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Math Help - Length and area #3

  1. #1
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    Length and area #3

    2. A circle is inscribed in square. One point of a circle is 1 and 2 away from the two closest sides of the square. Find the area of the square.

    3. A certain city block is in the form of a parallelogram. Two of it's sides measure 32ft and 41ft. If the area of the land in the block is 656ft^2, what is the length if it's longer diagonal?

    Thanks for helping!
    Last edited by mr fantastic; October 14th 2010 at 04:47 AM. Reason: Questins moved to new thread.
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  2. #2
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    Hello, ejaykasai!

    2. A circle is inscribed in square. One point on the circle is 1 unit and 2 units away
    from the two closest sides of the square. Find the area of the square.
    Code:
        O         r
        - * - - - - - - - * -
          | *             *
      r-2 |   * r         | r-2
          |     *        *|
          |       *       |
          |         * A * |
        B + - - - - - * - + -
          |    r-1    :   |
        2 |        *  :   | 2
        - * -*- - - - + - * -
          :    r-1    : 1 :

    The circle has center \,O and radius \,r.
    Point \,A is 1 unit and 2 units from the circumscribed square.

    In right triangle OBA\!:\;\;(r-1)^2 + (r-2)^2 \:=\:r^2

    . . r^2 - 6r + 5 \:=\:0 \quad\Rightarrow\quad (r-5)(r-1) \:=\:0

    Hence: . r = 5,\;\rlap{/////}r = 1


    The side of the square is 10 units.

    Therefore, the area of the square is: . 10^2 = 100 units ^2.
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  3. #3
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    Hello again, ejaykasai!

    3. A certain city block is in the form of a parallelogram.
    Two of it's sides measure 32 ft and 41 ft.
    If the area of the block is 656 ft^2, what is the length if its longer diagonal?
    Code:
               A *--------------------* D
                /                    /|
               /                    / |
           32 /                    /  |
             /                  32/   | h
            /                    /    |
           /                    /     |
          *--------------------*- - - *
          B - - - - 41 - - - - C  - - E

    We have parallelogram ABCD\!:\;\;AB = 32,\:BC = 41.
    Let the height be: h \:=\:DE.

    Since \text{(Area)} \:=\:\text{(base)} \times \text{(height)}
    . . we have: . 41h \:=\:656 \quad\Rightarrow\quad h \:=\:16


    Code:
               A *--------------------* D
                /                 *  /|
               /              *     / |
           32 /           *        /  |
             /        *         32/   | 16
            /     *              /    |
           /  *                 /     |
          *--------------------*- - - *
          B - - - - 41 - - - - C  - - E

    In right triangle DEC\!:\;CE^2 + 16^2 \:=\:32^2 \quad\Rightarrow\quad CE^2 = 768

    Hence: . CE = 16\sqrt{3} \quad\Rightarrow\quad BE \:=\:41 + 16\sqrt{3}


    In right triangle DEB\!:\;\;BD^2 \:=\:(41+16\sqrt{3})^2 + (16)^2

    . . BD^2 \;=\;1681 + 1312\sqrt{3} + 768 + 256 \;=\;2705 + 1312\sqrt{3}


    Therefore: . BD \;=\;\sqrt{2705 + 1312\sqrt{3}} \;\approx\;70.55\text{ ft}
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