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Math Help - Area of a quadrilateral solved with trig

  1. #1
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    Area of a quadrilateral solved with trig

    So this is another one from Bostock and Chandlers Core A-Level Maths:

    The Question

    In a quadrilateral PQRS, PQ=6cm, QR = 7cm, RS = 9cm, PQR=115 and PRS=80.

    Find the length of PR, then considering it split into two separate triangles find the area of PQRS.

    My Solution

    <br />
PR^2 = 6^2-7^2-2(6)(7)cos(115)= 108.12<br />

    so

     <br />
PR =sqrt(108.12) = 10.98<br />

    The area of PRQ

    <br />
0.5*6*7*sin(115)=19.03cm^2<br />

    So then I look for the angle QRP

    <br />
6sin(115)/10.98 = 0.495 thus 29.68<br />

    This means the angle PRS = 80-29.68 = 50.32

    I then use this angle to find the area of the other half PRS

    <br />
0.5*10.98*9*sin(50.32) = 38.02cm^2<br />

    finally add the two halves together giving:

    <br />
 19+38 = 57cm^2 <br />

    The book says the answer is 67.7 or 67.8 (allowing for rounding)

    So the question is why am I so far out... I've went through my workings a couple of times now and cannot see where I am going wrong.

    Any help appreciated
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  2. #2
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    Lexington, MA (USA)
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    Hello, Alice!

    In a quadrilateral PQRS\!:\; PQ\,=\,6\text{ cm},\;QR\,=\,7\text{ cm},\;RS \,=\,9\text{ cm},\;\angle PQR\,=\,115^o,\;\angle PRS\,=\,80^o

    (a) Find the length of PR.

    (b) Then find the area of PQRS.
    Code:
                 Q        7         R
                  o   *   *   *   O 
                 * 115        *    *    
                *           *   80   *
               *         *              *
            6 *       *                   * 9
             *     *                        *
            *   *                             *
           * *                                  *
          o   *   *   *   *   *   *   *   *   *   o
         P                                          S

    PR^2 \;=\;6^2 + 7^2 - 2(6)(67)\cos115^o \;=\;120.499934

    PR \:=\:10.977724619 \quad\Rightarrow\quad PR \;\approx\;10.98\text{ cm} .(a)


    Area \Delta PQR \:=\:\tfrac{1}{2}(6)(7)\sin115^o \:=\:19.03246353 \;\approx\;19.03\text{ cm}^2

    Area \Delta PRS \;=\;\tfrac{1}{2}(10.98)(9)\sin80^o \;=\;48.65935108 \;\approx\;48.66\text{ cm}^2


    Therefore: .area PQRS \;=\;19.03 + 48.66 \;=\;67.69\text{ cm}^2 .(b)

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  3. #3
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    Thanks for that, seems I drew my 80 degree angle as QRS!

    I really should avoid doing this sort of thing at 1 in the morning
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