Hello, Alice!

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

(a) Find the length of $\displaystyle PR$.

(b) Then find the area of $\displaystyle PQRS. $ Code:

Q 7 R
o * * * O
* 115° * *
* * 80° *
* * *
6 * * * 9
* * *
* * *
* * *
o * * * * * * * * * o
P S

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

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

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

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

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