter2)/18 = Area (here I squared and then divided both sides of the Perimeter = ... equation by 18 and substituted)
    The time you enjoy wasting is not wasted time.
    Bertrand Russell

    <- Remember to rate posts you find helpful.
    Reply With Quote Reply With Quote
  3. Nov 4th, 2003, 07:38 PM #3
    Acidic
    Acidic is offline
    Frenzied Member Acidic's Avatar
    Join Date
    Sep 2003
    Location
    UK
    Posts
    1,090
    I think I'm right, not 100% sure though.
    ok, lets say you have a perimeter of P.
    P/6 is then the number of arbitury unit lengths (it's 12:33 am here so I'm tired). Now call that L.
    2L*L should be A
    so 2(P/6) * (P/6)
    0.5A = (P/6)^2
    6sqrt[.5A] = P

    I think thats right
    Have I helped you? Please Rate my posts.
    Reply With Quote Reply With Quote
  4. Nov 4th, 2003, 09:14 PM #4
    dolor
    dolor is offline

    Thread Starter
    Addicted Member dolor's Avatar
    Join Date
    Sep 2002
    Location
    in a place seldom found on a map
    Posts
    220
    acidic: i haven't done that problem myself yet, but i think that's right
    - you've been privileged to read a post by Miz
    Reply With Quote Reply With Quote
  6. Nov 5th, 2003, 12:17 PM #5
    alkatran
    alkatran is offline
    Fanatic Member alkatran's Avatar
    Join Date
    Apr 2002
    Location
    Canada
    Posts
    860
    width =2x, height = x
    area = width*height
    area = 2x*x = 2x^2

    perimeter = width + height
    perimeter = 2x+x = 3x

    soooo

    assuming
    P(3x) = 2x^2
    P(x) = 2(x/3)^2
    p(x) = 2x^2/9
    p(x) = x^2 / 4.5
    Don't pay attention to this signature, it's contradictory.
    Reply With Quote Reply With Quote
  7. Nov 5th, 2003, 12:50 PM #6
    Acidic
    Acidic is offline
    Frenzied Member Acidic's Avatar
    Join Date
    Sep 2003
    Location
    UK
    Posts
    1,090
    OK, I think I was right earlier but here is some working out easier to follow:
    A = 1/6P * 1/3P
    A = 1/18P^2
    18A = P^2

    sqrt[18A] = P
    or:

    3*sqrt[6A] = P
    which ever you prefer.
    Have I helped you? Please Rate my posts.
    Reply With Quote Reply With Quote
  8. Nov 5th, 2003, 05:26 PM #7
    DiGiTaIErRoR
    DiGiTaIErRoR is offline
    So Unbanned DiGiTaIErRoR's Avatar
    Join Date
    Apr 1999
    Location
    /dev/null
    Posts
    4,111
    Well...

    w=p/3
    h=p/6

    Thus... a=p^2/18

    so...

    p=3*sqr(2a)
    Reply With Quote Reply With Quote
  9. Nov 5th, 2003, 05:28 PM #8
    DiGiTaIErRoR
    DiGiTaIErRoR is offline
    So Unbanned DiGiTaIErRoR's Avatar
    Join Date
    Apr 1999
    Location
    /dev/null
    Posts
    4,111
    Originally posted by Acidic
    OK, I think I was right earlier but here is some working out easier to follow:
    A = 1/6P * 1/3P
    A = 1/18P^2
    18A = P^2

    sqrt[18A] = P
    or:

    3*sqrt[6A] = P
    which ever you prefer.
    sqrt(18a) = 3*sqrt(2a)

    3*3 = 9 * 2 = 18

    Reply With Quote Reply With Quote
  11. Nov 5th, 2003, 05:56 PM #9
    dolor
    dolor is offline

    Thread Starter
    Addicted Member dolor's Avatar
    Join Date
    Sep 2002
    Location
    in a place seldom found on a map
    Posts
    220
    Originally posted by Acidic

    sqrt[18A] = P
    or:

    3*sqrt[6A] = P
    which ever you prefer.
    wouldn't that be sqrt(54A) instead of sqrt(18A)?

    ...because 3 * sqrt(6A) = sqrt(9) * sqrt(6A) = sqrt(54A)
    - you've been privileged to read a post by Miz
    Reply With Quote Reply With Quote
  12. Nov 5th, 2003, 06:13 PM #10
    DiGiTaIErRoR
    DiGiTaIErRoR is offline
    So Unbanned DiGiTaIErRoR's Avatar
    Join Date
    Apr 1999
    Location
    /dev/null
    Posts
    4,111
    Originally posted by dolor
    wouldn't that be sqrt(54A) instead of sqrt(18A)?

    ...because 3 * sqrt(6A) = sqrt(9) * sqrt(6A) = sqrt(54A)
    He took the 3 out correctly, but forget it's sqrt(9)*sqrt(2)=sqrt(18), not... what he did.
    Reply With Quote Reply With Quote
  13. Nov 5th, 2003, 06:30 PM #11
    dolor
    dolor is offline

    Thread Starter
    Addicted Member dolor's Avatar
    Join Date
    Sep 2002
    Location
    in a place seldom found on a map
    Posts
    220
    Originally posted by DiGiTaIErRoR
    He took the 3 out correctly, but forget it's sqrt(9)*sqrt(2)=sqrt(18), not... what he did.
    ahh, i see now. thx
    - you've been privileged to read a post by Miz
    Reply With Quote Reply With Quote
  14. Nov 5th, 2003, 08:13 PM #12
    Acidic
    Acidic is offline
    Frenzied Member Acidic's Avatar
    Join Date
    Sep 2003
    Location
    UK
    Posts
    1,090
    Yes, my error, sorry.
    Have I helped you? Please Rate my posts.
    Reply With Quote Reply With Quote
  15. Nov 5th, 2003, 09:01 PM #13
    dolor
    dolor is offline

    Thread Starter
    Addicted Member dolor's Avatar
    Join Date
    Sep 2002
    Location
    in a place seldom found on a map
    Posts
    220
    Originally posted by Acidic
    Yes, my error, sorry.
    dont be..its no biggie
    - you've been privileged to read a post by Miz
    Reply With Quote Reply With Quote
« Previous Thread | Next Thread »

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  

Forum Rules


Click Here to Expand Forum to Full Width