Area of a quadrilateral solved with trig

**AliceFisher** · April 24th 2010, 01:03 PM

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

$ PR^2 = 6^2-7^2-2(6)(7)cos(115)= 108.12 $

so

$ PR =sqrt(108.12) = 10.98 $

The area of PRQ

$ 0.5*6*7*sin(115)=19.03cm^2 $

So then I look for the angle QRP

$ 6sin(115)/10.98 = 0.495 thus 29.68 $

This means the angle PRS = 80-29.68 = 50.32

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

$ 0.5*10.98*9*sin(50.32) = 38.02cm^2 $

finally add the two halves together giving:

$ 19+38 = 57cm^2 $

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

**Soroban** · April 24th 2010, 04:44 PM

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)

**AliceFisher** · April 25th 2010, 12:49 AM

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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