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Math Help - Geometry problem help!?!

  1. #1
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    Geometry problem help!?!

    http://i78.photobucket.com/albums/j95/ashree101/lmk.jpg

    it's a picture of a problem i need help solving.
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  2. #2
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    Smile Solution

    Please see here ...

    Please specify what exactly is 74 degree. Then only first three parts can be solved.

    Now for next two parts...

    a) To find the area of tr. ABC
    -> As ABC is isosceles, therefore, its median can be dropped from A. This median would be the perpendicular bisector of BC. Let this median be AD.
    So, BD = 1/2 x BC = 10 cm. So AD is 12.5 cm (approx) (Pythagoras theorem)
    So area of tr. ABD = 1/2 x BD x AD = 1/2 x 10 x 12.5= 62.5 cm sq. and, so area of tr. ABC = 2 x ABD = 125 cm sq.

    b) to find the area of circle
    -> This circle is the circumcircle of ABC. So, O lies on AD as 2:1. Now, OA = 2/3 x AD = 8.3 cm (approx.)
    area of circle = 22/7 x OA x OA = 216.3 sq. cm. (approx.)
    Attached Thumbnails Attached Thumbnails Geometry problem help!?!-picture3.jpg  
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  3. #3
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    Hello, freebrd01!

    Code:
                  o o o    C
              o          ♠
            o     16  *   * o
           o       *       * o 74
                *           *
          o  * 37           *o
        B ♠   *   *   *   *   ♠ A
          o        20         o
    
           o                 o
            o               o
              o           o
                  o o o

    Find:
    . . (a)\;\angle C\quad (b)\;\angle A \quad (c)\;AC\quad (d)\:\text{area of }\Delta ABC \quad (e)\;\text{area of circle}

    Since \text{arc}(CA) \,=\, 74^o,\;\angle B \,=\, 37^o


    Law of Cosines: . AC^2 \:=\:16^2 + 20^2 - 2(16)(20)\cos37^o \;=\;144.8732736

    Hence: . AC \;=\;\sqrt{144.8732736} \;=\;12.0363314 \quad\Rightarrow\quad AC\;\approx\;12 .(c)


    We have \Delta ABC with: . AC = 12,\;BC = 16,\;AB = 20

    Check it out! . \Delta ABC is a right triangle!

    (I assume that is the intent of the problem.)

    Hence: . \angle C \,=\,90^o (a)

    . .And: . \angle A \,=\,53^o (b)


    \text{Area }\Delta ABC \:=\:\tfrac{1}{2}(BC)(AC) \:=\:\tfrac{1}{2}(16)(12) \:=\:96 (d)



    Since the diameter is AB = 20, the radius is 10.

    \text{Area of circle} \:=\:\pi(10^2) \:=\:100\pi (e)

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