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Find the area of the parallelogram $ABCD$ with side $AB=10\sqrt{3}$ $cm$ and diagonals $BD=10\sqrt{3}$ $cm$ and $BC=10$ $cm.$

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

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    $\begingroup$ Are you sure of the condition? $AM + MB = 5 + 5\sqrt{3} < 10\sqrt{3} = AB$. Such parallelogram doesn't exist $\endgroup$ – Evgeniy May 22 at 18:47
  • $\begingroup$ For which triangle? Which triangle does not exist? $\endgroup$ – Knowledge Greedy May 22 at 18:55
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    $\begingroup$ $\triangle AMB$. The triangle inequality doesn't hold for it $\endgroup$ – Evgeniy May 22 at 18:58
  • $\begingroup$ Yep, I have just seen that. My mistake! $\endgroup$ – Knowledge Greedy May 22 at 19:21
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Using the fact that triangle $ABD$ is isosceles: $$[ABCD]=10\sqrt{(10\sqrt3)^2-5^2}=50\sqrt{11}.$$

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