In the quadrilateral $A B C D$, we have:

$$
\angle D A B=\angle A B C=\angle B C D=30^{\circ}, A B=4 \text{~cm}, B C=2 \sqrt{3} \text{~cm}
$$

a) Determine the value of the angle $\angle D C A$.

b) Determine the length of $C D$.

c) Find the area of the quadrilateral $A B C D$.

![](https://cdn.mathpix.com/cropped/2024_05_01_82072e76ed187b592b1cg-17.jpg?height=363&width=716&top_left_y=860&top_left_x=742)
Solution

a) Since $\frac{B C}{A B}=\cos 30^{\circ}$ and $\angle A B C=30^{\circ}$, it follows that $\angle A C B=90^{\circ}$. Therefore, $\angle B A C=60^{\circ}$ and $\frac{A C}{A B}=$ $\sin 30^{\circ}=\frac{1}{2}$, so $A C=\frac{A B}{2}=2 \mathrm{~cm}$. Consequently, $\angle D A C=30^{\circ}$ and $\angle D C A=60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_82072e76ed187b592b1cg-17.jpg?height=370&width=703&top_left_y=1753&top_left_x=759)

b) From the previous item and the trigonometric relationships in triangle $A D C$, it follows that $\frac{A D}{A C}=$ $\sin 60^{\circ}$ and $\frac{C D}{A C}=\sin 30^{\circ}$. Therefore,

$$
A D=\sqrt{3} \mathrm{~cm} \text { and } C D=1 \mathrm{~cm}
$$

c) The area of quadrilateral $A B C D$ is

$$
\frac{A C \cdot B C}{2}-\frac{C D \cdot A D}{2}=\frac{2 \cdot 2 \sqrt{3}}{2}-\frac{1 \cdot \sqrt{3}}{2}=\frac{3 \sqrt{3}}{2}
$$