The diagonals of a parallelogram are $8$ and $4$. They meet at $60°$. Find the sides and area of the parallelogram.
I tried to use the cosine rule here having $4$ and $2$ as the sides and $60°$ as the angle. But the answer I got is different than the answer key. Also, what about the area and the other two sides?
One side = $\sqrt{20-16\cos60°} = 3.46$
The other side I got is $5.29$
And, what should we use to calculate the area? $ab\sin x$?
You can calculate the area directly by noticing the diagonals bisect each other, as shown in the image above.
Then notice that the triangles opposite each other are also congruent, again by ASA. One pair of the triangles has an angle of $60º$, and the other pair has an angle of $120º$. Thus the area of the parallelogram is $2 \cdot \frac{1}{2}(4)(2) \sin 60º + 2 \cdot \frac{1}{2} (4)(2) \sin 120º = 8\sqrt{3}$.