# Solve for the area of a parallelogram; given diagonals and a side

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 $$S=\dfrac{AC\cdot BD}{2}$$?

• Are you sure of the condition? $AM + MB = 5 + 5\sqrt{3} < 10\sqrt{3} = AB$. Such parallelogram doesn't exist – Evgeniy May 22 at 18:47
• For which triangle? Which triangle does not exist? – Knowledge Greedy May 22 at 18:55
• $\triangle AMB$. The triangle inequality doesn't hold for it – Evgeniy May 22 at 18:58
• Yep, I have just seen that. My mistake! – Knowledge Greedy May 22 at 19:21

Using the fact that triangle $$ABD$$ is isosceles: $$[ABCD]=10\sqrt{(10\sqrt3)^2-5^2}=50\sqrt{11}.$$