# [SOLVED] parallelograms

• June 16th 2008, 01:12 PM
majik92
[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.
• June 16th 2008, 01:55 PM
Moo
Hello,

See the sketch below...

Hint : use trigonometry in triangle OAD. Everything else is there (Wink)
• June 16th 2008, 02:11 PM
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
• June 16th 2008, 02:15 PM
Moo
Quote:

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 ? :p
• June 16th 2008, 02:59 PM
majik92
oops, the height's 2√(3) -> A of triangle 5√(3) -> A of parallelogram 10√(3)
• June 16th 2008, 03:02 PM
Moo
Quote:

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 :D
• June 16th 2008, 03:42 PM
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 $4\cdot\mbox{area}(\Delta AOD)$.
Now $\mbox{area}(\Delta AOD) = \frac{1}{2}\left( 5 \right)\left( 4 \right)\sin \left( {60^ \circ } \right)$.
• June 16th 2008, 04:26 PM
majik92
Quote:

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

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
• June 16th 2008, 05:19 PM
Plato
Quote:

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.
$\mathcal{A}_{ADB}=\frac 12 \cdot AH \cdot DB=\frac 12 (\cos 60 \cdot OA) \cdot DB$
$\mathcal{A}_{ADB}=\frac 12 \cdot \frac{\sqrt{3}}{2} \cdot 4 \cdot 10=10\sqrt{3}$