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Math Help - rectangle and square

  1. #1
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    rectangle and square

    I need help in the below problem

    The figure is made up of a square and a rectangle. The ratio of the area of the square to the area of the rectangle is 2 : 5. I/4 of the square is shaded.
    What fraction of the figure is unshaded?
    Attached Thumbnails Attached Thumbnails rectangle and square-square-rectangle.gif  
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: rectangle and square

    Quote Originally Posted by kingman View Post
    I need help in the below problem

    The figure is made up of a square and a rectangle. The ratio of the area of the square to the area of the rectangle is 2 : 5. I/4 of the square is shaded.
    What fraction of the figure is unshaded?
    Let the areas of the square and the rectangle be 2 and 5 respectively.

    Area of the shaded area = 1/4 * 2 = 0.5

    Total area = 2 + 5 - 0.5 = 6.5

    Area of the unshaded area = 6.5 - 0.5 = 6

    Fraction unshaded = 6 / 6.5 = 12 / 13
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  3. #3
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    Re: rectangle and square

    Hello, kingman!

    Another approach . . .


    The figure is made up of a square and a rectangle.
    The ratio of the area of the square to the area of the rectangle is 2:5.
    I/4 of the square is shaded.
    What fraction of the figure is unshaded?

    Code:
          * - - - - - - - *
          |               |
          |               |
          |     C     * - + - *
          |           |:::|   |
          |           |:A:|   |
          * - - - - - + - *   |
                      |       |
                      |   B   |
                      * - - - *
    Let S = area of the square.
    Let R = area of the rectangle.

    We have: . A = \tfrac{1}{4}S,\;B = \tfrac{3}{4}S

    We have: . \frac{S}{R} = \frac{2}{5} \quad\Rightarrow\quad R \,=\,\tfrac{5}{2}S

    Then: . C \:=\:R - A \:=\:\tfrac{5}{2}S - \tfrac{1}{4}S \:=\:\tfrac{9}{4}S


    The unshaded area is: . B+C \:=\:\tfrac{3}{4}S + \tfrac{9}{4}S \:=\:3S

    The total area is: . A+B+C \:=\:\tfrac{1}{4}S + \tfrac{3}{4}S + \tfrac{9}{4}S \:=\:\tfrac{13}{4}S


    Therefore: . \text{Fraction unshaded} \;=\; \frac{3S}{\frac{13}{4}S} \;=\;\frac{12}{13}

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  4. #4
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    Re: rectangle and square

    Quote Originally Posted by Soroban View Post
    Hello, kingman!

    Another approach . . .


    Code:
        * - - - - - - - *
        |               |
        |               |
        |     C     * - + - *
        |           |:::|   |
        |           |:A:|   |
        * - - - - - + - *   |
                    |       |
                    |   B   |
                    * - - - *
    Let S = area of the square.
    Let R = area of the rectangle.

    We have: . A = \tfrac{1}{4}S,\;B = \tfrac{3}{4}S

    We have: . \frac{S}{R} = \frac{2}{5} \quad\Rightarrow\quad R \,=\,\tfrac{5}{2}S

    Then: . C \:=\:R - A \:=\:\tfrac{5}{2}S - \tfrac{1}{4}S \:=\:\tfrac{9}{4}S


    The unshaded area is: . B+C \:=\:\tfrac{3}{4}S + \tfrac{9}{4}S \:=\:3S

    The total area is: . A+B+C \:=\:\tfrac{1}{4}S + \tfrac{3}{4}S + \tfrac{9}{4}S \:=\:\tfrac{13}{4}S


    Therefore: . \text{Fraction unshaded} \;=\; \frac{3S}{\frac{13}{4}S} \;=\;\frac{12}{13}

    I wonder why all the numbers and text displayed are blurred; so cannot figure out what is wriiten on the solution page.
    thanks
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