## Task B-1.7.

A quadrilateral $A B C D$ is given. If $|A B|=6 \mathrm{~cm},|A D|=4 \mathrm{~cm}, \angle D A B=\angle A B C=$ $60^{\circ}$ and $\angle A D C=90^{\circ}$, calculate the lengths of the diagonals and the area of this quadrilateral.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_5ec97620b8cc23e4b297g-06.jpg?height=466&width=725&top_left_y=475&top_left_x=220)

At vertex $C$ of quadrilateral $ABCD$, we draw a line parallel to side $\overline{AD}$. Then, at point $E$, where this parallel line intersects side $\overline{AB}$, we draw a line parallel to side $\overline{CD}$. The point where this parallel line intersects side $\overline{AD}$ is marked as $G$. This divides quadrilateral $ABCD$ into rectangle $GEC D$, equilateral triangle $EBC$, and right triangle $AEG$, as shown in the figure.

Sketch of the quadrilateral divided in this manner.

Using the labels from the figure, from right triangle $AEG$, we have $\cos 60^{\circ} = \frac{4 - b}{6 - b}$, which simplifies to $b = 2 \text{ cm}$.

Then, $|AG| = 4 - 2 = 2 \text{ cm}$, $|AE| = 6 - 2 = 4 \text{ cm}$,

$|DC| = |GE| = \sqrt{4^2 - 2^2} = 2\sqrt{3} \text{ cm}$.

The length of diagonal $\overline{AC}$ is calculated using the Pythagorean theorem for triangle $ACD$: $|AC| = 2\sqrt{7} \text{ cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_5ec97620b8cc23e4b297g-06.jpg?height=472&width=754&top_left_y=1760&top_left_x=203)

The length of diagonal $\overline{BD}$ is calculated from right triangle $NBD$, where point $N$ is the foot of the perpendicular from vertex $D$ to side $\overline{AB}$.

From right triangle $AND$, we have:

$|DN| = 4 \sin 60^{\circ} = 2\sqrt{3} \text{ cm}$, $|AN| = 4 \cos 60^{\circ} = 2 \text{ cm}$.

1 point

Then, $|NB| = 6 - 2 = 4 \text{ cm}$, and $|BD| = \sqrt{|DN|^2 + |NB|^2} = \sqrt{28} = 2\sqrt{7} \text{ cm}$.

The area of the given quadrilateral can be calculated as the sum of the areas of rectangle $GEC D$, triangle $EBC$, and triangle $AEG$.

$P = P_{AEG} + P_{EBC} + P_{GEC D} = \frac{2 \cdot 2\sqrt{3}}{2} + \frac{4\sqrt{3}}{4} + 2 \cdot 2\sqrt{3} = 7\sqrt{3} \text{ cm}^2$.