Find the area of the parallelogram $ABCD$ with side $AB=10\sqrt{3}$ $cm$ and diagonals $BD=10\sqrt{3}$ $cm$ and $BC=10$ $cm.$
Using the fact that $AC^2+BD^2=2(AB^2+BC^2)$ we can find the other side of the parallelogram $\Rightarrow BC=10\sqrt{3}$ $cm.$ What is the fast way to solve for the area of $ABCD$ from here? Maybe using $S=\dfrac{AC\cdot BD}{2}$?
Using the fact that triangle $ABD$ is isosceles: $$[ABCD]=10\sqrt{(10\sqrt3)^2-5^2}=50\sqrt{11}.$$