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    A rectangle has a diagonal of sqrt45. The length is twice the length of the width, find the length and width of this rectangle.

    Unfortunately, exhaustion seems to have got to me and I cannot work out this question when I should normally be able to within a matter of minutes.

    Any help is appreciated.
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    Give the sides the value of x to the width and work from there
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    Illuminati? BTW I have no idea what you're on about
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    (Original post by Loyale)
    Give the sides the value of x to the width and work from there
    I've done that but I can't seem to work it out any further than, x^2 + 2x^2 = sqrt45^2
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    Use Pythagoras once you've used algebra for the length and the width and create an equation out of this so,

    Width is x
    Length is 2x

    X^2+2x^2=sqrt45
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    Simply the sqrt45^2?
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    You get x^2+2x^2=45 if I'm not wrong and solve the qudratic
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    (Original post by Loyale)
    You get x^2+2x^2=45 if I'm not wrong and solve the qudratic
    Thanks, I did think that but i'll have another go and see what I get
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    I seem to get very long decimal answers, but the actual answer is width is 3cm but I don't know how to get there?
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    (Original post by Loyale)
    Use Pythagoras once you've used algebra for the length and the width and create an equation out of this so,

    Width is x
    Length is 2x

    X^2+2x^2=sqrt45
    But if the length is 2x, then surely the square of the length is (2x)^2=4x^2, not 2x^2.

    The equation would then be \displaystyle x^2+4x^2=45.
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    (Original post by Dingooose)
    But if the length is 2x, then surely the square of the length is (2x)^2=4x^2, not 2x^2.

    The equation would then be \displaystyle x^2+4x^2=45.
    Oh Yh my bad, just very tired too
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    (Original post by joepiekos)
    I seem to get very long decimal answers, but the actual answer is width is 3cm but I don't know how to get there?
    This is because the equation that Loyale derived is incorrect. Check my previous post.
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    (Original post by Loyale)
    Oh Yh my bad, just very tired too
    It's ok, everyone makes mistakes.
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    (Original post by Dingooose)
    This is because the equation that Loyale derived is incorrect. Check my previous post.
    Thank you so much for your help, I don't know why I didn't get this earlier XD
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    wait, if it's x^2 + 4x^2 = 45, I can't solve that like a normal quadratic can I?
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    (Original post by joepiekos)
    wait, if it's x^2 + 4x^2 = 45, I can't solve that like a normal quadratic can I?
    Oh wait I know what to do now, factorise to get X^2(1+4)=45 then X^2=45/5 which is x^2 =9 so x= 9?
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    (Original post by joepiekos)
    Oh wait I know what to do now, factorise to get X^2(1+4)=45 then X^2=45/5 which is x^2 =9 so x= 9?
    If x^2=9, then x=3. Recall what x represents and also keep in mind that you're being asked for the width AND length of the rectangle.
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    Yeah so the length is then 6 because it's just double the width (x)
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    (Original post by joepiekos)
    Yeah so the length is then 6 because it's just double the width (x)
    Indeed.
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    (Original post by Dingooose;[url="tel:59381661")
    59381661[/url]]Indeed.
    I greatly appreciate your help thank you Very much
 
 
 

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