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Math Help - How can i solve this problem! rectangles

  1. #1
    Newbie miladceph's Avatar
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    How can i solve this problem! rectangles

    Hi, How's everything guys?!
    Guys i have a problem with a basic math question

    Question: Length of a rectangle is equal to [2x+1] and Width of the rectangle is equal to [x+5] Calculate the perimeter and area of ​​the rectangle in terms of x.

    Please help me guys I Solve it but i think my answer is wrong!

    please cheack it:x:x:x

    Length : 2x+1
    Width : x+5


    Perimiter : 2(2x+1+x+5) => 2(3x +6) => 6x + 12 => x= 6/12 => x=2
    Area: (2x+1).(x+5) => (3x+ 10x + x + 5) => (14x + 5) => x = 5/14
    Last edited by miladceph; May 29th 2013 at 10:49 AM.
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  2. #2
    MHF Contributor ebaines's Avatar
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    Illinois
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    Re: How can i solve this problem! rectangles

    Quote Originally Posted by miladceph View Post
    please cheack it:x:x:x

    Length : 2x+1
    Width : x+5


    Perimiter : 2(2x+1+x+5) => 2(3x +6) => 6x + 12 => x= 6/12 => x=2
    Area: (2x+1).(x+5) => (3x+ 10x + x + 5) => (14x + 5) => x = 5/14
    For the perimeter calculation you were doing great for both of these right up til the last statement. You have:

    Perimeter = 6x + 12.

    That's the answer right there, as it defines the perimter in terms of x, so no need to go on! Your next move was to say that x = 12/6, but that's wrong. Applying basic algebra you might have said this:

    Perimter = 6x+12
    6x = Perimeter - 12
    x = Perimter/6 - 2

    But that simply gives x in terms of the perimter, but that's not what they're asking.

    As for the area calculation you didn't multiply the two terms correctly. Area = (2x+1)(x+5) = 2x(x+5) + 1(x+5) = 2x^2 +10x + x + 5 = 2x^2 +11x + 5.
    Thanks from miladceph, topsquark and HallsofIvy
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