1. ## [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.

2. Hello,

See the sketch below...

Hint : use trigonometry in triangle OAD. Everything else is there

3. 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

4. Originally Posted by majik92
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
Hmmm, there is a slight error

How do you find 5√(3) ? Can you check it ?

5. oops, the height's 2√(3) -> A of triangle 5√(3) -> A of parallelogram 10√(3)

6. Originally Posted by majik92
oops, the height's 2√(3) -> A of triangle 5√(3) -> A of parallelogram 10√(3)
Yes, that's better ^^
But see the red part

7. 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 $\displaystyle \mbox{area}(\Delta AOD) = \frac{1}{2}\left( 5 \right)\left( 4 \right)\sin \left( {60^ \circ } \right)$.

8. Originally Posted by Plato
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 $\displaystyle \mbox{area}(\Delta AOD) = \frac{1}{2}\left( 5 \right)\left( 4 \right)\sin \left( {60^ \circ } \right)$.
but i found the area of triangle DAB (as per Moo's graphic) and doubled it b/c triangle DAB is congruent to triangle BCD

9. Originally Posted by majik92
but i found the area of triangle DAB (as per Moo's graphic) and doubled it b/c triangle DAB is congruent to triangle BCD
But what you did may be wrong.
10. $\displaystyle \mathcal{A}_{ADB}=\frac 12 \cdot AH \cdot DB=\frac 12 (\cos 60 \cdot OA) \cdot DB$
$\displaystyle \mathcal{A}_{ADB}=\frac 12 \cdot \frac{\sqrt{3}}{2} \cdot 4 \cdot 10=10\sqrt{3}$