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Math Help - Area of rectangle

  1. #1
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    Exclamation Area of rectangle

    The diagonal of a rectangle is 17c.m. long and the perimeter of the rectangle is 46 c.m. Find the area of the rectangle.

    Please tell me how to solve this problem.
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Lets say , a is length and b is breadth of rectangle

    Perimeter = 2(a+b) = 46
    a+b = 23
    b = 23 - a
    Diagonal = \sqrt{a^2 +b^2 }=17

    a^2 + b^2 = 289


    a^2 + a^2 + 529 - 46a = 289

    2a^2 - 46a + 240 = 0

    a^2 - 23a + 120 = 0

    The value that we will get as a can be either length or breadth (as even if you solved for b you would have recieved same equaion)

    Hence area = Lengt*Breadth = Multiplication of roots of this quadratic equation

    -----------------------
    For an equation

    lx^2 +mx +n = 0

    Multiplication of roots = n/l

    -----------------

    So here it will be = 120
    Last edited by ADARSH; April 3rd 2009 at 10:54 PM. Reason: blasphemy, I wrote 240 as 220 and solved , thanks Masters!
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by rickymylv View Post
    The diagonal of a rectangle is 17c.m. long and the perimeter of the rectangle is 46 c.m. Find the area of the rectangle.

    Please tell me how to solve this problem.
    Hi rickymylv,

    Let's assume the sides are integers.

    2l + 2w = 46

    [1] l + w = 23

    [2] l^2 + w^2 = 289

    Solve the above system.

    [1]  l = 23-w

    [2]  (23-w)^2+w^2=289

     529-46w+w^2+w^2-289=0

    2w^2-46w+240=0

    w^2-23w+120

    (w-8)(w-15)=0

    w=8 \ \ or \ \ w=15

    Therefore, l=8 \ \ or \ \ l=15

    The width is 8 and the length is 15.
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