1. Consider right triangle ABK. In this triangle AB is the hypotenuse, BK and AK are legs. By the Pythagorean theorem,
[tex]AB^2=AK^2+BK^2,\\\\6^2=3^2+BK^2,\\\\BK^2=36-9=27,\\\\BK=3\sqrt{3}\ un.[/tex]
2. Use the definition of [tex]\cos \angle A:[/tex]
[tex]\cos \angle A=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{AK}{AB}=\dfrac{3}{6}=\dfrac{1}{2}.[/tex]
Then [tex]m\angle A=60^{\circ}.[/tex]
3. Consider right triangle ABD. In this triangle AD is the hypotenuse, AB and BD are legs. Since [tex]m\angle A=60^{\circ},[/tex] then
[tex]m\angle BDA=180^{\circ}-90^{\circ}-60^{\circ}=30^{\circ}.[/tex]
The leg that is opposite to the angle of 30° is half of the hypotenuse, so
[tex]AD=2AB=12\ un.[/tex]
4. The area of parallelogram aBCD is
[tex]A_{ABCD}=AD\cdot BK=12\cdot 3\sqrt{3}=36\sqrt{3}\ sq. un.[/tex]
