area of ​​a quadrilateral Get the area of ​​a quadrilateral?
$‎\angle ‎A‎‎‎_{1}‎+‎\angle ‎C‎_{3}‎=30‎^{‎\circ‎}‎‎‎‎‎$‎
$\angle ‎A‎‎‎_{2}‎+‎\angle ‎C‎_{4}‎=90‎^{‎\circ‎}‎‎‎$
$CD=9, DA=5, BC=8 , AB=4$

 A: This can follow directly from Bretschneider's formula:
$$K = \sqrt {(s-a)(s-b)(s-c)(s-d) - abcd  \cdot \cos^2 \left(\frac{A+C}{2}\right)}$$
A: You have
$$
\angle ‎A‎‎‎_{1}‎+‎\angle ‎C‎_{3}‎=30‎^{‎\circ‎}‎‎‎‎‎\tag1
$$
and
$$
\angle ‎A‎‎‎_{2}‎+‎\angle ‎C‎_{4}‎=90‎^{‎\circ‎}‎‎‎‎‎\tag2
$$
Adding $(1)$ and $(2)$, you will get
$$
(\angle ‎A‎‎‎_{1}‎+\angle ‎A‎‎‎_{2})‎+(‎\angle ‎C‎_{3}+‎\angle ‎C‎_{4})‎=\angle ‎A‎‎‎+\angle ‎C‎‎‎=120‎^{‎\circ‎}.\tag3
$$
Now, divide the quadrilateral into two triangles, $\Delta ABD$ and $\Delta BCD$. Using cosine formula, determine $BD$. From $\Delta ABD$ you will get
$$
\begin{align}
BD^2&=AB^2+BD^2-2AB\cdot BD\cdot\cos\angle ‎A\\
&=4^2+5^2-2\cdot4\cdot5\cdot\cos\angle ‎A\\
&=41-20\cos\angle ‎A.\tag4
\end{align}
$$
From $\Delta BCD$ you will get
$$
\begin{align}
BD^2&=BC^2+CD^2-2BC\cdot CD\cdot\cos\angle ‎C\\
&=8^2+9^2-2\cdot8\cdot9\cdot\cos(120^\circ-\angle ‎A)\\
&=145-144\cos(120^\circ-\angle ‎A).\tag5
\end{align}
$$
Using $(4)$ and $(5)$, you will get
$$
41-20\cos\angle ‎A=145-144\cos(120^\circ-\angle ‎A).\tag6
$$
Solve for $\angle ‎A$ from $(6)$ and use $\angle ‎A$ to obtain $\angle ‎C$ using $(3)$. Thus, you can obtain the area of quadrilateral $ABCD$ by adding area of $\Delta ABD$ and $\Delta BCD$.
$$
\begin{align}
[ABCD]&=[ABD]+[BCD]\\
&=\frac{1}{2}AB\cdot BD\cdot\sin\angle ‎A+\frac{1}{2}BC\cdot CD\cdot\sin\angle ‎C.
\end{align}
$$
$$\\$$

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