The diagonals of a parallelogram have measures of 8 and 10 and intersect at a 60 degree angle. Find the area of the parallelogram.

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- Jun 16th 2008, 01:12 PMmajik92[SOLVED] parallelograms
The diagonals of a parallelogram have measures of 8 and 10 and intersect at a 60 degree angle. Find the area of the parallelogram.

- Jun 16th 2008, 01:55 PMMoo
Hello,

See the sketch below...

Hint : use trigonometry in triangle OAD. Everything else is there (Wink) - Jun 16th 2008, 02:11 PMmajik92
thank you

then that means the height of the triangle is 5√(3) and its base is 10. so the area is 25√(3) making the whole parallelogram's area to be 50√(3). i think that's right - Jun 16th 2008, 02:15 PMMoo
- Jun 16th 2008, 02:59 PMmajik92
oops, the height's 2√(3) -> A of triangle 5√(3) -> A of parallelogram 10√(3)

- Jun 16th 2008, 03:02 PMMoo
- Jun 16th 2008, 03:42 PMPlato
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 $\displaystyle \mbox{area}(\Delta AOD) = \frac{1}{2}\left( 5 \right)\left( 4 \right)\sin \left( {60^ \circ } \right)$. - Jun 16th 2008, 04:26 PMmajik92
- Jun 16th 2008, 05:19 PMPlato
- Jun 16th 2008, 11:06 PMMoo
$\displaystyle \mathcal{A}_{ADB}=\frac 12 \cdot AH \cdot DB=\frac 12 (\cos 60 \cdot OA) \cdot DB$

$\displaystyle \mathcal{A}_{ADB}=\frac 12 \cdot \frac{\sqrt{3}}{2} \cdot 4 \cdot 10=10\sqrt{3}$

Area of the parallelogram is twice this one...

Same result as Plato's, slight different method... This was going to be the correct result ~