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Math Help - Rectangles

  1. #1
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    Exclamation Rectangles

    Can someone help me with these, please

    1. The length of a rectangle is 6cm less than twice its width. Find the dimensions of a rectangle if its area is 108cm^2 (squared)?

    2. Find the dimensions of a rectangle whose perimeter is 46m and whose area is 126m^2 (squared).

    3. Find the dimensions of a rectangle whose perimeter is 42m and whose area is 104^2 (squared).

    Please help

    EDIT: Please help, I don't understand it at all. Our teacher always starts a new chapter on the day of the previous test and between periods we don't get anytime to comprehend much.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    What's with the shouting?

    Quote Originally Posted by doomdesire34 View Post
    Can someone help me with these, please

    1. The length of a rectangle os 6cm less than twice its width. Find the dimensions of a rectangle if its area is 108cm^2 (squared)?
    let the width be x
    then the length is 2x - 6

    now area is length times width, and we are told that is 108
    => length*width = 108
    => (2x - 6)x = 108
    => 2x^2 - 6x = 108
    => 2x^2 - 6x - 108 = 0
    => x^2 - 3x - 54 = 0
    => (x - 9)(x + 6) = 0
    so x - 9 = 0 => x = 9
    or x + 6 = 0 => x = -6

    x cannot be -6, so x = 9

    so the dimensions of the rectangle is: width = 9cm, length = 2x - 6 = 12cm

    2. Find the dimensions of a rectangle whose perimeter is 46m and whose area is 126m^2 (squared).
    let the sides of the rectangle be x and y
    let the perimeter be P
    let the area be A

    now P = 2x + 2y = 46
    and A = xy = 126

    so to find x and y, we need to solve the simultaneous equations:

    2x + 2y = 46 ...................(1)
    xy = 126 ........................(2)

    x + y = 23 ..................(3) = (1)/2
    xy = 126 .....................(2)

    From (2), we see that x = 126/y, substitute 126/y for x in (3), we get:

    126/y + y = 23
    => 126 + y^2 = 23y ...........i multiplied through by y
    => y^2 - 23y + 126 = 0
    => (y - 14)(y - 9) = 0
    => y - 14 = 0 => y = 14
    or y - 9 = 0 => y = 9

    but x + y = 23
    => x = 14 ...........when y = 9
    or x = 9 .............when y = 14

    x and y represent arbitrary sides, and the numbers are the same, so just pick one

    so one side is 14 and the other is 9
    you can say x = 14 and y = 9 or x = 9 and y = 14, same thing

    3. Find the dimensions of a rectangle whose perimeter is 42m and whose area is 104^2 (squared).
    this is exactly like the one above. why don't you try it and post your solution?
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  3. #3
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    Smile One minute

    sorry, just gimme a few more minutes, thanks for the reply
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  4. #4
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    ok I have the sides as x and y and the
    P = 2x + 2y = 42
    A = xy = 104

    2x + 2y =42
    xy = 104

    104y + y = 21
    104 + y^2 = 21y
    y^2 - 21y + 104 = 0
    (y - 13) (y - 8) = 0
    y - 13 = 0 y = 13
    y - 8 = 0 y = 8

    Dimensions: 13m, 8m
    Last edited by doomdesire34; April 24th 2007 at 04:27 PM.
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  5. #5
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    ok I have the sides as x and y and the
    P = 2x + 2y = 42
    A = xy = 104

    2x + 2y =42
    xy = 104

    104y + y = 21
    104 + y^2 = 21y
    y^2 - 21y + 104 = 0
    (y - 13) (y - 8) = 0
    y - 13 = 0 y = 13
    y - 8 = 0 y = 8

    Dimensions: 13m, 8mits from above, sorry I just edited the "sorry I'm stuck" one to the correct problem. Is it right?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by doomdesire34 View Post
    ok I have the sides as x and y and the
    P = 2x + 2y = 42
    A = xy = 104

    2x + 2y =42
    xy = 104

    104y + y = 21
    104 + y^2 = 21y
    y^2 - 21y + 104 = 0
    (y - 13) (y - 8) = 0
    y - 13 = 0 y = 13
    y - 8 = 0 y = 8

    Dimensions: 13m, 8m

    its from above, sorry I just edited the "sorry I'm stuck" one to the correct problem. Is it right?
    Yup, that seems ok. good job
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  7. #7
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    Talking

    Thank you very much for helping, I have a feeling I'm going to be the only person with these done for tommorrow
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