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    How do you find the area of a rectangle with a given diagonal, width and length but the width and length is in algebra so x and 2x?
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    Write the diagonal in terms of x, then solve for x. Finally, calculate the area.
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    (Original post by RogerOxon)
    Write the diagonal in terms of x, then solve for x. Finally, calculate the area.
    My diagonal is 25 so how do I put that in terms of x? I've never done diagonal before so that's why am confused.
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    (Original post by Daydreamer3)
    My diagonal is 25 so how do I put that in terms of x? I've never done diagonal before so that's why am confused.
    Try and form an equation using Pythagoras' Theorem. Please post your thoughts/working if you're unsure.
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    (Original post by notnek)
    Try and form an equation using Pythagoras' Theorem. Please post your thoughts/working if you're unsure.
    Ohhh Okay so
    2x^2+x^2=25?
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    (Original post by Daydreamer3)
    Ohhh Okay so
    2x^2+x^2=25?
    Nearly but (2x)^2 = 4x^2 not 2x^2.
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    (Original post by notnek)
    Nearly but (2x)^2 = 4x^2 not 2x^2.
    Oh right I didnt know you had to put it into brackets....
    So does x = 2.236067977?
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    (Original post by Daydreamer3)
    Oh right I didnt know you had to put it into brackets....
    So does x = 2.236067977?
    You haven't said if the diagonal length is 5 but if it is then that is correct. You can write is as \sqrt{5} though - no need to use your calculator.

    Also this looks like it wouldn't be a calculator question in the GCSE exam.

    Can you find the area?
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    (Original post by notnek)
    You haven't said if the diagonal length is 5 but if it is then that is correct. You can write is as \sqrt{5} though - no need to use your calculator.

    Also this looks like it wouldn't be a calculator question in the GCSE exam.

    Can you find the area?
    The diagonal is 25....
    I did
    5x^2=25
    x^2=5
    x= square root of 5
    Is that wrong?
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    (Original post by Daydreamer3)
    The diagonal is 25....
    I did
    5x^2=25
    x^2=5
    x= square root of 5
    Is that wrong?
    If the diagonal is 25 then you've made a mistake because you have to square the hypotenuse when using Pythagoras. So your equation should be:

    (2x)^2 + x^2 = 25^2
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    (Original post by notnek)
    If the diagonal is 25 then you've made a mistake because you have to square the hypotenuse when using Pythagoras. So your equation should be:

    (2x)^2 + x^2 = 25^2
    Oh right so x=11.18033989 or square root 125
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    (Original post by Daydreamer3)
    Oh right so x=11.18033989 or square root 125
    That's correct. Can you now find the area?

    Please post future maths questions in the maths help forum
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    (Original post by notnek)
    That's correct. Can you now find the area?

    Please post future maths questions in the maths help forum
    I got 250
    Thank you for the help!
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    (Original post by Daydreamer3)
    I got 250
    Thank you for the help!
    Me too. Here's my working - note that a calculator is not required:

    (2x)^2+x^2=25^2

    5x^2=25^2

    x^2=\frac{({5^2})^2}{5}=\frac{5^  4}{5}=5^3

    A=2x^2=2.5^3=250
 
 
 
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