Thread: [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.

Hello,

See the sketch below...

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

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

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 ?

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

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

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 .
Now .

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 .
Now .

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

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.
Go over your work.

Area of the parallelogram is twice this one...


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