A Quadrilateral's area given four sides and a diagonal Assume there exists a quadrilateral called ABCD and AB=5cm,BC=13cm,CD=16cm, DA=20cm and diagonal AC=12cm. The exercise now states that I should calculate the area of a quadrilateral. Thank you for your time!!! :)
 A: Hint:
The quadrilateral is composed by two triangles for which you know the three sides. So you can calculate the area of each triangle with the Heron's formula.
A: Observe that $5^2+12^2=13^2$ and $12^2+16^2=20^2.$ Therefore in $\triangle A B C, \;$ $\angle B A C=90^o$ and in $\triangle A C D,  \;$ $\angle A C D=90^o$. So the area of $\triangle A B C$ is $5.12/2=30$ and the area of $\triangle A C D$ is $12.16/2=96$, for a total area of $30+96=126.$ 
A: HINT.-There is the formula $$A=\frac{\sqrt{4(l\cdot l_1)^2-(a^2-b^2+c^2-d^2)^2}}{2}$$ where $A$ is the area, $a,b,c,d$ the sides, $l$ and $l_1$ the two diagonals. You have to calculate the second diagonal and apply the formula.
A: By the Law of Cosines:
$$AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cdot \cos \angle ABC$$
$$\iff \cos \angle ABC=\frac{5^2+13^2-12^2}{2\cdot 5\cdot 13}=\frac{5}{13}$$
Analogously: $$\cos \angle CDA=\frac{16^2+20^2-12^2}{2\cdot 16\cdot 20}=\frac{4}{5}$$
Now, we know that $\angle ABC$ and $\angle CDA$ are both acute, because $\angle ABC<\angle BAC$ and $\angle CDA<\angle ACD$. Therefore:
$$\sin \angle ABC=\sqrt{1-\cos^2 \angle ABC}=\frac{12}{13}$$
$$\sin\angle CDA=\sqrt{1-\cos^2\angle CDA}=\frac{3}{5}$$
$$A_{ABCD}=A_{\triangle ABC}+A_{\triangle CDA}$$
$$=\frac{\sin \angle ABC\cdot 5\cdot 13}{2}+\frac{\sin\angle CDA\cdot 16\cdot 20}{2}=126$$
