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Math Help - finding the perimeter

  1. #1
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    Lightbulb finding the perimeter

    finding the perimeter-untitled.jpg
    Seven congruent rectangles form a bigger rectangle as shown in the attachment.If the area of the bigger attachment is 336 square units,what is the perimeter of the bigger rectangle.(Sorry fig. not accurate)
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  2. #2
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by anshulbshah View Post
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    Seven congruent rectangles form a bigger rectangle as shown in the attachment.If the area of the bigger attachment is 336 square units,what is the perimeter of the bigger rectangle.(Sorry fig. not accurate)
    You can get an equation relating the lengths of the sides of one of the smaller rectangles straight away, which will help.

    Let the sides of the small rects be a and b.

    Then 3a = 4b.

    Then the perimeter is 4b + 2 (a+b) + 3a.

    The area is given by (a+b)(3a) = 336.

    Then I expect you just need to eliminate either a or b from this system of simultaneous equations and the solution should drop into your lap.
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  3. #3
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    area of 7 rectangles=336 square unit
    area of 1 rectangle=48 square units
    lets take the breadth of smaller rectangle as B and length as L
    [B will be the smaller side as per the figure]

    As per the figure
    4. B= 3. L
    B= 3L/4

    area of smaller rectangle=48square units
    so,
    L.B=48
    now we substitute the value of B in this equation
    L. 3L/4=48
    so
    L=8
    B=6

    as per the figure
    area of bigger rectangle=
    5L+6B
    => 5X8+6X6
    =>40 + 36
    = 76 units
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  4. #4
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    Hello, anshulbshah!

    Seven congruent rectangles form a bigger rectangle as shown in the diagram.
    If the area of the bigger rectangle is 336 square units,
    what is the perimeter of the bigger rectangle?
    Code:
                y           y           y           y
          * - - - - - * - - - - - * - - - - - * - - - - - *
          |           |           |           |           |
          |           |           |           |           |
          |           |           |           |           |
        x |           |           |           |           | x
          |           |           |           |           |
          |           |           |           |           |
          |           |           |           |           |
          * - - - - - * - * - - - * - - - * - * - - - - - *
          |               |               |               |
          |               |               |               |
        y |               |               |               | y
          |               |               |               |
          |               |               |               |
          * - - - - - - - * - - - - - - - * - - - - - - - *
                  x               x               x
    Let: . \begin{array}{ccc}x &=& \text{length of a small rectangle} \\ y &=& \text{width of a small rectangle} \end{array}


    From the top and bottom edges of the rectangle,
    . . we see that: . 4y \:=\:3x \quad\Rightarrow\quad y \:=\:\tfrac{3}{4}x .[1]


    The length of the big rectangle is: L \,=\,3x
    The width of the big rectanble is: W \,=\,x + y
    . . Hence, its area is: . L\cdot W \:=\:(3x)(x+y) \:=\:3x^2 + 3xy

    We are told that the area is 336 unitsē.
    . . So we have: . 3x^2 + 3xy \:=\:336 .[2]

    Substitute [1] into [2]: . 3x^2 + 3x\left(\tfrac{3}{4}x\right) \:=\:336 \quad\Rightarrow\quad 3x^2 + \tfrac{9}{4}x^2 \:=\:336

    . . \tfrac{21}{4}x^2 \:=\:336 \quad\Rightarrow\quad x^2 \:=\:64 \quad\Rightarrow\quad \boxed{x \:=\:8}

    Substitute into [1]: . y \:=\:\tfrac{3}{4}(8) \quad\Rightarrow\quad \boxed{y \:=\:6}


    The length of the big rectangle is: . L \:=\:3x \:=\:3(8) \quad\Rightarrow\quad L\:=\:24
    The width of the big rectangle is: . W \:=\:x+y \:=\:8+6 \quad\Rightarrow\quad W\:=\:14


    Therefore, the perimeter is: . 2L + 2W \;=\;2(24) + 2(14) \;=\;76 units.


    Edit: Too slow again . . .
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  5. #5
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by Soroban View Post
    Edit: Too slow again . . .
    ... too complete as well ...
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