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Math Help - length of the bigger square..

  1. #1
    rcs
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    length of the bigger square..

    The ratio of the sides of 2 squares is 4:5. If the sum of their areas is 180 sq.cm, what is the length of the bigger square?

    please do me a favor on this...
    as i thought , can this be 4x + 5x = 180?

    can anybody let me understand on this...
    thanks a lot
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: length of the bigger square..

    Let x be a side of the smaller square and let y be a side of the bigger square.
    The area of the smaller square is: x^2
    The area of the bigger square is: y^2
    So: x^2+y^2=180 (1)

    You know the ratio of the sides of 2 square is: \frac{4}{5}, try to express this in function of x and y and then you can substitute this given into equation (1).
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  3. #3
    rcs
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    Re: length of the bigger square..

    4/5 as x how about for y? we have only 1 equation... i cannot perform elimination in here.

    thanks..
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    Re: length of the bigger square..

    No, he did not say x was 4/5. Your problem says that the ratio of lengths is 4 to 5: x=(4/5)y.
    x^2+ y^2= 180 becomes \frac{16}{25}y^2+ y^2= 180.
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    Re: length of the bigger square..

    Hello, rcs!

    The ratio of the sides of two squares is 4:5.
    If the sum of their areas is 180 sq.cm,
    what is the length of the bigger square?

    You had a good start . . .

    Let 4x = side of smaller square.
    Let 5x = side of larger square.

    Their areas are: . \begin{Bmatrix}(4x)^2 &=& 16x^2 \\ (5x)^2 &=& 25x^2 \end{Bmatrix}

    The sum of their areas is 180: . 16x^2 + 25x^2 \:=\:180

    Go for it!

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    rcs
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    Re: length of the bigger square..

    Sir Soroban, HallsofIvy you guys are an angel !
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