The diagonals of a parallelogram have measures of 8 and 10 and intersect at a 60 degree angle. Find the area of the parallelogram.
I notice that this thread is marked ‘solved’.
But in fact it is not. There is no correct solution yet.
What should have been noted from the beginning is: the area of the parallelogram is $\displaystyle 4\cdot\mbox{area}(\Delta AOD)$.
Now $\displaystyle \mbox{area}(\Delta AOD) = \frac{1}{2}\left( 5 \right)\left( 4 \right)\sin \left( {60^ \circ } \right)$.
$\displaystyle \mathcal{A}_{ADB}=\frac 12 \cdot AH \cdot DB=\frac 12 (\cos 60 \cdot OA) \cdot DB$
$\displaystyle \mathcal{A}_{ADB}=\frac 12 \cdot \frac{\sqrt{3}}{2} \cdot 4 \cdot 10=10\sqrt{3}$
Area of the parallelogram is twice this one...
Same result as Plato's, slight different method... This was going to be the correct result ~